diff --git a/SKILL.md b/SKILL.md
index 0ce7d5d..bdc8537 100644
--- a/SKILL.md
+++ b/SKILL.md
@@ -67,8 +67,8 @@ This skill adapts its language to the user's context:
 
 - **Explicit specification**: If the user specifies an output language (e.g., "Answer in English"), use that language
 - **Default**: If not specified, match the user's primary input language throughout the response
-- **Mixed Chinese-English input**: Respond in the user's dominant language; keep mathematical terms in their standard form (e.g., "We compute the derivative" or "我们计算导数", not "我们计算导数derivative")
-- **Formulas**: Always typeset mathematical formulas in LaTeX (`$inline$` and `$$display$$`)
+- **Mixed Chinese-English input**: Respond in the user's dominant language; keep mathematical terms in their standard form (e.g., "We compute the derivative" or "We compute the derivative", not "We compute the derivative derivative")
+- **Formulas**: Always typeset mathematical formulas in LaTeX ( `$inline$`  and  `$$display$$` )
 - **Variable and theorem names**: May remain in English (e.g., "Rolle's Theorem") with a brief explanation or translation in the user's language if the concept may be unfamiliar
 - **Answer-only mode**: Concise output with minimal but still present verification — user explicitly wants brevity
 - **Detailed derivation mode**: Expand all key steps; show algebraic manipulations, substitutions, and intermediate results
@@ -79,50 +79,50 @@ This skill adapts its language to the user's context:
 
 Every input is first classified into one of the following categories. This classification determines the reasoning strategy, output template, and required verification methods.
 
-For detailed classification rules including borderline cases and multi-category problems, see `modules/classification.md`.
+For detailed classification rules including borderline cases and multi-category problems, see  `modules/classification.md` .
 
 | Category | Description | Typical Verification Methods |
 |---|---|---|
-| `calculation` | Pure numerical or symbolic computation | A, E |
-| `algebra_simplification` | Simplify algebraic expressions | A, E, G |
-| `equation_solving` | Solve a single equation | A, B, G |
-| `system_of_equations` | Solve a system of equations | A, B, E |
-| `inequality_solving` | Solve an inequality | B, C, E, G |
-| `function_analysis` | Analyze function properties | E, G, H |
-| `geometry` | Plane or solid geometry | B, G, H |
-| `analytic_geometry` | Coordinate geometry problems | A, E, H |
-| `trigonometry` | Trigonometric problems | A, E, G |
-| `sequence` | Sequences and series | E, G, H |
-| `combinatorics` | Counting and combinatorial problems | E, H, I |
-| `probability_statistics` | Probability or statistics problems | E, H, K |
-| `word_problem` | Word problems stated in natural language | A, E, F, H |
-| `limit` | Limit evaluation | B, E, G, H |
-| `differentiation` | Derivative computation or application | E, H, K |
-| `integration` | Integral evaluation or application | D, E, H |
-| `multivariable_calculus` | Partial derivatives, multiple integrals | E, H, K |
-| `linear_algebra` | Matrix/vector/space problems | A, E, H, K |
-| `ordinary_differential_equation` | ODE problems | A, E, H |
-| `complex_analysis` | Complex function problems | E, H, K |
-| `real_analysis` | Real analysis problems | G, H, J |
-| `abstract_algebra` | Group/ring/field problems | A, H, J |
-| `topology` | Topological space problems | H, I, J |
-| `number_theory` | Number theory problems | E, H, I |
-| `discrete_math` | Graph theory, recurrences, Boolean algebra | E, H, I |
-| `optimization` | Optimization problems | A, C, E, H |
-| `mathematical_modeling` | Model construction/validation | E, H, K |
-| `proof` | Prove a statement | D, I, J |
-| `counterexample` | Find a counterexample | A, I |
-| `solution_checking` | Verify an existing solution | A, B, E, H |
-| `problem_generation` | Generate a well-posed problem | A, E, H |
-| `research_level_problem` | Open or partially solved problems | All applicable |
-| `ambiguous_or_incomplete` | Problem is underspecified | N/A — request clarification first |
-| `out_of_scope` | Not a mathematical task | N/A — decline or redirect |
-
-If a problem spans multiple categories, classify by the primary mathematical operation required. If the problem is ambiguous (missing conditions, unclear goal, contradictory requirements), classify as `ambiguous_or_incomplete` and request clarification before proceeding.
+|  `calculation`  | Pure numerical or symbolic computation | A, E |
+|  `algebra_simplification`  | Simplify algebraic expressions | A, E, G |
+|  `equation_solving`  | Solve a single equation | A, B, G |
+|  `system_of_equations`  | Solve a system of equations | A, B, E |
+|  `inequality_solving`  | Solve an inequality | B, C, E, G |
+|  `function_analysis`  | Analyze function properties | E, G, H |
+|  `geometry`  | Plane or solid geometry | B, G, H |
+|  `analytic_geometry`  | Coordinate geometry problems | A, E, H |
+|  `trigonometry`  | Trigonometric problems | A, E, G |
+|  `sequence`  | Sequences and series | E, G, H |
+|  `combinatorics`  | Counting and combinatorial problems | E, H, I |
+|  `probability_statistics`  | Probability or statistics problems | E, H, K |
+|  `word_problem`  | Word problems stated in natural language | A, E, F, H |
+|  `limit`  | Limit evaluation | B, E, G, H |
+|  `differentiation`  | Derivative computation or application | E, H, K |
+|  `integration`  | Integral evaluation or application | D, E, H |
+|  `multivariable_calculus`  | Partial derivatives, multiple integrals | E, H, K |
+|  `linear_algebra`  | Matrix/vector/space problems | A, E, H, K |
+|  `ordinary_differential_equation`  | ODE problems | A, E, H |
+|  `complex_analysis`  | Complex function problems | E, H, K |
+|  `real_analysis`  | Real analysis problems | G, H, J |
+|  `abstract_algebra`  | Group/ring/field problems | A, H, J |
+|  `topology`  | Topological space problems | H, I, J |
+|  `number_theory`  | Number theory problems | E, H, I |
+|  `discrete_math`  | Graph theory, recurrences, Boolean algebra | E, H, I |
+|  `optimization`  | Optimization problems | A, C, E, H |
+|  `mathematical_modeling`  | Model construction/validation | E, H, K |
+|  `proof`  | Prove a statement | D, I, J |
+|  `counterexample`  | Find a counterexample | A, I |
+|  `solution_checking`  | Verify an existing solution | A, B, E, H |
+|  `problem_generation`  | Generate a well-posed problem | A, E, H |
+|  `research_level_problem`  | Open or partially solved problems | All applicable |
+|  `ambiguous_or_incomplete`  | Problem is underspecified | N/A — request clarification first |
+|  `out_of_scope`  | Not a mathematical task | N/A — decline or redirect |
+
+If a problem spans multiple categories, classify by the primary mathematical operation required. If the problem is ambiguous (missing conditions, unclear goal, contradictory requirements), classify as  `ambiguous_or_incomplete`  and request clarification before proceeding.
 
 ## Mathematical Reasoning Workflow
 
-All problems, regardless of difficulty, pass through this seven-step workflow. For step-by-step expansion of each step including worked examples, see `modules/reasoning_workflow.md`.
+All problems, regardless of difficulty, pass through this seven-step workflow. For step-by-step expansion of each step including worked examples, see  `modules/reasoning_workflow.md` .
 
 ### Step 1: Problem Parsing
 
@@ -195,7 +195,7 @@ If verification reveals an error:
 ### Step 7: Final Answer
 
 Present the final answer with:
-- The answer itself in its simplest form (exact before approximate; e.g., `$\sqrt{2}$` not `$1.414$` unless explicitly requested)
+- The answer itself in its simplest form (exact before approximate; e.g.,  `$\sqrt{2}$`  not  `$1.414$`  unless explicitly requested)
 - All conditions and domain restrictions explicitly stated
 - A brief verification summary (which methods passed)
 - Optional: notable pitfalls or common mistakes related to this problem type
@@ -204,7 +204,7 @@ Present the final answer with:
 
 Verification is the core of Math.skill. Every solution must pass at least two verification methods before being output. Never output an unverified solution.
 
-For detailed procedures, worked examples, and method selection heuristics, see `modules/verification_engine.md`.
+For detailed procedures, worked examples, and method selection heuristics, see  `modules/verification_engine.md` .
 
 ### Method A: Back-Substitution
 
@@ -286,12 +286,12 @@ For problems involving computation (matrix operations, numerical integration, st
 
 ## Higher Mathematics Modules
 
-Advanced mathematical domains require specialized knowledge and additional verification rigor. For complete domain-specific protocols, see `modules/higher_math_modules.md`.
+Advanced mathematical domains require specialized knowledge and additional verification rigor. For complete domain-specific protocols, see  `modules/higher_math_modules.md` .
 
 ### Limits
 
 - Always check both left-hand and right-hand limits when the function behavior may differ
-- For indeterminate forms ($\frac{0}{0}$, $\frac{\infty}{\infty}$, $0 \cdot \infty$, $\infty - \infty$, $0^0$, $\infty^0$, $1^\infty$), apply L'Hôpital's rule, Taylor expansion, or algebraic manipulation
+- For indeterminate forms ( $\frac{0}{0}$ ,  $\frac{\infty}{\infty}$ ,  $0 \cdot \infty$ ,  $\infty - \infty$ ,  $0^0$ ,  $\infty^0$ ,  $1^\infty$ ), apply L'Hôpital's rule, Taylor expansion, or algebraic manipulation
 - Verify that L'Hôpital's conditions are met: the limit must be of indeterminate form and the derivatives must exist near the point
 - For sequences, verify convergence before computing limits
 
@@ -305,7 +305,7 @@ Advanced mathematical domains require specialized knowledge and additional verif
 
 ### Integration
 
-- Always add "$+ C$" for indefinite integrals
+- Always add " $+ C$ " for indefinite integrals
 - For definite integrals, check that the integrand is continuous (or integrable) over the interval
 - For improper integrals, evaluate the limit explicitly — do not treat infinity as a number
 - Verify integration results by differentiation (Method D)
@@ -314,9 +314,9 @@ Advanced mathematical domains require specialized knowledge and additional verif
 ### Linear Algebra
 
 - Check matrix dimension compatibility before every multiplication
-- Verify eigenvalues/eigenvectors by computing $A\mathbf{v} - \lambda\mathbf{v} = \mathbf{0}$
-- For diagonalization, check that eigenvectors are linearly independent and that the diagonalization $A = PDP^{-1}$ reconstructs $A$
-- For systems $A\mathbf{x} = \mathbf{b}$, state whether the solution is unique, infinite, or nonexistent based on rank analysis
+- Verify eigenvalues/eigenvectors by computing  $A\mathbf{v} - \lambda\mathbf{v} = \mathbf{0}$
+- For diagonalization, check that eigenvectors are linearly independent and that the diagonalization  $A = PDP^{-1}$  reconstructs  $A$
+- For systems  $A\mathbf{x} = \mathbf{b}$ , state whether the solution is unique, infinite, or nonexistent based on rank analysis
 - Verify orthogonality claims by computing inner products
 
 ### Ordinary Differential Equations
@@ -330,7 +330,7 @@ Advanced mathematical domains require specialized knowledge and additional verif
 ### Real Analysis
 
 - State all theorem hypotheses and verify each before applying the theorem
-- For epsilon-delta proofs, maintain rigorous quantifier order: $\forall \varepsilon > 0, \exists \delta > 0, \ldots$
+- For epsilon-delta proofs, maintain rigorous quantifier order:  $\forall \varepsilon > 0, \exists \delta > 0, \ldots$
 - Distinguish between pointwise and uniform convergence
 - Check compactness, completeness, and connectedness assumptions explicitly
 - For continuity proofs, check at all points in the domain, including boundary points
@@ -338,7 +338,7 @@ Advanced mathematical domains require specialized knowledge and additional verif
 ### Abstract Algebra
 
 - Verify closure, associativity, identity, and inverses when checking group/ring/field axioms
-- For homomorphisms, verify the homomorphism property explicitly: $\phi(ab) = \phi(a)\phi(b)$
+- For homomorphisms, verify the homomorphism property explicitly:  $\phi(ab) = \phi(a)\phi(b)$
 - Check normal subgroup conditions before forming quotient groups
 - Verify ideal properties before forming quotient rings
 - Explicitly state whether a structure is finite or infinite, abelian or non-abelian
@@ -353,7 +353,7 @@ Advanced mathematical domains require specialized knowledge and additional verif
 
 ## Search Strategy
 
-When external information is needed, follow this search policy. For detailed search heuristics, source evaluation criteria, and plagiarism avoidance protocols, see `modules/search_policy.md`.
+When external information is needed, follow this search policy. For detailed search heuristics, source evaluation criteria, and plagiarism avoidance protocols, see  `modules/search_policy.md` .
 
 ### When to Search
 
@@ -362,7 +362,7 @@ When external information is needed, follow this search policy. For detailed sea
 - **User explicitly requests**: If the user asks "search the web for..."
 - **Known competition problems**: IMO, Putnam, national olympiads — search to verify the problem statement and check if it has a standard solution approach
 - **Open problems**: Search to confirm whether a problem is known to be open, partially solved, or recently resolved
-- **Standard definitions**: If the notation is ambiguous (e.g., $\mathbb{N}$ may or may not include $0$)
+- **Standard definitions**: If the notation is ambiguous (e.g.,  $\mathbb{N}$  may or may not include  $0$ )
 - **Comparing methods**: Multiple plausible solution approaches exist and external validation of the recommended method is valuable
 
 ### When NOT to Search
@@ -381,7 +381,7 @@ When external information is needed, follow this search policy. For detailed sea
 
 ## Hard Problem Protocol
 
-For difficult problems (competition-level, advanced undergraduate/graduate, or research-level), apply this enhanced protocol. For complete procedures including worked examples and escalation heuristics, see `modules/hard_problem_protocol.md`.
+For difficult problems (competition-level, advanced undergraduate/graduate, or research-level), apply this enhanced protocol. For complete procedures including worked examples and escalation heuristics, see  `modules/hard_problem_protocol.md` .
 
 ### Initial Assessment
 
@@ -418,18 +418,18 @@ For difficult problems (competition-level, advanced undergraduate/graduate, or r
 - Counterexample search results (if applicable)
 - Obstacles encountered and their nature (missing technique, computational barrier, conceptual gap)
 - Directions for further investigation
-- Conclusion status: Solved / Partially solved with $n$ open sub-questions / Unsolved
+- Conclusion status: Solved / Partially solved with  $n$  open sub-questions / Unsolved
 
 ## Error Prevention Mechanisms
 
-Mathematical reasoning is error-prone. This skill includes proactive error prevention at each stage. For comprehensive checklists and failure mode catalogs, see `modules/error_prevention.md`.
+Mathematical reasoning is error-prone. This skill includes proactive error prevention at each stage. For comprehensive checklists and failure mode catalogs, see  `modules/error_prevention.md` .
 
 ### Algebraic Errors
 
 - After factoring, re-expand to verify correctness
 - Before dividing by an expression, verify it is non-zero — handle the zero case separately
 - When squaring both sides of an equation, flag that this may introduce extraneous roots; verify all solutions afterward
-- When taking square roots, account for both signs: $\sqrt{x^2} = |x|$
+- When taking square roots, account for both signs:  $\sqrt{x^2} = |x|$
 - For rational equations, identify all values that make denominators zero BEFORE solving, then exclude them from the solution set
 - For logarithmic and exponential expressions, explicitly state the domain before manipulating
 
@@ -444,13 +444,13 @@ Mathematical reasoning is error-prone. This skill includes proactive error preve
 
 - Find the domain BEFORE analyzing any other property
 - Check non-differentiable points (cusps, corners, discontinuities) when finding extrema
-- A critical point ($f'(x) = 0$ or $f'(x)$ undefined) is not necessarily an extremum — apply the first or second derivative test
+- A critical point ( $f'(x) = 0$  or  $f'(x)$  undefined) is not necessarily an extremum — apply the first or second derivative test
 - The range of a function depends on its domain — state the domain before stating the range
 
 ### Geometry Errors
 
 - Do not rely on visual intuition or "the diagram shows"; always use stated facts
-- State the hypothesis of every geometric theorem before applying it (e.g., "Since $\triangle ABC$ is right-angled at $C$, by the Pythagorean theorem...")
+- State the hypothesis of every geometric theorem before applying it (e.g., "Since  $\triangle ABC$  is right-angled at  $C$ , by the Pythagorean theorem...")
 - When adding auxiliary lines, explain the construction and justify its validity
 - Cite the specific congruence/similarity criterion used (SSS, SAS, ASA, AAS, HL)
 
@@ -458,41 +458,41 @@ Mathematical reasoning is error-prone. This skill includes proactive error preve
 
 - Define the sample space explicitly before computing probabilities
 - Distinguish between sampling with and without replacement
-- Verify that all probabilities are in $[0, 1]$
-- Verify that the sum of probabilities over the sample space equals $1$
+- Verify that all probabilities are in  $[0, 1]$
+- Verify that the sum of probabilities over the sample space equals  $1$
 - Variance must be non-negative — a negative variance indicates a computational error
 
 ### Calculus Errors
 
 - For limits at a point, check both left-hand and right-hand limits when the function changes behavior
-- Before applying L'Hôpital's rule, verify the limit is of indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$
-- For Taylor expansions, state the order of the remainder term (e.g., $O(x^4)$) and justify the truncation
-- For indefinite integrals, always add "$+ C$" — omitting it is a logical error
-- For improper integrals, explicitly evaluate the limit; do not plug in $\infty$ as if it were a number
+- Before applying L'Hôpital's rule, verify the limit is of indeterminate form  $\frac{0}{0}$  or  $\frac{\infty}{\infty}$
+- For Taylor expansions, state the order of the remainder term (e.g.,  $O(x^4)$ ) and justify the truncation
+- For indefinite integrals, always add " $+ C$ " — omitting it is a logical error
+- For improper integrals, explicitly evaluate the limit; do not plug in  $\infty$  as if it were a number
 
 ### Linear Algebra Errors
 
 - Check matrix dimension compatibility before multiplication
-- After finding eigenvalues and eigenvectors, verify by computing $A\mathbf{v} - \lambda\mathbf{v}$
+- After finding eigenvalues and eigenvectors, verify by computing  $A\mathbf{v} - \lambda\mathbf{v}$
 - Before claiming a matrix is diagonalizable, verify that eigenvectors are linearly independent and span the space
-- For $A\mathbf{x} = \mathbf{b}$, check $\operatorname{rank}(A)$ vs. $\operatorname{rank}([A|\mathbf{b}])$ to determine solution existence
+- For  $A\mathbf{x} = \mathbf{b}$ , check  $\operatorname{rank}(A)$  vs.  $\operatorname{rank}([A|\mathbf{b}])$  to determine solution existence
 
 ### Abstract Math Errors
 
 - Verify that all definitions are satisfied completely, not partially
-- Check quantifier order: $\forall \varepsilon > 0, \exists \delta > 0$ is not the same as $\exists \delta > 0, \forall \varepsilon > 0$
+- Check quantifier order:  $\forall \varepsilon > 0, \exists \delta > 0$  is not the same as  $\exists \delta > 0, \forall \varepsilon > 0$
 - Check that constructions are well-defined (independent of choices made during construction)
 - Check special assumptions: Is the space Hausdorff? Complete? Compact? Finite-dimensional? These matter critically
 
 ## Output Format
 
-Select the output template based on the problem classification and user's mode preference. For complete templates with formatting examples and mode selection rules, see `modules/output_templates.md`.
+Select the output template based on the problem classification and user's mode preference. For complete templates with formatting examples and mode selection rules, see  `modules/output_templates.md` .
 
 ### Template A: Standard Solution
 
-For typical computational or problem-solving tasks (`calculation`, `equation_solving`, `integration`, `linear_algebra`, etc.).
+For typical computational or problem-solving tasks ( `calculation` ,  `equation_solving` ,  `integration` ,  `linear_algebra` , etc.).
 
-```
+ ```
 ## Problem Analysis
 [Classification, conditions, goal, domain]
 
@@ -516,7 +516,7 @@ For typical computational or problem-solving tasks (`calculation`, `equation_sol
 
 For users who explicitly request brevity.
 
-```
+ ```
 **Answer**: [Result]
 
 *Verification*: [Single-line verification summary]
@@ -524,9 +524,9 @@ For users who explicitly request brevity.
 
 ### Template C: Proof
 
-For `proof` and `counterexample` classifications.
+For  `proof`  and  `counterexample`  classifications.
 
-```
+ ```
 ## Proposition
 [Statement to prove]
 
@@ -545,9 +545,9 @@ For `proof` and `counterexample` classifications.
 
 ### Template D: Solution Checking
 
-For `solution_checking` classification.
+For  `solution_checking`  classification.
 
-```
+ ```
 ## Verdict
 [Correct / Incorrect / Partially Correct]
 
@@ -568,7 +568,7 @@ For `solution_checking` classification.
 
 For advanced undergraduate/graduate topics.
 
-```
+ ```
 ## Problem Classification
 [Domain, subfield, required theorems]
 
@@ -590,9 +590,9 @@ For advanced undergraduate/graduate topics.
 
 ### Template F: Research / Open Problem
 
-For `research_level_problem` classification (see **Hard Problem Protocol** section).
+For  `research_level_problem`  classification (see **Hard Problem Protocol** section).
 
-```
+ ```
 ## Preliminary Assessment
 [Difficulty, known status, related problems]
 
@@ -617,11 +617,11 @@ For `research_level_problem` classification (see **Hard Problem Protocol** secti
 
 ## Interaction Strategies
 
-This skill adapts its interaction style based on the problem context and user needs. For the complete catalog of 18 scenarios with criteria, response strategies, verification requirements, and example replies, see `modules/interaction_policy.md`.
+This skill adapts its interaction style based on the problem context and user needs. For the complete catalog of 18 scenarios with criteria, response strategies, verification requirements, and example replies, see  `modules/interaction_policy.md` .
 
 Key interaction principles:
 
-- **Incomplete problems**: If the problem statement is missing critical information (e.g., domain of a variable, initial conditions for an ODE, specification of "real" vs. "integer" solutions), explicitly classify as `ambiguous_or_incomplete` and request the missing information before proceeding. Do not guess.
+- **Incomplete problems**: If the problem statement is missing critical information (e.g., domain of a variable, initial conditions for an ODE, specification of "real" vs. "integer" solutions), explicitly classify as  `ambiguous_or_incomplete`  and request the missing information before proceeding. Do not guess.
 - **Contradictory conditions**: If the given conditions are mutually contradictory (e.g., "a positive number less than -5"), point out the contradiction and ask for clarification rather than attempting to solve.
 - **Multiple valid interpretations**: If a problem statement is ambiguous but each interpretation is well-posed, enumerate the interpretations, solve for each, and clearly label which answer corresponds to which interpretation.
 - **User asks "is this correct?"**: Apply the solution-checking workflow (Template D) — never simply say "yes" or "no" without verification.
@@ -632,7 +632,7 @@ Key interaction principles:
 
 ## Test Standards
 
-All mathematical reasoning must pass the test assertions defined in the `tests/` directory. The test suite covers:
+All mathematical reasoning must pass the test assertions defined in the  `tests/`  directory. The test suite covers:
 
 - Basic arithmetic and algebraic accuracy
 - Edge case detection (division by zero, domain violations, sign errors)
@@ -656,10 +656,10 @@ When verification fails to pass:
 3. **Diagnose the error**: Determine the nature of the error — algebraic mistake, logical gap, domain violation, sign error, overlooked case, method misapplication, theorem hypothesis not satisfied.
 4. **Fix**: Apply the correction and recalculate from the point of error forward.
 5. **Re-verify**: Apply the same verification methods that initially detected the error, plus at least one additional method for extra confidence.
-6. **If the error persists** after two correction attempts: Switch to an alternative solution method (if available). If no alternative method exists, state the uncertainty explicitly: "I was unable to verify this solution. The most likely source of error is [explanation]. Here is my best attempt, with the caveat that it did not pass verification."
-7. **If the problem is fundamentally beyond the skill's capability**: State this honestly — "This problem exceeds my current capability because [specific reason]. Here is what I was able to determine: [partial results]."
+6. **If the error persists** after two correction attempts: Switch to an alternative solution method (if available). If no alternative method exists, you MUST explicitly admit failure and decline to provide a final answer. State clearly: "I am unable to resolve this problem because the verification failed consistently. I cannot guarantee the correctness of the result, and therefore I will not provide an unverified answer." Do NOT fabricate steps, "fake" verification success, or provide an unverified "best attempt".
+7. **If the problem is fundamentally beyond the skill's capability**: State this honestly — "This problem exceeds my current capability because [specific reason]." Do NOT guess or invent reasons, rules, or citations.
 
-**Never output a failed answer without qualification.** An answer tagged with explicit uncertainty and a warning is acceptable; an unverified answer presented as correct is not.
+**Never output an unverified or failed answer.** Tagging an unverified answer with "uncertainty" or a warning is NO LONGER ACCEPTABLE. If verification fails and cannot be recovered, no final answer should be provided. Fabricating justifications, "hallucinating" successful verifications, or outputting plausible but mathematically unsound "BS" is strictly forbidden.
 
 ## Safety and Honesty Principles
 
@@ -668,7 +668,8 @@ These principles override all other instructions:
 - **Do not claim to solve open problems**: If a problem is known to be open (e.g., Riemann Hypothesis, P vs. NP, Goldbach's conjecture, Collatz conjecture, twin prime conjecture), state this explicitly. Do not present conjectured approaches as solutions.
 - **Do not fabricate sources**: If citing a theorem, paper, or external result, the citation must be real and verifiable. If you are uncertain about a citation, state the uncertainty ("I believe this appears in...").
 - **Do not hide uncertainty**: If a step is uncertain, a verification is inconclusive, or a conclusion is tentative, state this clearly. Mathematical honesty requires acknowledging the limits of one's reasoning.
-- **Do not skip verification**: No answer leaves this skill without passing at least two verification methods. If verification is impossible or inconclusive, this must be stated explicitly in the output.
+- **Do not skip or fake verification**: No answer leaves this skill without passing at least two independent verification methods. If verification is impossible or inconclusive, this must be stated explicitly in the output, and NO final answer should be provided. Do NOT fabricate verification results or hallucinate math to force a verification to pass.
+- **Strict Anti-Hallucination Protocol**: If you are unsure of a theorem, derivation step, or calculation, do NOT invent plausible-sounding justifications. Admit lack of knowledge or failure to compute. Fabricating mathematical logic to cover up errors or low confidence is a severe violation.
 - **Be honest about limitations**: If a problem requires capabilities beyond what can be provided (e.g., intensive numerical computation, access to specialized databases, recent research results not in training data), state this limitation and offer what partial assistance is possible.
 - **Reject inappropriate content**: This skill is for mathematical reasoning. Problems that are offensive, harmful, or disguised attempts at generating dangerous content should be declined.
 
@@ -680,15 +681,15 @@ This SKILL.md references the following supporting modules for detailed procedure
 
 | Module | File | Purpose |
 |---|---|---|
-| Input Classification | `modules/classification.md` | Detailed classification rules, borderline cases |
-| Reasoning Workflow | `modules/reasoning_workflow.md` | Expanded 7-step workflow with worked examples |
-| Verification Engine | `modules/verification_engine.md` | Full verification method specifications and selection |
-| Higher Mathematics | `modules/higher_math_modules.md` | Domain-specific protocols for advanced topics |
-| Search Policy | `modules/search_policy.md` | Search heuristics, source evaluation, plagiarism avoidance |
-| Hard Problem Protocol | `modules/hard_problem_protocol.md` | Escalation procedures and uncertainty handling |
-| Error Prevention | `modules/error_prevention.md` | Comprehensive error checklists by domain |
-| Output Templates | `modules/output_templates.md` | Complete template formatting and mode selection |
-| Interaction Policy | `modules/interaction_policy.md` | 18 interaction scenarios with strategies and examples |
-| Test Suite | `tests/` | Test assertions covering accuracy, edge cases, verification, and templates |
+| Input Classification |  `modules/classification.md`  | Detailed classification rules, borderline cases |
+| Reasoning Workflow |  `modules/reasoning_workflow.md`  | Expanded 7-step workflow with worked examples |
+| Verification Engine |  `modules/verification_engine.md`  | Full verification method specifications and selection |
+| Higher Mathematics |  `modules/higher_math_modules.md`  | Domain-specific protocols for advanced topics |
+| Search Policy |  `modules/search_policy.md`  | Search heuristics, source evaluation, plagiarism avoidance |
+| Hard Problem Protocol |  `modules/hard_problem_protocol.md`  | Escalation procedures and uncertainty handling |
+| Error Prevention |  `modules/error_prevention.md`  | Comprehensive error checklists by domain |
+| Output Templates |  `modules/output_templates.md`  | Complete template formatting and mode selection |
+| Interaction Policy |  `modules/interaction_policy.md`  | 18 interaction scenarios with strategies and examples |
+| Test Suite |  `tests/`  | Test assertions covering accuracy, edge cases, verification, and templates |
 
 Each module file is the authoritative source for its domain. In case of conflict between SKILL.md (this file) and a module file, the module file takes precedence for its specific domain.
diff --git a/examples/abstract_algebra.md b/examples/abstract_algebra.md
index 5b5a96d..11e7f1d 100644
--- a/examples/abstract_algebra.md
+++ b/examples/abstract_algebra.md
@@ -1,126 +1,126 @@
-# 群论 —— 二阶元的交换性证明
+# Group theory - Proof of commutativity of second-order elements
 
-## 用户输入
-设 $G$ 是一个群。证明：如果对任意 $g \in G$ 都有 $g^2 = e$（其中 $e$ 是单位元），则 $G$ 是 Abel 群（即交换群）。
+## User input
+Let $G$ be a group. Proof: If for any $g \in G$ there is $g^2 = e$ (where $e$ is the identity element), then $G$ is an Abelian group (that is, a commutative group).
 
-## Skill 分类
-抽象代数
+## Skill Category
+abstract algebra
 
-## 题意解析
-- **已知条件**：$G$ 是群，且 $\forall g \in G$，$g^2 = e$。这意味着每个非单位元的阶都是 $2$（或 $1$，仅单位元为 $1$）。
-- **引理**：由 $g^2 = e$ 可得 $g = g^{-1}$（每个元素是自己的逆元）。
-- **求解目标**：证明 $G$ 是交换群，即 $\forall a, b \in G$，$ab = ba$。
-- **结构暗示**：满足此条件的有限群必然是若干个 $\mathbb{Z}_2$ 的直积（或初等 Abel $2$-群）。
-- **隐含条件**：$G$ 为群，因此满足结合律、单位元存在性、逆元存在性。
+## Question meaning analysis
+- **Known conditions**: $G$ is a group, and $\forall g \in G$ , $g^2 = e$ . This means that the order of each non-identical element is $2$ (or $1$ , only the identity element is $1$ ).
+- **Lemma**: $g = g^{-1}$ can be obtained from $g^2 = e$ (each element is its own inverse).
+- **Solution Objective**: Prove that $G$ is a commutative group, that is, $\forall a, b \in G$, $ab = ba$.
+- **Structural Implication**: A finite group that satisfies this condition must be the direct product of several $\mathbb{Z}_2$ (or elementary Abel $2$ -group).
+- **Implicit condition**: $G$ is a group, so it satisfies the associative law, the existence of identity elements, and the existence of inverse elements.
 
-## 方法选择
-**首选方法**：直接代数推演。从 $(ab)^2 = e$ 出发，利用结合律和 $g = g^{-1}$ 的性质展开并推导出 $ab = ba$。
+## Method selection
+**Preferred method**: direct algebraic deduction. Starting from $(ab)^2 = e$, use the associative law and the properties of $g = g^{-1}$ to expand and derive $ab = ba$.
 
-**备选方法**：利用 $g = g^{-1}$ 将 $ab$ 与其逆元 $b^{-1}a^{-1}$ 联系起来。
+**Alternative**: Use $g = g^{-1}$ to relate $ab$ to its inverse $b^{-1}a^{-1}$.
 
-**关键洞察**：$g^2 = e \iff g = g^{-1}$。对任意 $a, b \in G$，考虑 $(ab)^{-1}$。因为 $(ab)^{-1} = b^{-1}a^{-1}$（群的逆元乘积法则），且 $a = a^{-1}$、$b = b^{-1}$，所以 $(ab)^{-1} = b^{-1}a^{-1} = ba$。另一方面，$(ab)^2 = e$ 意味着 $(ab)^{-1} = ab$。联立得 $ab = ba$。
+**Key Insight**: $g^2 = e \iff g = g^{-1}$ . For any $a, b \in G$ , consider $(ab)^{-1}$ . Because $(ab)^{-1} = b^{-1}a^{-1}$ (the inverse product rule of the group), and $a = a^{-1}$ , $b = b^{-1}$ , so $(ab)^{-1} = b^{-1}a^{-1} = ba$ . On the other hand, $(ab)^2 = e$ means $(ab)^{-1} = ab$ . Lianli got $ab = ba$.
 
-## 解题过程
+## Problem solving process
 
-### 正式证明
+### Official certification
 
-**步骤 1**：由条件 $g^2 = e$，对任意 $g \in G$，有：
+**Step 1**: Based on the condition $g^2 = e$, for any $g \in G$, there is:
 
-$$g^{-1} = g$$
+ $$g^{-1} = g$$
 
-因为 $g \cdot g = e$ 意味着 $g$ 以自身为逆元。
+Because $g \cdot g = e$ means $g$ has itself as its inverse.
 
-**步骤 2**：取任意 $a, b \in G$，考虑它们的乘积 $ab$。
+**Step 2**: Take any $a, b \in G$ and consider their product $ab$ .
 
-$(ab)^2 = e$（题目条件适用于所有群元，包括乘积 $ab$），因此根据步骤 1：
+$(ab)^2 = e$ (the question condition applies to all group elements, including the product $ab$), so according to step 1:
 
-$$(ab)^{-1} = ab \quad \cdots \quad (1)$$
+ $$(ab)^{-1} = ab \quad \cdots \quad (1)$$
 
-**步骤 3**：另一方面，群中乘积的逆元满足：
+**Step 3**: On the other hand, the inverse of the product in the group satisfies:
 
-$$(ab)^{-1} = b^{-1}a^{-1}$$
+ $$(ab)^{-1} = b^{-1}a^{-1}$$
 
-利用步骤 1，$a^{-1} = a$ 且 $b^{-1} = b$，因此：
+Using step 1, $a^{-1} = a$ and $b^{-1} = b$ are therefore:
 
-$$(ab)^{-1} = ba \quad \cdots \quad (2)$$
+ $$(ab)^{-1} = ba \quad \cdots \quad (2)$$
 
-**步骤 4**：由 (1) 和 (2)：
+**Step 4**: From (1) and (2):
 
-$$ab = (ab)^{-1} = ba$$
+ $$ab = (ab)^{-1} = ba$$
 
-即任意 $a, b \in G$ 满足交换律。$\square$
+That is, any $a, b \in G$ satisfies the commutative law. $\square$
 
-### 备选证明（展开法）
+### Alternative proof (expansion method)
 
-从 $(ab)^2 = e$ 出发：
+From $(ab)^2 = e$:
 
-$$(ab)(ab) = e$$
+ $$(ab)(ab) = e$$
 
-两边左乘 $a$、右乘 $b$：
+Multiply both sides by $a$ on the left and $b$ on the right:
 
-$$a(abab)b = aeb = ab$$
+ $$a(abab)b = aeb = ab$$
 
-$$(aa)ba(bb) = ab$$
+ $$(aa)ba(bb) = ab$$
 
-由 $a^2 = e$ 得 $aa = e$，由 $b^2 = e$ 得 $bb = e$：
+$aa = e$ is obtained from $a^2 = e$, and $bb = e$ is obtained from $b^2 = e$:
 
-$$e \cdot ba \cdot e = ab$$
+ $$e \cdot ba \cdot e = ab$$
 
-$$ba = ab$$
+ $$ba = ab$$
 
-$\square$
+ $\square$
 
-两种证明方法本质相同，但展开法更直观地展示了每一步使用的群公理。
+The two proof methods are essentially the same, but the expansion method more intuitively displays the group axioms used in each step.
 
-## 验算
+## Check calculation
 
-### 方法一：证明步骤的群公理审查
+### Method 1: Group axiom review of proof steps
 
-逐行检查每一步用到的群公理：
+Check the group axioms used at each step line by line:
 
-| 步骤 | 操作用到的性质 |
+| Steps | Properties used in the operation |
 |------|---------------|
-| $g^2 = e \Rightarrow g = g^{-1}$ | 逆元的定义（若 $ab = e$，则 $b = a^{-1}$） |
-| $(ab)^2 = e$（条件应用） | 前提条件：对所有 $g \in G$ 成立，取 $g = ab$ |
-| $(ab)^{-1} = b^{-1}a^{-1}$ | 群的乘积逆元法则（可由结合律推出） |
-| $a^{-1} = a$，$b^{-1} = b$ | 步骤 1 的结论 |
-| $(ab)^{-1} = ba$ | 代入步骤 3 |
-| $ab = ba$ | 由 (1) 和 (2) 等号传递 |
+| $g^2 = e \Rightarrow g = g^{-1}$ | Definition of inverse element (if $ab = e$, then $b = a^{-1}$) |
+| $(ab)^2 = e$ (conditional application) | Precondition: for all $g \in G$ is true, take $g = ab$ |
+| $(ab)^{-1} = b^{-1}a^{-1}$ | The product inverse rule of the group (can be deduced from the associative law) |
+| $a^{-1} = a$, $b^{-1} = b$ | Conclusion of step 1 |
+| $(ab)^{-1} = ba$ | Substitute into step 3 |
+| $ab = ba$ | Passed by (1) and (2) equal sign |
 
-每一步都基于群公理或已证的结论，逻辑链条完整。✓
+Each step is based on group axioms or proven conclusions, and the logical chain is complete. ✓
 
-### 方法二：具体实例验证
+### Method 2: Verification with specific examples
 
-**例 1**：$G = \mathbb{Z}_2 = \{0, 1\}$（模 2 加法群）。
+**Example 1**: $G = \mathbb{Z}_2 = \{0, 1\}$ (modulo 2 additive group).
 
-$0 + 0 = 0$，$1 + 1 = 0 \pmod{2}$，满足 $g + g = 0$。且 $\mathbb{Z}_2$ 是交换群。✓
+$0 + 0 = 0$, $1 + 1 = 0 \pmod{2}$, satisfy $g + g = 0$. And $\mathbb{Z}_2$ is a swap group. ✓
 
-**例 2**：$G = \mathbb{Z}_2 \times \mathbb{Z}_2 = \{(0,0), (0,1), (1,0), (1,1)\}$（Klein 四元群）。
+**Example 2**: $G = \mathbb{Z}_2 \times \mathbb{Z}_2 = \{(0,0), (0,1), (1,0), (1,1)\}$ (Klein Quaternary Group).
 
-每个元素加自身得 $(0,0)$（单位元），满足条件。验证：
+Each element plus itself gets $(0,0)$ (unit yuan), which satisfies the condition. verify:
 
-$$(0,1) + (1,0) = (1,1),\quad (1,0) + (0,1) = (1,1)$$
+ $$(0,1) + (1,0) = (1,1),\quad (1,0) + (0,1) = (1,1)$$
 
-交换性成立。✓
+Commutativity is established. ✓
 
-**例 3**：$G = S_3$（三次对称群）。$(12)^2 = e$ 成立但 $(123)^2  = (132) \neq e$，不满足条件，交换性也不成立。✓
+**Example 3**: $G = S_3$ (cubic symmetry group). $(12)^2 = e$ is established but $(123)^2  = (132) \neq e$ , does not satisfy the condition and commutativity is not established. ✓
 
-### 方法三：检查是否隐含假设了交换性
+### Method 3: Check whether commutativity is implicitly assumed
 
-关键检查点：在展开法证明中，步骤
+Key checkpoint: In the expansion proof, steps
 
-$$a(abab)b = (aa)ba(bb)$$
+ $$a(abab)b = (aa)ba(bb)$$
 
-使用了**结合律**将括号重排：$a((ab)(ab))b = (a(ab))((ab)b) = ((aa)b)(a(bb))$。这完全合法，不涉及交换律。
+Use **associative law** to rearrange the brackets: $a((ab)(ab))b = (a(ab))((ab)b) = ((aa)b)(a(bb))$ . This is perfectly legal and does not involve commutativity.
 
-证明中没有出现类似 $ab = ba$ 的假设被提前使用。✓
+No assumptions like $ab = ba$ are used in the proof earlier. ✓
 
-## 最终答案
+## Final answer
 
-$$\boxed{\forall a, b \in G: ab = ba \quad \text{（即 $G$ 是 Abel 群）}}$$
+$$\boxed{\forall a, b \in G: ab = ba \quad \text{(i.e. $G$ is an Abelian group)}}$$
 
-## 易错点
-1. **$(ab)^{-1}$ 的两种表示**：关键等式 $(ab)^{-1} = ab$ 来自条件 $g^2 = e$，而 $(ab)^{-1} = b^{-1}a^{-1}$ 来自群的一般性质。将两者联系起来是本题的核心技巧。
-2. **逆元顺序**：$(ab)^{-1} = b^{-1}a^{-1}$，不是 $a^{-1}b^{-1}$（除非群已经是交换群）。初学者经常弄反顺序。
-3. **$g^2 = e$ 不意味着 $g = e$**：$g^2 = e$ 只说明 $g$ 的阶是 $1$ 或 $2$，并不意味着 $g = e$。例如 Klein 四元群中除了单位元外还有 3 个满足 $g^2 = e$ 的非平凡元素。
-4. **有限性的推断**：从条件可以推出若 $G$ 有限，则 $|G|$ 是 2 的幂。但证明交换性时不需要 $G$ 有限——该结论对任意群（包括无限群）成立。
+## Easy to make mistakes
+1. **Two representations of $(ab)^{-1}$**: The key equation $(ab)^{-1} = ab$ comes from the condition $g^2 = e$, while $(ab)^{-1} = b^{-1}a^{-1}$ comes from the general properties of the group. Connecting the two is the core skill of this question.
+2. **Reverse element order**: $(ab)^{-1} = b^{-1}a^{-1}$, not $a^{-1}b^{-1}$ (unless the group is already a commutative group). Beginners often reverse the order.
+3. ** $g^2 = e$ does not mean $g = e$ **: $g^2 = e$ only indicates that the order of $g$ is $1$ or $2$, and does not mean $g = e$. For example, in addition to the identity element, there are three non-trivial elements satisfying $g^2 = e$ in the Klein quaternion group.
+4. **Inference of finiteness**: It can be deduced from the condition that if $G$ is finite, then $|G|$ is a power of 2. But the proof of commutativity does not require $G$ to be finite - the result holds for any group, including infinite groups.
diff --git a/examples/algebra.md b/examples/algebra.md
index b3640c2..7e2f7c8 100644
--- a/examples/algebra.md
+++ b/examples/algebra.md
@@ -1,150 +1,150 @@
-# 代数式化简与因式分解
+# Algebraic simplification and factorization
 
-## 用户输入
-化简：$\displaystyle \frac{x^3-8}{x^2-4} \times \frac{x+2}{x^2+2x+4}$
+## User input
+Simplify: $\displaystyle \frac{x^3-8}{x^2-4} \times \frac{x+2}{x^2+2x+4}$
 
-## Skill 分类
-代数（Algebra）— 多项式因式分解与分式化简
+## Skill Category
+Algebra—polynomial factorization and fractional simplification
 
-## 题意解析
+## Question meaning analysis
 
-**已知条件：**
-- 分式表达式：$\frac{x^3-8}{x^2-4} \times \frac{x+2}{x^2+2x+4}$
+**Known conditions:**
+- Fraction expression: $\frac{x^3-8}{x^2-4} \times \frac{x+2}{x^2+2x+4}$
 
-**化简目标：**
-- 通过因式分解将分子分母的公因式约去，得到最简分式。
+**Simplification goal:**
+- Remove the common factors of the numerator and denominator through factorization to obtain the simplest fraction.
 
-**变量与定义域：**
-分式有意义的条件为所有分母 $\neq 0$：
+**Variables and Domains:**
+A fraction is meaningful if all denominators are $\neq 0$ :
 
-- $x^2-4 \neq 0$ 即 $x \neq \pm 2$
-- $x^2+2x+4$ 的判别式 $\Delta = 2^2 - 4 \times 1 \times 4 = 4 - 16 = -12 < 0$，对一切实数 $x \neq 0$ 均不为零（恒正）。
-- 综合：$x \in \mathbb{R} \setminus \{-2, 2\}$
+- $x^2-4 \neq 0$ is $x \neq \pm 2$
+- The discriminant $\Delta = 2^2 - 4 \times 1 \times 4 = 4 - 16 = -12 < 0$ of $x^2+2x+4$ is non-zero (constantly positive) for all real numbers $x \neq 0$.
+- General: $x \in \mathbb{R} \setminus \{-2, 2\}$
 
-**隐式条件：**
-- $x^2+2x+4 = \frac{x^3-8}{x-2}$，这是立方差公式的推论，将在因式分解中用到。
+**Implicit condition:**
+- $x^2+2x+4 = \frac{x^3-8}{x-2}$, which is a corollary of the cubic variance formula and will be used in factorization.
 
-**解的个数：**
-- 化简结果是唯一确定的，定义域内恒成立。
+**Number of solutions:**
+- The simplification result is uniquely certain and always true within the definition domain.
 
-## 方法选择
+## Method selection
 
-**选择方法：** 因式分解 + 约分法
+**Selection method:** Factorization + Reduction method
 
-**理由：**
-- 分子包含 $x^3-8$，是立方差形式 $a^3 - b^3$，可用公式 $a^3-b^3 = (a-b)(a^2+ab+b^2)$ 分解。
-- 分母包含 $x^2-4$，是平方差形式 $a^2 - b^2$，可用公式 $a^2-b^2 = (a-b)(a+b)$ 分解。
-- 两个分式相乘时，分子乘分子、分母乘分母后，公共因子将直接抵消。
+**reason:**
+- The numerator contains $x^3-8$, which is the cubic difference form $a^3 - b^3$, and can be decomposed by the formula $a^3-b^3 = (a-b)(a^2+ab+b^2)$.
+- The denominator contains $x^2-4$, which is the square difference form $a^2 - b^2$, and can be decomposed by the formula $a^2-b^2 = (a-b)(a+b)$.
+- When two fractions are multiplied together, the common factors will cancel directly after multiplying the numerator by the numerator and the denominator by the denominator.
 
-**备选方法：**
-- 直接分子分母展开再因式分解——步骤冗余，且展开 $x^3-8$ 后无法直接看出因式。
-- 长除法验证——适合验证结果，但不适合作为主要化简手段。
+**Alternative method:**
+- Direct numerator and denominator expansion and then factorization - the steps are redundant, and the factors cannot be directly seen after expanding $x^3-8$.
+- Long division verification - suitable for verifying results, but not suitable as a primary means of simplification.
 
-## 解题过程
+## Problem solving process
 
-### 第一步：因式分解各多项式
+### Step 1: Factor each polynomial
 
-**分子 $M_1 = x^3 - 8$：**
+**Numerator $M_1 = x^3 - 8$:**
 
-将 $8$ 写为 $2^3$，应用立方差公式 $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$：
+Write $8$ as $2^3$ and apply the cubic difference formula $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ :
 
-$$
+ $$
 x^3 - 2^3 = (x-2)(x^2 + x \cdot 2 + 2^2) = (x-2)(x^2 + 2x + 4)
 $$
 
-**分母 $D_1 = x^2 - 4$：**
+**Denominator $D_1 = x^2 - 4$:**
 
-将 $4$ 写为 $2^2$，应用平方差公式 $a^2 - b^2 = (a-b)(a+b)$：
+Write $4$ as $2^2$ and apply the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$ :
 
-$$
+ $$
 x^2 - 2^2 = (x-2)(x+2)
 $$
 
-**分子 $M_2 = x+2$：** 已为最简形式，无需分解。
+**Numerator $M_2 = x+2$:** is in its simplest form and does not need to be decomposed.
 
-**分母 $D_2 = x^2+2x+4$：** 判别式 $\Delta = -12 < 0$，在实数范围内不可再分解。
+**Denominator $D_2 = x^2+2x+4$:** Discriminant $\Delta = -12 < 0$, which cannot be decomposed in the range of real numbers.
 
-### 第二步：代入并组成连乘
+### Step 2: Substitute and form a continuous multiplication
 
-原式 = $\displaystyle \frac{M_1}{D_1} \times \frac{M_2}{D_2} = \frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)} \times \frac{x+2}{x^2+2x+4}$
+Original = $\displaystyle \frac{M_1}{D_1} \times \frac{M_2}{D_2} = \frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)} \times \frac{x+2}{x^2+2x+4}$
 
-### 第三步：约分
+### Step 3: Approximate points
 
-将分子分母写成一行：
+Write the numerator and denominator on one line:
 
-$$
+ $$
 \frac{(x-2)(x^2+2x+4)(x+2)}{(x-2)(x+2)(x^2+2x+4)}
 $$
 
-约去公因式（定义域内这些因子均不为零）：
+Reducing the common factors (none of these factors is zero in the domain):
 
-- 约去 $(x-2)$（因为 $x \neq 2$）
-- 约去 $(x+2)$（因为 $x \neq -2$）
-- 约去 $(x^2+2x+4)$（恒不为零）
+- Make an appointment with $(x-2)$ (because $x \neq 2$ )
+- Make an appointment with $(x+2)$ (because $x \neq -2$ )
+- Make an appointment with $(x^2+2x+4)$ (always not zero)
 
-所有因子全部消去，结果为：
+All factors are eliminated, and the result is:
 
-$$
+ $$
 = 1
 $$
 
-（在 $x \neq \pm 2$ 的条件下。）
+(Conditional to $x \neq \pm 2$.)
 
-## 验算
+## Check calculation
 
-### 验算方法一：数值代入法
+### Verification method one: Numerical substitution method
 
-在定义域内选取三个不同的 $x$ 值，分别代入原式和化简结果 $1$，验证它们相等。
+Select three different $x$ values ​​in the domain, substitute the original formula and the simplified result $1$ respectively, and verify that they are equal.
 
-**取 $x = 3$：**
+**Take $x = 3$:**
 
-- 原式 = $\frac{3^3-8}{3^2-4} \times \frac{3+2}{3^2+2 \times 3+4} = \frac{27-8}{9-4} \times \frac{5}{9+6+4} = \frac{19}{5} \times \frac{5}{19} = 1$ ✓
+- Original = $\frac{3^3-8}{3^2-4} \times \frac{3+2}{3^2+2 \times 3+4} = \frac{27-8}{9-4} \times \frac{5}{9+6+4} = \frac{19}{5} \times \frac{5}{19} = 1$ ✓
 
-**取 $x = 0$：**
+**Take $x = 0$:**
 
-- 原式 = $\frac{0^3-8}{0^2-4} \times \frac{0+2}{0^2+0+4} = \frac{-8}{-4} \times \frac{2}{4} = 2 \times \frac{1}{2} = 1$ ✓
+- Original = $\frac{0^3-8}{0^2-4} \times \frac{0+2}{0^2+0+4} = \frac{-8}{-4} \times \frac{2}{4} = 2 \times \frac{1}{2} = 1$ ✓
 
-**取 $x = -1$：**
+**Take $x = -1$:**
 
-- 原式 = $\frac{(-1)^3-8}{(-1)^2-4} \times \frac{-1+2}{(-1)^2+2\times(-1)+4} = \frac{-1-8}{1-4} \times \frac{1}{1-2+4} = \frac{-9}{-3} \times \frac{1}{3} = 3 \times \frac{1}{3} = 1$ ✓
+- Original = $\frac{(-1)^3-8}{(-1)^2-4} \times \frac{-1+2}{(-1)^2+2\times(-1)+4} = \frac{-1-8}{1-4} \times \frac{1}{1-2+4} = \frac{-9}{-3} \times \frac{1}{3} = 3 \times \frac{1}{3} = 1$ ✓
 
-三个不同的数值验证全部通过。
+Three different numerical verifications all passed.
 
-### 验算方法二：展开还原法
+### Calculation method two: expansion reduction method
 
-将化简结果 $1$ 还原为：化简后的分式应为 $\frac{1}{1}$，这等价于说原分式的分子恒等于分母（在定义域内）。
+Restore the simplified result $1$ to: the simplified fraction should be $\frac{1}{1}$ , which is equivalent to saying that the numerator of the original fraction is always equal to the denominator (within the domain of definition).
 
-验证：原式分子 = $(x^3-8)(x+2)$，分母 = $(x^2-4)(x^2+2x+4)$。
+Verification: The numerator of the original formula = $(x^3-8)(x+2)$, the denominator = $(x^2-4)(x^2+2x+4)$.
 
-展开分子：
+Expand the numerator:
 
-$$
+ $$
 (x^3-8)(x+2) = x^3 \cdot x + x^3 \cdot 2 - 8 \cdot x - 8 \cdot 2 = x^4 + 2x^3 - 8x - 16
 $$
 
-展开分母：
+Expand the denominator:
 
-$$
+ $$
 (x^2-4)(x^2+2x+4) = x^2(x^2+2x+4) - 4(x^2+2x+4)
 $$
-$$
+ $$
 = x^4 + 2x^3 + 4x^2 - 4x^2 - 8x - 16
 $$
-$$
+ $$
 = x^4 + 2x^3 - 8x - 16
 $$
 
-分子 $=$ 分母 $= x^4 + 2x^3 - 8x - 16$，因此在 $x \neq \pm 2$ 时原式 $= 1$。✓
+The numerator $=$ and the denominator $= x^4 + 2x^3 - 8x - 16$ , so at $x \neq \pm 2$ the original formula is $= 1$ . ✓
 
-## 最终答案
+## Final answer
 
-$$
+ $$
 \boxed{1} \quad (x \neq 2,\ x \neq -2)
 $$
 
-## 易错点
-1. **忘记定义域限制：** 化简后结果虽然是 $1$，但原分式在 $x=2$ 和 $x=-2$ 处无定义。答案必须注明 $x \neq \pm 2$。
-2. **立方差公式记错：** $a^3-b^3 = (a-b)(a^2+ab+b^2)$，注意中间 $+ab$ 项，不要误写为 $-ab$。如果是 $a^3+b^3 = (a+b)(a^2-ab+b^2)$。
-3. **平方差公式记错：** $a^2-b^2 = (a-b)(a+b)$，不要误写为 $(a-b)^2$。
-4. **约分合法性：** 只有当因子确实不为零时才能约去。虽然 $x^2+2x+4$ 在实数范围内恒不为零，但做题时要养成检查约去因子是否可能为零的习惯。
-5. **过度约分：** 有人可能在使用立方差公式分解后遗漏 $(x^2+2x+4)$ 项，导致约分不彻底。三个因子 $(x-2)$, $(x+2)$, $(x^2+2x+4)$ 都要约去。
+## Easy to make mistakes
+1. **Forgot domain restriction:** Although the result after simplification is $1$, the original fraction is not defined at $x=2$ and $x=-2$. Answers must be marked with $x \neq \pm 2$.
+2. **Mirror in remembering the cubic difference formula:** $a^3-b^3 = (a-b)(a^2+ab+b^2)$, pay attention to the $+ab$ item in the middle, do not mistakenly write it as $-ab$. if $a^3+b^3 = (a+b)(a^2-ab+b^2)$ .
+3. **The square difference formula is incorrectly remembered:** $a^2-b^2 = (a-b)(a+b)$, do not write it as $(a-b)^2$ by mistake.
+4. **Legality of reduction:** It can be reduced only when the factor is indeed not zero. Although $x^2+2x+4$ is always non-zero in the range of real numbers, you should develop the habit of checking whether the reduction factors may be zero when solving problems.
+5. **Over-reduction:** Someone may miss the $(x^2+2x+4)$ term after using the cubic variance formula to decompose, resulting in incomplete reduction. The three factors $(x-2)$ , $(x+2)$ , $(x^2+2x+4)$ are all required.
diff --git a/examples/arithmetic.md b/examples/arithmetic.md
index 92367fa..552cb74 100644
--- a/examples/arithmetic.md
+++ b/examples/arithmetic.md
@@ -1,163 +1,163 @@
-# 分数四则混合运算
+#Four mixed operations on fractions
 
-## 用户输入
-计算：$( \frac{2}{3} + \frac{5}{6} ) \times ( 1 - \frac{3}{8} ) \div \frac{7}{12}$
+## User input
+Calculation: $( \frac{2}{3} + \frac{5}{6} ) \times ( 1 - \frac{3}{8} ) \div \frac{7}{12}$
 
-## Skill 分类
-算术（Arithmetic）— 分数四则混合运算
+## Skill Category
+Arithmetic—Four mixed arithmetic operations on fractions
 
-## 题意解析
-本题是一道分数四则混合运算题，涉及分数的加法、减法、乘法和除法。
+## Question meaning analysis
+This question is a mixed operation question on four fractions, involving addition, subtraction, multiplication and division of fractions.
 
-**已知条件：**
-- 表达式：$( \frac{2}{3} + \frac{5}{6} ) \times ( 1 - \frac{3}{8} ) \div \frac{7}{12}$
+**Known conditions:**
+- Expression: $( \frac{2}{3} + \frac{5}{6} ) \times ( 1 - \frac{3}{8} ) \div \frac{7}{12}$
 
-**运算顺序：**
-- 先算括号内的加法和减法
-- 再按从左到右的顺序算乘法和除法
+**Order of operations:**
+- Do the addition and subtraction in parentheses first
+- Then calculate multiplication and division in order from left to right
 
-**变量与定义域：**
-- 所有分母均不为零：$3$, $6$, $8$, $12$ 均为非零常数，定义域无限制。
+**Variables and Domains:**
+- All denominators are non-zero: $3$ , $6$ , $8$ , $12$ are all non-zero constants, and the domain is unlimited.
 
-**解的个数：**
-- 算术表达式的值是唯一确定的。
+**Number of solutions:**
+- The value of an arithmetic expression is uniquely determined.
 
-## 方法选择
-**选择方法：** 通分 + 约分法（标准分数运算方法）
+## Method selection
+**Selection method:** Common fraction + Reduction method (standard fraction operation method)
 
-**理由：**
-- 分数运算是算术中的基础内容，通分和约分是最直接、最标准的方法。
-- 将整数 $1$ 转换为分母为 $8$ 的分数是处理整数与分数混合运算的常规技巧。
-- 除法转换为乘以倒数，可以统一为乘法运算。
+**reason:**
+- Fraction operations are the basic content in arithmetic, and general division and reduction are the most direct and standard methods.
+- Converting an integer $1$ to a fraction with a denominator of $8$ is a common technique for working with mixed integers and fractions.
+- Division is converted to multiplication by the reciprocal, which can be unified into multiplication operations.
 
-**备选方法：**
-- 可以先全部转换为小数计算，但会产生舍入误差，不如分数运算精确。
-- 可以通过通分将整个表达式写成一个分数再约分，但步骤更多。
+**Alternative method:**
+- You can convert everything to decimal calculations first, but this will cause rounding errors and is not as accurate as fractional calculations.
+- It is possible to write the entire expression as a fraction and then reduce it by universal division, but there are more steps.
 
-## 解题过程
+## Problem solving process
 
-### 第一步：计算第一个括号 $A = \frac{2}{3} + \frac{5}{6}$
+### Step 1: Calculate the first bracket $A = \frac{2}{3} + \frac{5}{6}$
 
-通分，分母取 $3$ 和 $6$ 的最小公倍数 $6$：
+Common denominator, the denominator is the least common multiple of $3$ and $6$ $6$:
 
-$$
+ $$
 \frac{2}{3} = \frac{2 \times 2}{3 \times 2} = \frac{4}{6}
 $$
 
-$$
+ $$
 A = \frac{4}{6} + \frac{5}{6} = \frac{4+5}{6} = \frac{9}{6}
 $$
 
-约分（分子分母同时除以 $3$）：
+Reduction (the numerator and denominator are divided by $3$ at the same time):
 
-$$
+ $$
 A = \frac{9 \div 3}{6 \div 3} = \frac{3}{2}
 $$
 
-### 第二步：计算第二个括号 $B = 1 - \frac{3}{8}$
+### Step 2: Calculate the second bracket $B = 1 - \frac{3}{8}$
 
-将整数 $1$ 化为分母为 $8$ 的分数：
+Convert the integer $1$ into a fraction with the denominator $8$:
 
-$$
+ $$
 1 = \frac{8}{8}
 $$
 
-$$
+ $$
 B = \frac{8}{8} - \frac{3}{8} = \frac{8-3}{8} = \frac{5}{8}
 $$
 
-### 第三步：计算乘法 $C = A \times B$
+### Step 3: Calculate multiplication $C = A \times B$
 
-$$
+ $$
 C = \frac{3}{2} \times \frac{5}{8} = \frac{3 \times 5}{2 \times 8} = \frac{15}{16}
 $$
 
-分子分母互质，不需要再约分。
+The numerator and denominator are mutually prime, so no further reduction is needed.
 
-### 第四步：计算除法 $= C \div \frac{7}{12}$
+### Step 4: Calculate division $= C \div \frac{7}{12}$
 
-除以一个分数等于乘以它的倒数：
+Dividing a fraction is equivalent to multiplying by its reciprocal:
 
-$$
+ $$
 \frac{15}{16} \div \frac{7}{12} = \frac{15}{16} \times \frac{12}{7} = \frac{15 \times 12}{16 \times 7} = \frac{180}{112}
 $$
 
-约分：分子分母的最大公约数为 $4$：
+Reduction: The greatest common divisor of the numerator and denominator is $4$:
 
-$$
+ $$
 \frac{180 \div 4}{112 \div 4} = \frac{45}{28}
 $$
 
-$45$ 和 $28$ 的最大公约数为 $1$（$28=2^2\times 7$, $45=3^2\times 5$，无公共质因子），已是最简分数。
+The greatest common divisor of $45$ and $28$ is $1$ ( $28=2^2\times 7$ , $45=3^2\times 5$ , no common prime factors), which is the simplest fraction.
 
-## 验算
+## Check calculation
 
-### 验算方法一：小数近似计算
+### Calculation method one: decimal approximate calculation
 
-将每个分数转换为小数进行验算：
+Convert each fraction to a decimal for verification:
 
-- $\frac{2}{3} \approx 0.666667$
-- $\frac{5}{6} \approx 0.833333$
-- $\frac{2}{3} + \frac{5}{6} \approx 1.500000$
+-  $\frac{2}{3} \approx 0.666667$
+-  $\frac{5}{6} \approx 0.833333$
+-  $\frac{2}{3} + \frac{5}{6} \approx 1.500000$
 
-- $1 - \frac{3}{8} = 0.625000$
+-  $1 - \frac{3}{8} = 0.625000$
 
-- $1.5 \times 0.625 = 0.937500$
+-  $1.5 \times 0.625 = 0.937500$
 
-- $0.9375 \div \frac{7}{12} = 0.9375 \times \frac{12}{7} = 0.9375 \times 1.714286 \approx 1.607143$
+-  $0.9375 \div \frac{7}{12} = 0.9375 \times \frac{12}{7} = 0.9375 \times 1.714286 \approx 1.607143$
 
-计算 $\frac{45}{28}$ 的小数值：
+Compute the decimal value of $\frac{45}{28}$:
 
-$$
+ $$
 \frac{45}{28} = 45 \div 28 \approx 1.607143
 $$
 
-两者一致，验算通过。
+If both are consistent, the calculation passes.
 
-### 验算方法二：逆向验算
+### Verification method two: reverse verification
 
-从答案出发，逆向运算验证是否回到原题给定的数据。
+Starting from the answer, the reverse operation verifies whether it returns to the data given in the original question.
 
-**反推步骤一：** 将最终结果 $\frac{45}{28}$ 乘以 $\frac{7}{12}$ 应得到 $\frac{15}{16}$：
+**Back step one:** Multiply the final result $\frac{45}{28}$ by $\frac{7}{12}$ and you should get $\frac{15}{16}$:
 
-$$
+ $$
 \frac{45}{28} \times \frac{7}{12} = \frac{45 \times 7}{28 \times 12} = \frac{315}{336}
 $$
 
-约分：分子分母同时除以 $3$：
+Reduction: divide both the numerator and denominator by $3$:
 
-$$
+ $$
 \frac{315 \div 3}{336 \div 3} = \frac{105}{112}
 $$
 
-再同时除以 $7$：
+And then divide by $7$ at the same time:
 
-$$
+ $$
 \frac{105 \div 7}{112 \div 7} = \frac{15}{16} \quad ✓
 $$
 
-**反推步骤二：** 将 $\frac{15}{16}$ 除以 $\frac{5}{8}$ 应得到 $\frac{3}{2}$：
+**Reverse Step 2:** Divide $\frac{15}{16}$ by $\frac{5}{8}$ and you should get $\frac{3}{2}$:
 
-$$
+ $$
 \frac{15}{16} \div \frac{5}{8} = \frac{15}{16} \times \frac{8}{5} = \frac{15 \times 8}{16 \times 5} = \frac{120}{80} = \frac{3}{2} \quad ✓
 $$
 
-**反推步骤三：** 将 $\frac{3}{2}$ 减去 $\frac{5}{6}$ 应得到 $\frac{2}{3}$：
+**Back step three:** Subtract $\frac{5}{6}$ from $\frac{3}{2}$ and you should get $\frac{2}{3}$:
 
-$$
+ $$
 \frac{3}{2} - \frac{5}{6} = \frac{9}{6} - \frac{5}{6} = \frac{4}{6} = \frac{2}{3} \quad ✓
 $$
 
-三次逆向运算全部吻合，验证通过。
+The three reverse operations all match and the verification passes.
 
-## 最终答案
+## Final answer
 
-$$
+ $$
 \boxed{\frac{45}{28}}
 $$
 
-## 易错点
-1. **运算顺序错误：** 容易忽略括号内的运算优先于括号外的乘除，直接从左到右算。正确做法是**先算括号，再从左到右算乘除**。
-2. **约分时机：** 有些人习惯在每一步都约分，有些人在最后一步才约分。两种做法都对，但**中间约分可以减小数字，降低计算难度**。例如第一步中 $\frac{9}{6}$ 应约分为 $\frac{3}{2}$，否则后续乘法会得到 $\frac{9 \times 5}{6 \times 8} = \frac{45}{48}$，虽然也能最终化简，但数字更大。
-3. **除法处理：** 除以分数时忘记乘以倒数，或者分子分母颠倒是常见错误。$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$。
-4. **整数化分数：** $1 - \frac{3}{8}$ 中忘记将 $1$ 通分为 $\frac{8}{8}$，直接算成 $\frac{1-3}{8}$ 导致错误。
+## Easy to make mistakes
+1. **Wrong order of operations:** It is easy to ignore that operations within parentheses take precedence over multiplication and division outside parentheses, and are calculated directly from left to right. The correct approach is to calculate the parentheses first, then calculate multiplication and division from left to right.
+2. **Time to make an appointment:** Some people are used to making an appointment at every step, while some people only make an appointment at the last step. Both approaches are correct, but **intermediate reduction can reduce the number and reduce the difficulty of calculation**. For example, in the first step, $\frac{9}{6}$ should be reduced to $\frac{3}{2}$, otherwise subsequent multiplication will result in $\frac{9 \times 5}{6 \times 8} = \frac{45}{48}$. Although it can be simplified eventually, the number is larger.
+3. **Division Processing:** Forgetting to multiply by the reciprocal when dividing by a fraction, or reversing the numerator and denominator is a common mistake. $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$ .
+4. **Integer fraction:** In $1 - \frac{3}{8}$, we forget to divide $1$ into $\frac{8}{8}$, and directly calculate it as $\frac{1-3}{8}$, resulting in an error.
diff --git a/examples/calculus.md b/examples/calculus.md
index c05db1a..8133036 100644
--- a/examples/calculus.md
+++ b/examples/calculus.md
@@ -1,89 +1,89 @@
-# 极限计算 —— Taylor 展开法
+# Limit calculation - Taylor expansion method
 
-## 用户输入
-求极限：
+## User input
+Find the limit:
 
-$$\lim_{x \to 0} \frac{e^x - 1 - \sin x}{x^2}$$
+ $$\lim_{x \to 0} \frac{e^x - 1 - \sin x}{x^2}$$
 
-## Skill 分类
-微积分
+## Skill Category
+Calculus
 
-## 题意解析
-- **极限类型**：$x \to 0$ 时分子分母均趋于 $0$，为 $\frac{0}{0}$ 型不定式。
-- **分子结构**：$e^x - 1 - \sin x$。当 $x \to 0$ 时，$e^x \to 1$，$\sin x \to 0$，所以分子 $\to 1 - 1 - 0 = 0$。
-- **分母结构**：$x^2 \to 0$。
-- **关键观察**：$e^x$ 和 $\sin x$ 在 $x = 0$ 处均有 Taylor 展开式，展开到 $x^2$ 的阶数即可消去低阶项。
-- **展开阶数的选择**：分母为 $x^2$，展开到 $x^2$ 项（即 $O(x^3)$）以捕获分子中与 $x^2$ 同阶的项。
+## Question meaning analysis
+- **Limit type**: When $x \to 0$, both the numerator and denominator tend to $0$, which is the $\frac{0}{0}$ type infinitive.
+- **Molecular Structure**: $e^x - 1 - \sin x$. When $x \to 0$ , $e^x \to 1$ , $\sin x \to 0$ , so the numerator is $\to 1 - 1 - 0 = 0$ .
+- **Denominator Structure**: $x^2 \to 0$ .
+- **Key Observation**: $e^x$ and $\sin x$ both have Taylor expansions at $x = 0$, which can be expanded to the order of $x^2$ to eliminate the lower-order terms.
+- **Choice of expansion order**: Denominator is $x^2$, expand to $x^2$ terms (i.e. $O(x^3)$) to capture terms in the numerator with the same order as $x^2$.
 
-## 方法选择
-**首选方法**：Taylor 展开法。将 $e^x$ 和 $\sin x$ 分别展开到足够高阶，代入后化简，低阶项相互抵消，余下主导项即极限值。
+## Method selection
+**Preferred method**: Taylor expansion. Expand $e^x$ and $\sin x$ to a sufficiently high order and simplify after substitution. The low-order terms cancel each other out, and the remaining dominant terms are the limit values.
 
-**备选方法**：洛必达法则。本题也可用洛必达法则求解，但需要求导两次，计算量较大且容易出错，Taylor 展开更直接。数值逼近可作为数值验证。
+**Alternative Method**: Lópida's Law. This problem can also be solved by L'Hôpital's rule, but it requires two derivatives, which requires a large amount of calculation and is prone to errors. Taylor expansion is more direct. Numerical approximations serve as numerical verification.
 
-**理由**：形如 $e^x - \text{多项式}$ 的极限，Taylor 展开通常比洛必达法则更高效，尤其是当分子含多个函数时。
+**Rationale**: As a limit of the form $e^x - \text{polynomial}$, Taylor's expansion is often more efficient than L'Hobida's rule, especially when the molecule contains multiple functions.
 
-## 解题过程
+## Problem solving process
 
-### 第一步：写出 Taylor 展开式
+### Step 1: Write the Taylor expansion
 
-$e^x$ 在 $x = 0$ 处的展开（到 $x^3$ 项）：
+Expansion of $e^x$ at $x = 0$ (to $x^3$ item):
 
-$$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)$$
+ $$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)$$
 
-$\sin x$ 在 $x = 0$ 处的展开（到 $x^3$ 项）：
+Expansion of $\sin x$ at $x = 0$ (to $x^3$ item):
 
-$$\sin x = x - \frac{x^3}{6} + O(x^5)$$
+ $$\sin x = x - \frac{x^3}{6} + O(x^5)$$
 
-### 第二步：代入分子
+### Step 2: Substitute the numerator
 
-$$e^x - 1 - \sin x = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)\right) - 1 - \left(x - \frac{x^3}{6} + O(x^5)\right)$$
+ $$e^x - 1 - \sin x = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)\right) - 1 - \left(x - \frac{x^3}{6} + O(x^5)\right)$$
 
-消去常数项和一次项：
+Eliminate constant and linear terms:
 
-$$= x + \frac{x^2}{2} + \frac{x^3}{6} - x + \frac{x^3}{6} + O(x^4)$$
+ $$= x + \frac{x^2}{2} + \frac{x^3}{6} - x + \frac{x^3}{6} + O(x^4)$$
 
-$$= \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)$$
+ $$= \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)$$
 
-### 第三步：求极限
+### Step 3: Find the limit
 
-$$\frac{e^x - 1 - \sin x}{x^2} = \frac{\frac{x^2}{2} + \frac{x^3}{3} + O(x^4)}{x^2} = \frac{1}{2} + \frac{x}{3} + O(x^2)$$
+ $$\frac{e^x - 1 - \sin x}{x^2} = \frac{\frac{x^2}{2} + \frac{x^3}{3} + O(x^4)}{x^2} = \frac{1}{2} + \frac{x}{3} + O(x^2)$$
 
-令 $x \to 0$：
+Let $x \to 0$ :
 
-$$\lim_{x \to 0} \frac{e^x - 1 - \sin x}{x^2} = \frac{1}{2}$$
+ $$\lim_{x \to 0} \frac{e^x - 1 - \sin x}{x^2} = \frac{1}{2}$$
 
-## 验算
+## Check calculation
 
-### 方法一：洛必达法则
+### Method 1: Lópida’s Law
 
-第一次应用（$\frac{0}{0}$ 型，条件满足：$x \to 0$，$e^x - 1 - \sin x$ 和 $x^2$ 均可导）：
+First application ($\frac{0}{0}$ type, conditions are met: $x \to 0$, $e^x - 1 - \sin x$ and $x^2$ can all be derived):
 
-$$\lim_{x \to 0} \frac{e^x - 1 - \sin x}{x^2} = \lim_{x \to 0} \frac{e^x - \cos x}{2x}$$
+ $$\lim_{x \to 0} \frac{e^x - 1 - \sin x}{x^2} = \lim_{x \to 0} \frac{e^x - \cos x}{2x}$$
 
-仍为 $\frac{0}{0}$ 型（$\frac{1 - 1}{0}$），第二次应用洛必达：
+Still of type $\frac{0}{0}$ ( $\frac{1 - 1}{0}$ ), apply Lupida for the second time:
 
-$$= \lim_{x \to 0} \frac{e^x + \sin x}{2} = \frac{1 + 0}{2} = \frac{1}{2}$$
+ $$= \lim_{x \to 0} \frac{e^x + \sin x}{2} = \frac{1 + 0}{2} = \frac{1}{2}$$
 
-结果一致。✓
+The results are consistent. ✓
 
-（注意：虽然洛必达在此题可行，但当分子含更多函数时，求导将变得繁琐；Taylor 展开的优势在于统一处理。）
+(Note: Although Lupida is feasible in this problem, when the molecule contains more functions, the derivation will become cumbersome; the advantage of Taylor expansion is that it can be processed uniformly.)
 
-### 方法二：数值逼近
+### Method 2: Numerical approximation
 
-| $x$ | $\frac{e^x - 1 - \sin x}{x^2}$ |
+|  $x$  |  $\frac{e^x - 1 - \sin x}{x^2}$  |
 |-----|--------------------------------|
-| 0.1 | $\frac{1.1051709 - 1 - 0.0998334}{0.01} = \frac{0.0053375}{0.01} \approx 0.5338$ |
-| 0.01 | $\frac{1.0100502 - 1 - 0.0099998}{0.0001} = \frac{0.0000504}{0.0001} \approx 0.504$ |
-| 0.001 | $\frac{1.0010005 - 1 - 0.0010000}{0.000001} = \frac{0.0000005}{0.000001} \approx 0.500$ |
+| 0.1 |  $\frac{1.1051709 - 1 - 0.0998334}{0.01} = \frac{0.0053375}{0.01} \approx 0.5338$  |
+| 0.01 |  $\frac{1.0100502 - 1 - 0.0099998}{0.0001} = \frac{0.0000504}{0.0001} \approx 0.504$  |
+| 0.001 |  $\frac{1.0010005 - 1 - 0.0010000}{0.000001} = \frac{0.0000005}{0.000001} \approx 0.500$  |
 
-当 $x \to 0$ 时，数值逼近 $0.5$。✓
+When $x \to 0$, the value approaches $0.5$. ✓
 
-## 最终答案
+## Final answer
 
-$$\boxed{\lim_{x \to 0} \frac{e^x - 1 - \sin x}{x^2} = \frac{1}{2}}$$
+ $$\boxed{\lim_{x \to 0} \frac{e^x - 1 - \sin x}{x^2} = \frac{1}{2}}$$
 
-## 易错点
-1. **展开阶数不足**：本题只展开到 $x$ 次是不够的（$e^x$ 和 $\sin x$ 的一次项相互抵消），至少要展开到 $x^2$ 项才能得到非零结果。
-2. **O 记号的处理**：$O(x^4)/x^2 = O(x^2) \to 0$（当 $x \to 0$），但不能简单写为 0 而丢失信息。关键是确保主导项被正确保留。
-3. **洛必达法则适用条件**：每次应用前必须确认分子分母在 $x \to 0$ 时均趋于 $0$（或均趋于 $\infty$），且导数存在、导数之比极限存在。
-4. **$\sin x$ 的展开**：$\sin x = x - x^3/6 + \cdots$，注意是减去 $x^3/6$ 而非加。
+## Easy to make mistakes
+1. **Insufficient expansion order**: It is not enough to expand this question to $x$ times (the linear terms of $e^x$ and $\sin x$ cancel each other out). It must be expanded to at least $x^2$ items to obtain a non-zero result.
+2. **O token processing**: $O(x^4)/x^2 = O(x^2) \to 0$ (when $x \to 0$), but cannot simply be written as 0 without losing information. The key is to ensure that the dominant term is preserved correctly.
+3. **Applicable conditions for Lópida's law**: Before each application, it must be confirmed that the numerator and denominator tend to $0$ (or both tend to $\infty$) when they are at $x \to 0$, and the derivative exists and the ratio limit of the derivative exists.
+4. **Expansion of $\sin x$**: $\sin x = x - x^3/6 + \cdots$. Note that $x^3/6$ is subtracted instead of added.
diff --git a/examples/counterexample.md b/examples/counterexample.md
index 4fee0dc..53565cd 100644
--- a/examples/counterexample.md
+++ b/examples/counterexample.md
@@ -1,120 +1,120 @@
-# 反例构造：可导但导数不连续
+# Counterexample construction: Differentiable but the derivative is discontinuous
 
-## 用户输入
-有人断言："如果函数 $f$ 在开区间 $(a, b)$ 上可导，则其导函数 $f'$ 在 $(a, b)$ 上连续。" 请判断该断言是否正确。
+## User input
+Someone asserts: "If function $f$ is differentiable on the open interval $(a, b)$, then its derivative function $f'$ is continuous on $(a, b)$." Please determine whether this assertion is correct.
 
-## Skill 分类
-实分析 / 反例构造
+## Skill Category
+Real analysis/counterexample construction
 
-## 题意解析
-- **断言**：可导 $\implies$ 导数连续
-- **目标**：判断真伪；若为伪，构造反例
-- **需要检查的条件**：
-  1. 函数在定义域每点可导
-  2. 导函数在某点不连续（极限不存在或极限不等于函数值）
-- **常见认知偏差**：许多初学者误认为可导函数的导数"应该"是连续的，但可导性只保证函数本身连续，不保证导数的连续性
-- **解的多重性**：反例不唯一，但经典反例最具教学价值
+## Question meaning analysis
+- **Assertion**: Differentiable $\implies$ derivative is continuous
+- **Goal**: Determine authenticity; if false, construct a counterexample
+- **Conditions to be checked**:
+1. The function is differentiable at every point in the domain
+2. The derivative function is discontinuous at a certain point (the limit does not exist or the limit is not equal to the function value)
+- **Common cognitive bias**: Many beginners mistakenly believe that the derivative of a differentiable function "should" be continuous, but differentiability only guarantees the continuity of the function itself, not the continuity of the derivative.
+- **Multiplicity of Solutions**: Counterexamples are not unique, but classic counterexamples have the most teaching value
 
-## 方法选择
-采用**构造反例法**，构造一个处处可导但导数在一点不连续的函数。经典反例为：
+## Method selection
+Use the **Construction of Counterexample Method** to construct a function that is differentiable everywhere but has a discontinuous derivative at one point. The classic counterexample is:
 
-$$f(x) = \begin{cases} x^2 \sin(1/x), & x \neq 0 \\ 0, & x = 0 \end{cases}$$
+ $$f(x) = \begin{cases} x^2 \sin(1/x), & x \neq 0 \\ 0, & x = 0 \end{cases}$$
 
-选择该例的理由：$x^2$ 提供可导性（平方项在零点可导），$\sin(1/x)$ 提供导数振荡（导致不连续），两者乘积极为巧妙。
+The reason for choosing this example: $x^2$ provides differentiability (the square term is differentiable at the zero point), $\sin(1/x)$ provides derivative oscillation (leading to discontinuity), and the positive multiplication of the two is clever.
 
-## 解题过程
+## Problem solving process
 
-**断言：错误。可导不蕴含导数连续。**
+**Assertion: False. Differentiability does not imply continuity of derivatives. **
 
-### 构造反例
+### Constructing counterexamples
 
-定义函数 $f: \mathbb{R} \to \mathbb{R}$：
+Define function $f: \mathbb{R} \to \mathbb{R}$:
 
-$$f(x) = \begin{cases} x^2 \sin\left(\dfrac{1}{x}\right), & x \neq 0 \\[12pt] 0, & x = 0 \end{cases}$$
+ $$f(x) = \begin{cases} x^2 \sin\left(\dfrac{1}{x}\right), & x \neq 0 \\[12pt] 0, & x = 0 \end{cases}$$
 
-### Step 1：证明 $f$ 在 $x \neq 0$ 处可导
+### Step 1: Prove that $f$ is derivable at $x \neq 0$
 
-当 $x \neq 0$ 时，$f(x)$ 是可导函数 $x^2$ 和 $\sin(1/x)$ 的乘积（复合函数可导，因 $\sin$ 处处可导，$1/x$ 在 $x \neq 0$ 可导），由乘积法则：
+When $x \neq 0$, $f(x)$ is the product of differentiable functions $x^2$ and $\sin(1/x)$ (the composite function is differentiable, because $\sin$ is differentiable everywhere and $1/x$ is differentiable at $x \neq 0$), according to the product rule:
 
-$$f'(x) = 2x \cdot \sin\left(\frac{1}{x}\right) + x^2 \cdot \cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)$$
+ $$f'(x) = 2x \cdot \sin\left(\frac{1}{x}\right) + x^2 \cdot \cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)$$
 
-$$f'(x) = 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right), \quad x \neq 0$$
+ $$f'(x) = 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right), \quad x \neq 0$$
 
-### Step 2：证明 $f$ 在 $x = 0$ 处可导
+### Step 2: Prove that $f$ is derivable at $x = 0$
 
-由导数定义：
+Defined by the derivative:
 
-$$f'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h \to 0} h \cdot \sin\left(\frac{1}{h}\right)$$
+ $$f'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h \to 0} h \cdot \sin\left(\frac{1}{h}\right)$$
 
-由于 $\left|\sin(1/h)\right| \leq 1$，有 $0 \leq |h \cdot \sin(1/h)| \leq |h|$，由夹逼定理（squeeze theorem），
+Since $\left|\sin(1/h)\right| \leq 1$ , there is $0 \leq |h \cdot \sin(1/h)| \leq |h|$ , by squeeze theorem,
 
-$$\lim_{h \to 0} h \cdot \sin\left(\frac{1}{h}\right) = 0$$
+ $$\lim_{h \to 0} h \cdot \sin\left(\frac{1}{h}\right) = 0$$
 
-故 $f'(0) = 0$。$f$ 在 $x = 0$ 处可导。
+Hence $f'(0) = 0$ . $f$ is derivable at $x = 0$.
 
-### Step 3：证明 $f'$ 在 $x = 0$ 处不连续
+### Step 3: Prove that $f'$ is discontinuous at $x = 0$
 
-要证明 $f'$ 在 $x = 0$ 不连续，需证 $\lim_{x \to 0} f'(x)$ 不存在（或存在但不等于 $f'(0) = 0$）。
+To prove that $f'$ is discontinuous at $x = 0$, you need to prove that $\lim_{x \to 0} f'(x)$ does not exist (or exists but is not equal to $f'(0) = 0$).
 
-考察 $f'(x) = 2x\sin(1/x) - \cos(1/x)$（$x \neq 0$）：
+Examine $f'(x) = 2x\sin(1/x) - \cos(1/x)$ ( $x \neq 0$ ):
 
-- 第一项 $\lim_{x \to 0} 2x\sin(1/x) = 0$（夹逼定理）
-- 第二项 $\lim_{x \to 0} \cos(1/x)$ **不存在**
+- The first term $\lim_{x \to 0} 2x\sin(1/x) = 0$ (pinch theorem)
+- The second item $\lim_{x \to 0} \cos(1/x)$ **does not exist**
 
-当 $x \to 0$ 时，$1/x \to \infty$，$\cos(1/x)$ 在 $[-1, 1]$ 之间无限振荡，极限不存在。
+When $x \to 0$, $1/x \to \infty$, $\cos(1/x)$ oscillate infinitely between $[-1, 1]$, the limit does not exist.
 
-例如：取序列 $x_n = \frac{1}{2n\pi}$，则 $\cos(1/x_n) = \cos(2n\pi) = 1$，$f'(x_n) \to -1$。
-取序列 $x_n = \frac{1}{(2n+1)\pi}$，则 $\cos(1/x_n) = \cos((2n+1)\pi) = -1$，$f'(x_n) \to 1$。
+For example: take the sequence $x_n = \frac{1}{2n\pi}$, then $\cos(1/x_n) = \cos(2n\pi) = 1$, $f'(x_n) \to -1$.
+Take the sequence $x_n = \frac{1}{(2n+1)\pi}$, then $\cos(1/x_n) = \cos((2n+1)\pi) = -1$, $f'(x_n) \to 1$.
 
-两个子序列趋于不同极限，因此 $\lim_{x \to 0} f'(x)$ 不存在。
+The two subsequences tend to different limits, so $\lim_{x \to 0} f'(x)$ does not exist.
 
-**结论**：$f'$ 在 $x = 0$ 处不连续（甚至极限都不存在），但 $f$ 在 $x = 0$ 处可导。原断言被证伪。
+**Conclusion**: $f'$ is discontinuous at $x = 0$ (not even the limit exists), but $f$ is differentiable at $x = 0$. The original assertion was falsified.
 
-## 验算
+## Check calculation
 
-**验算方法 1：验证 $f'(0)$ 的计算**
+**Verification method 1: Verify the calculation of $f'(0)$**
 
-$$\lim_{h \to 0} \frac{h^2 \sin(1/h)}{h} = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right)$$
+ $$\lim_{h \to 0} \frac{h^2 \sin(1/h)}{h} = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right)$$
 
-夹逼定理：$-|h| \leq h\sin(1/h) \leq |h|$，当 $h \to 0$ 时两端均趋于 $0$，故极限为 $0$ ✓
+Pinch theorem: $-|h| \leq h\sin(1/h) \leq |h|$, when $h \to 0$, both ends tend to $0$, so the limit is $0$ ✓
 
-**验算方法 2：验证导数公式的求导**
+**Check calculation method 2: Verify the derivation of the derivative formula**
 
-对 $x \neq 0$：
-$$f'(x) = \frac{d}{dx}[x^2] \cdot \sin(1/x) + x^2 \cdot \frac{d}{dx}[\sin(1/x)]$$
-$$= 2x \cdot \sin(1/x) + x^2 \cdot \cos(1/x) \cdot (-1/x^2)$$
-$$= 2x\sin(1/x) - \cos(1/x)$$
+For $x \neq 0$ :
+ $$f'(x) = \frac{d}{dx}[x^2] \cdot \sin(1/x) + x^2 \cdot \frac{d}{dx}[\sin(1/x)]$$
+ $$= 2x \cdot \sin(1/x) + x^2 \cdot \cos(1/x) \cdot (-1/x^2)$$
+ $$= 2x\sin(1/x) - \cos(1/x)$$
 
-求导正确 ✓
+The derivation is correct ✓
 
-**验算方法 3：验证 $f'$ 的不连续性——两点序列法**
+**Check method 3: Verify the discontinuity of $f'$ - two-point sequence method**
 
-- 序列 $x_n^{(1)} = 1/(2n\pi)$：$f'(x_n^{(1)}) = \frac{2}{2n\pi}\sin(2n\pi) - \cos(2n\pi) = 0 - 1 = -1$
-- 序列 $x_n^{(2)} = 1/((2n+1)\pi)$：$f'(x_n^{(2)}) = \frac{2}{(2n+1)\pi}\sin((2n+1)\pi) - \cos((2n+1)\pi) = 0 - (-1) = 1$
+- Sequence $x_n^{(1)} = 1/(2n\pi)$ : $f'(x_n^{(1)}) = \frac{2}{2n\pi}\sin(2n\pi) - \cos(2n\pi) = 0 - 1 = -1$
+- Sequence $x_n^{(2)} = 1/((2n+1)\pi)$ : $f'(x_n^{(2)}) = \frac{2}{(2n+1)\pi}\sin((2n+1)\pi) - \cos((2n+1)\pi) = 0 - (-1) = 1$
 
-$\lim_{n \to \infty} f'(x_n^{(1)}) = -1$，$\lim_{n \to \infty} f'(x_n^{(2)}) = 1$，两极限不等，$\lim_{x \to 0} f'(x)$ 不存在 ✓
+$\lim_{n \to \infty} f'(x_n^{(1)}) = -1$, $\lim_{n \to \infty} f'(x_n^{(2)}) = 1$, the two limits are not equal, $\lim_{x \to 0} f'(x)$ does not exist ✓
 
-**验算方法 4：确认反例有效性的完整清单**
+**Check method 4: Complete list to confirm the validity of counterexamples**
 
-| 检查项 | 结果 |
+| Check items | Results |
 |--------|------|
-| $f$ 在 $x \neq 0$ 可导 | ✓ 初等函数的乘积和复合 |
-| $f$ 在 $x = 0$ 可导 | ✓ $f'(0) = 0$ |
-| $f'$ 在 $x \neq 0$ 的表达式正确 | ✓ 已验证求导 |
-| $\lim_{x \to 0} f'(x)$ 不存在 | ✓ 序列法证明 |
-| 因此 $f'$ 在 $x=0$ 不连续 | ✓ |
+| $f$ is differentiable in $x \neq 0$ | ✓ Product and composition of elementary functions |
+| $f$ is differentiable from $x = 0$ | ✓ $f'(0) = 0$ |
+| The expression of $f'$ in $x \neq 0$ is correct | ✓ Derivative verified |
+| $\lim_{x \to 0} f'(x)$ does not exist | ✓ Sequence method proof |
+| Therefore $f'$ is discontinuous at $x=0$ | ✓ |
 
-## 最终答案
+## Final answer
 
-原断言 **错误**。反例：函数
+The original assertion was **wrong**. Counterexample: function
 
-$$f(x) = \begin{cases} x^2 \sin(1/x), & x \neq 0 \\ 0, & x = 0 \end{cases}$$
+ $$f(x) = \begin{cases} x^2 \sin(1/x), & x \neq 0 \\ 0, & x = 0 \end{cases}$$
 
-处处可导，$f'(0) = 0$，但 $\lim_{x \to 0} f'(x)$ 不存在，故 $f'$ 在 $x=0$ 不连续。
+It is derivable everywhere, $f'(0) = 0$, but $\lim_{x \to 0} f'(x)$ does not exist, so $f'$ is not continuous in $x=0$.
 
-## 易错点
-1. **混淆条件与结论**：可导 $\implies$ $f$ 连续，但导数的连续性需要额外的条件（如 $f$ 属于 $C^1$ 类）
-2. **误以为反例的特殊性**：这个反例不是"病态"的边界情况，而是标准教材中的经典反例，深刻说明了"可导"与"导数连续"之间的差距
-3. **夹逼定理的条件**：使用夹逼定理求 $f'(0)$ 时，需要确认 $\sin(1/h)$ 确实有界（$|\sin(\cdot)| \leq 1$），这是正确使用的关键
-4. **极限不存在的判定**：不能仅凭"振荡"就断言极限不存在，需要严格构造两个趋于不同极限的子序列
-5. **不要过度推广**：虽然存在反例，但达布定理（Darboux's theorem）保证导数具有介值性质——导数即使不连续，也不能有第一类间断点
+## Easy to make mistakes
+1. **Confusing conditions and conclusions**: Differentiable $\implies$ $f$ is continuous, but the continuity of the derivative requires additional conditions (such as $f$ belongs to the $C^1$ class)
+2. **Mistake on the particularity of the counterexample**: This counterexample is not a "pathological" boundary situation, but a classic counterexample in standard textbooks, which profoundly illustrates the gap between "differentiable" and "derivative continuous"
+3. **Conditions of the pinch theorem**: When using the pinch theorem to find $f'(0)$, you need to confirm that $\sin(1/h)$ is indeed bounded ($|\sin(\cdot)| \leq 1$). This is the key to correct use.
+4. **Determination of the non-existence of the limit**: It cannot be concluded that the limit does not exist based on "oscillation" alone. It is necessary to strictly construct two subsequences tending to different limits.
+5. **Don’t overgeneralize**: Although there are counterexamples, Darboux’s theorem guarantees that the derivative has intermediate properties—even if the derivative is discontinuous, it cannot have the first type of discontinuity.
diff --git a/examples/definite_integral.md b/examples/definite_integral.md
index c590824..3e0dba7 100644
--- a/examples/definite_integral.md
+++ b/examples/definite_integral.md
@@ -1,128 +1,128 @@
-# 定积分：$\displaystyle \int_0^{\pi/2} \cos^2 x \, dx$ 的计算
+# Definite integral: Calculation of $\displaystyle \int_0^{\pi/2} \cos^2 x \, dx$
 
-## 用户输入
-计算定积分 $\displaystyle \int_0^{\pi/2} \cos^2 x \, dx$
+## User input
+Compute definite integrals $\displaystyle \int_0^{\pi/2} \cos^2 x \, dx$
 
-## Skill 分类
-微积分 / 定积分
+## Skill Category
+Calculus / Definite Integral
 
-## 题意解析
-- **被积函数**：$\cos^2 x$，是三角函数的平方
-- **积分区间**：$[0, \pi/2]$，余弦函数在该区间上非负
-- **目标**：计算精确的定积分值
-- **方法选择依据**：$\cos^2 x$ 需要通过三角恒等式降次
-- **解的约束**：定积分的值应为正实数（被积函数非负），且不超过区间长度 $\pi/2$（$\cos^2 x \leq 1$）
+## Question meaning analysis
+- **integrand**: $\cos^2 x$, which is the square of the trigonometric function
+- **Integral interval**: $[0, \pi/2]$, the cosine function is non-negative in this interval
+- **Goal**: Calculate the exact value of the definite integral
+- **Method selection basis**: $\cos^2 x$ needs to be reduced by trigonometric identity
+- **Constraints on the solution**: The value of the definite integral should be a positive real number (the integrand is non-negative) and not exceed the interval length $\pi/2$ ( $\cos^2 x \leq 1$ )
 
-## 方法选择
+## Method selection
 
-**主方法：三角恒等式降次 + 直接积分**
+**Main Method: Trigonometric Identity Reduction + Direct Integration**
 
-使用倍角公式 $\cos^2 x = \dfrac{1 + \cos 2x}{2}$，将二次三角函数转化为一次三角函数，然后直接积分。
+Use the double angle formula $\cos^2 x = \dfrac{1 + \cos 2x}{2}$ to convert the quadratic trigonometric function into a linear trigonometric function and then integrate directly.
 
-选择理由：对于 $\cos^2 x$ 或 $\sin^2 x$ 的形式，降次是最标准的处理方式，计算简单且不容出错。
+Reason for selection: For the form of $\cos^2 x$ or $\sin^2 x$, descending is the most standard processing method, the calculation is simple and error-free.
 
-**备选方法**：
-1. **Wallis 公式**：$\int_0^{\pi/2} \cos^n x \, dx$ 的递推公式
-2. **$\beta$ 函数**：化为 $B(1/2, 3/2)/2$ 计算
-3. **对称性**：$\int_0^{\pi/2} \cos^2 x \, dx = \int_0^{\pi/2} \sin^2 x \, dx$，且二者之和为 $\pi/2$，故各为 $\pi/4$
+**Alternative Method**:
+1. **Wallis formula**: Recursion formula of $\int_0^{\pi/2} \cos^n x \, dx$
+2. ** $\beta$ function**: converted into $B(1/2, 3/2)/2$ calculation
+3. **Symmetry**: $\int_0^{\pi/2} \cos^2 x \, dx = \int_0^{\pi/2} \sin^2 x \, dx$, and the sum of the two is $\pi/2$, so each is $\pi/4$
 
-本文展示主方法和对称性法，并用中点法做数值验证。
+This article shows the main method and symmetry method, and uses the midpoint method for numerical verification.
 
-## 解题过程
+## Problem solving process
 
-### 方法一：三角恒等式降次
+### Method 1: Reduction of trigonometric identities
 
-**Step 1：应用倍角公式**
+**Step 1: Apply the double angle formula**
 
-$$\cos 2x = 2\cos^2 x - 1 \implies \cos^2 x = \frac{1 + \cos 2x}{2}$$
+ $$\cos 2x = 2\cos^2 x - 1 \implies \cos^2 x = \frac{1 + \cos 2x}{2}$$
 
-验证恒等式：$2\cos^2 x - 1 = \cos^2 x - \sin^2 x = \cos 2x$ ✓
+Verify the identity: $2\cos^2 x - 1 = \cos^2 x - \sin^2 x = \cos 2x$ ✓
 
-**Step 2：代入积分**
+**Step 2: Substitute points**
 
-$$\int_0^{\pi/2} \cos^2 x \, dx = \int_0^{\pi/2} \frac{1 + \cos 2x}{2} \, dx$$
+ $$\int_0^{\pi/2} \cos^2 x \, dx = \int_0^{\pi/2} \frac{1 + \cos 2x}{2} \, dx$$
 
-$$= \frac{1}{2} \int_0^{\pi/2} 1 \, dx + \frac{1}{2} \int_0^{\pi/2} \cos 2x \, dx$$
+ $$= \frac{1}{2} \int_0^{\pi/2} 1 \, dx + \frac{1}{2} \int_0^{\pi/2} \cos 2x \, dx$$
 
-**Step 3：分别计算**
+**Step 3: Calculate separately**
 
-第一项：
-$$\frac{1}{2} \int_0^{\pi/2} 1 \, dx = \frac{1}{2} \cdot \left[x\right]_0^{\pi/2} = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$$
+First item:
+ $$\frac{1}{2} \int_0^{\pi/2} 1 \, dx = \frac{1}{2} \cdot \left[x\right]_0^{\pi/2} = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$$
 
-第二项：
-$$\frac{1}{2} \int_0^{\pi/2} \cos 2x \, dx = \frac{1}{2} \cdot \left[\frac{\sin 2x}{2}\right]_0^{\pi/2} = \frac{1}{4} \left[\sin 2x\right]_0^{\pi/2}$$
-$$= \frac{1}{4} (\sin \pi - \sin 0) = \frac{1}{4} (0 - 0) = 0$$
+Second item:
+ $$\frac{1}{2} \int_0^{\pi/2} \cos 2x \, dx = \frac{1}{2} \cdot \left[\frac{\sin 2x}{2}\right]_0^{\pi/2} = \frac{1}{4} \left[\sin 2x\right]_0^{\pi/2}$$
+ $$= \frac{1}{4} (\sin \pi - \sin 0) = \frac{1}{4} (0 - 0) = 0$$
 
-**Step 4：汇总**
+**Step 4: Summary**
 
-$$\int_0^{\pi/2} \cos^2 x \, dx = \frac{\pi}{4} + 0 = \frac{\pi}{4}$$
+ $$\int_0^{\pi/2} \cos^2 x \, dx = \frac{\pi}{4} + 0 = \frac{\pi}{4}$$
 
-### 方法二：对称性法（备选验证）
+### Method 2: Symmetry method (alternative verification)
 
-**Step 1：利用对称性**
+**Step 1: Take advantage of symmetry**
 
-在 $[0, \pi/2]$ 上，令 $u = \pi/2 - x$：
+On $[0, \pi/2]$, let $u = \pi/2 - x$:
 
-$$\int_0^{\pi/2} \cos^2 x \, dx = \int_{\pi/2}^0 \cos^2(\pi/2 - u) \cdot (-du)$$
-$$= \int_0^{\pi/2} \sin^2 u \, du$$
+ $$\int_0^{\pi/2} \cos^2 x \, dx = \int_{\pi/2}^0 \cos^2(\pi/2 - u) \cdot (-du)$$
+ $$= \int_0^{\pi/2} \sin^2 u \, du$$
 
-所以 $\int_0^{\pi/2} \cos^2 x \, dx = \int_0^{\pi/2} \sin^2 x \, dx$
+So $\int_0^{\pi/2} \cos^2 x \, dx = \int_0^{\pi/2} \sin^2 x \, dx$
 
-**Step 2：求和**
+**Step 2: Sum**
 
-$$\int_0^{\pi/2} \cos^2 x \, dx + \int_0^{\pi/2} \sin^2 x \, dx = \int_0^{\pi/2} (\cos^2 x + \sin^2 x) \, dx$$
-$$= \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2}$$
+ $$\int_0^{\pi/2} \cos^2 x \, dx + \int_0^{\pi/2} \sin^2 x \, dx = \int_0^{\pi/2} (\cos^2 x + \sin^2 x) \, dx$$
+ $$= \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2}$$
 
-由于两个积分相等，各为 $\pi/4$。即 $\int_0^{\pi/2} \cos^2 x \, dx = \frac{\pi}{2} \div 2 = \frac{\pi}{4}$ ✓
+Since the two integrals are equal, each is $\pi/4$ . That is $\int_0^{\pi/2} \cos^2 x \, dx = \frac{\pi}{2} \div 2 = \frac{\pi}{4}$ ✓
 
-## 验算
+## Check calculation
 
-**验算方法 1：中点法数值近似（$n = 10$）**
+**Check calculation method 1: Midpoint method numerical approximation ($n = 10$)**
 
-将 $[0, \pi/2]$ 等分为 $n = 10$ 个子区间，每个宽度 $\Delta x = \pi/20 \approx 0.15708$。
+Divide $[0, \pi/2]$ equally into $n = 10$ subranges, each width $\Delta x = \pi/20 \approx 0.15708$ .
 
-中点法公式：$\int_a^b f(x) \, dx \approx \Delta x \sum_{i=1}^n f\left(\frac{x_{i-1} + x_i}{2}\right)$
+Midpoint method formula: $\int_a^b f(x) \, dx \approx \Delta x \sum_{i=1}^n f\left(\frac{x_{i-1} + x_i}{2}\right)$
 
-| $i$ | 中点 $m_i$ | $\cos^2(m_i)$ |
+| $i$ | Midpoint $m_i$ | $\cos^2(m_i)$ |
 |-----|-----------|---------------|
-| 1 | $0.07854$ | $0.99384$ |
-| 2 | $0.23562$ | $0.94550$ |
-| 3 | $0.39270$ | $0.85355$ |
-| 4 | $0.54978$ | $0.72701$ |
-| 5 | $0.70686$ | $0.57725$ |
-| 6 | $0.86394$ | $0.41753$ |
-| 7 | $1.02102$ | $0.26278$ |
-| 8 | $1.17810$ | $0.12808$ |
-| 9 | $1.33518$ | $0.02802$ |
-| 10 | $1.49226$ | $0.00000$ |
+| 1 |  $0.07854$  |  $0.99384$  |
+| 2 |  $0.23562$  |  $0.94550$  |
+| 3 |  $0.39270$  |  $0.85355$  |
+| 4 |  $0.54978$  |  $0.72701$  |
+| 5 |  $0.70686$  |  $0.57725$  |
+| 6 |  $0.86394$  |  $0.41753$  |
+| 7 |  $1.02102$  |  $0.26278$  |
+| 8 |  $1.17810$  |  $0.12808$  |
+| 9 |  $1.33518$  |  $0.02802$  |
+| 10 |  $1.49226$  |  $0.00000$  |
 
-总和 $\approx 4.93357$
+Sum $\approx 4.93357$
 
-近似值 $= 0.15708 \times 4.93357 \approx 0.7750$
+Approximate value $= 0.15708 \times 4.93357 \approx 0.7750$
 
-精确值 $\pi/4 \approx 0.785398$，相对误差 $\approx 1.3\%$（$n=10$ 时中点法的预期精度）✓
+Exact value $\pi/4 \approx 0.785398$, relative error $\approx 1.3\%$ (expected accuracy of midpoint method at $n=10$)✓
 
-**验算方法 2：Wallis 公式验证**
+**Verification method 2: Wallis formula verification**
 
-$$\int_0^{\pi/2} \cos^{2} x \, dx = \frac{2-1}{2} \cdot \frac{\pi}{2} = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$$
+ $$\int_0^{\pi/2} \cos^{2} x \, dx = \frac{2-1}{2} \cdot \frac{\pi}{2} = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$$
 
-与结果一致 ✓
+Consistent with the results ✓
 
-**验算方法 3：区间合理性检查**
+**Check Calculation Method 3: Interval Reasonability Check**
 
-被积函数 $\cos^2 x$ 在 $[0, \pi/2]$ 上满足 $0 \leq \cos^2 x \leq 1$，因此：
-$$0 \leq \int_0^{\pi/2} \cos^2 x \, dx \leq \frac{\pi}{2}$$
+The integrand $\cos^2 x$ satisfies $0 \leq \cos^2 x \leq 1$ on $[0, \pi/2]$ , so:
+ $$0 \leq \int_0^{\pi/2} \cos^2 x \, dx \leq \frac{\pi}{2}$$
 
-$\pi/4 \approx 0.7854$ 在 $[0, 1.5708]$ 内 ✓
+$\pi/4 \approx 0.7854$ within $[0, 1.5708]$ ✓
 
-此外，$\cos^2 x$ 在 $[0, \pi/2]$ 上的平均值为 $\frac{\pi/4}{\pi/2} = \frac{1}{2}$，这与 $\cos^2 x + \sin^2 x = 1$ 且两函数图像对称的直观一致 ✓
+In addition, the average value of $\cos^2 x$ on $[0, \pi/2]$ is $\frac{\pi/4}{\pi/2} = \frac{1}{2}$, which is consistent with the intuition that $\cos^2 x + \sin^2 x = 1$ and the graphs of the two functions are symmetric ✓
 
-## 最终答案
-$$\int_0^{\pi/2} \cos^2 x \, dx = \frac{\pi}{4}$$
+## Final answer
+ $$\int_0^{\pi/2} \cos^2 x \, dx = \frac{\pi}{4}$$
 
-## 易错点
-1. **定积分上下限代入时漏算**：计算 $\frac{1}{4}[\sin 2x]_0^{\pi/2}$ 时注意 $\sin \pi = 0$（不是 $1$），$\sin 0 = 0$
-2. **倍角公式符号**：$\cos 2x = 2\cos^2 x - 1 = 1 - 2\sin^2 x$，注意变体不要混淆
-3. **降次后的系数**：$\cos^2 x = \frac{1 + \cos 2x}{2}$，积分后还要乘以 $\frac{1}{2}$ 的因子，容易漏掉
-4. **对称性法的前提**：$\int_0^{\pi/2} \cos^2 x \, dx = \int_0^{\pi/2} \sin^2 x \, dx$ 仅在 $[0, \pi/2]$ 上成立（变量代换 $u = \pi/2 - x$），不能随意推广到其他区间
-5. **数值验证的意义**：$n=10$ 的中点法误差约 1%，用于粗略验证而非精确证明；要确信结果正确应以解析推导为准
+## Easy to make mistakes
+1. **Missing calculation when substituting upper and lower limits of definite integral**: When calculating $\frac{1}{4}[\sin 2x]_0^{\pi/2}$, pay attention to $\sin \pi = 0$ (not $1$), $\sin 0 = 0$
+2. **Double angle formula symbol**: $\cos 2x = 2\cos^2 x - 1 = 1 - 2\sin^2 x$, be careful not to confuse the variants
+3. **Coefficient after reduction**: $\cos^2 x = \frac{1 + \cos 2x}{2}$. After integration, it must be multiplied by the factor of $\frac{1}{2}$, which is easy to miss.
+4. **Premise of the symmetry method**: $\int_0^{\pi/2} \cos^2 x \, dx = \int_0^{\pi/2} \sin^2 x \, dx$ is only established on $[0, \pi/2]$ (variable substitution $u = \pi/2 - x$), and cannot be extended to other intervals at will
+5. **Significance of numerical verification**: The midpoint method of $n=10$ has an error of about 1% and is used for rough verification rather than precise proof; to be sure that the results are correct, analytical derivation should prevail.
diff --git a/examples/equations.md b/examples/equations.md
index d2872aa..0a0818b 100644
--- a/examples/equations.md
+++ b/examples/equations.md
@@ -1,183 +1,183 @@
-# 分式方程的解法
+#Solution of fractional equations
 
-## 用户输入
-解方程：$\displaystyle \frac{x}{x-1} + \frac{1}{x+2} = \frac{3}{x^2+x-2}$
+## User input
+Solve the equation: $\displaystyle \frac{x}{x-1} + \frac{1}{x+2} = \frac{3}{x^2+x-2}$
 
-## Skill 分类
-方程（Equations）— 分式方程与增根检验
+## Skill Category
+Equations — Fractional equations and root-increasing tests
 
-## 题意解析
+## Question meaning analysis
 
-**已知条件：**
-- 分式方程：$\frac{x}{x-1} + \frac{1}{x+2} = \frac{3}{x^2+x-2}$
+**Known conditions:**
+- Fractional equation: $\frac{x}{x-1} + \frac{1}{x+2} = \frac{3}{x^2+x-2}$
 
-**求解目标：**
-- 求满足方程的实数 $x$。
+**Solution goal:**
+- Find the real number $x$ that satisfies the equation.
 
-**变量与定义域：**
-分式有意义的条件为所有分母 $\neq 0$：
+**Variables and Domains:**
+A fraction is meaningful if all denominators are $\neq 0$ :
 
-- $x-1 \neq 0$，即 $x \neq 1$
-- $x+2 \neq 0$，即 $x \neq -2$
-- $x^2+x-2 = (x-1)(x+2) \neq 0$，即 $x \neq 1$ 且 $x \neq -2$
+- $x-1 \neq 0$ , which is $x \neq 1$
+- $x+2 \neq 0$ , which is $x \neq -2$
+- $x^2+x-2 = (x-1)(x+2) \neq 0$ , that is, $x \neq 1$ and $x \neq -2$
 
-综合定义域：$x \in \mathbb{R} \setminus \{-2, 1\}$
+Comprehensive domain: $x \in \mathbb{R} \setminus \{-2, 1\}$
 
-**隐式条件：**
-- 观察第三个分母 $x^2+x-2$ 可分解为 $(x-1)(x+2)$，正好是前两个分母的乘积，这是本题设计的核心。
+**Implicit condition:**
+- Observe that the third denominator $x^2+x-2$ can be decomposed into $(x-1)(x+2)$, which is exactly the product of the first two denominators. This is the core of the design of this question.
 
-**解的个数：**
-- 去分母后得一元二次方程，最多两个根，需检验是否在定义域内。
+**Number of solutions:**
+- After removing the denominator, a quadratic equation of one variable is obtained, with up to two roots. It needs to be checked whether it is within the definition domain.
 
-## 方法选择
+## Method selection
 
-**选择方法：** 去分母法（乘以各分母的最小公倍式）+ 增根检验
+**Selection method:** Denominator removal method (multiply by the least common multiple of each denominator) + root-increasing test
 
-**理由：**
-- 分式方程的标准解法是两边同乘所有分母的 LCM（最小公倍式），将分式方程转化为整式方程。
-- 本题中 $x^2+x-2 = (x-1)(x+2)$，恰好是前两个分母的乘积，消元非常简洁。
+**reason:**
+- The standard solution to a fractional equation is to multiply both sides by the LCM (least common multiple) of all denominators, converting the fractional equation into an integer equation.
+- In this question, $x^2+x-2 = (x-1)(x+2)$ happens to be the product of the first two denominators, and elimination is very simple.
 
-**备选方法：**
-- 通分法：左边先通分后与右边比较——步骤等价，最终结果相同。
-- 换元法：不适用于本题，没有明显的可换元结构。
+**Alternative method:**
+- Common division method: first divide the left side and then compare it with the right side - the steps are equivalent and the final result is the same.
+- Substitution method: Not applicable to this question, there is no obvious substitutable structure.
 
-**风险评估：**
-- 去分母可能引入增根（使原分母为零的根），必须在解出后逐个代入原方程检验。
+**risk assessment:**
+- Removing the denominator may introduce additional roots (roots that make the original denominator zero), which must be substituted into the original equation one by one after solving.
 
-## 解题过程
+## Problem solving process
 
-### 第一步：确定定义域
+### Step 1: Determine the domain of definition
 
-解出使分母为零的 $x$ 值：
+Solve for the value of $x$ such that the denominator is zero:
 
-- $x-1=0 \Rightarrow x=1$
-- $x+2=0 \Rightarrow x=-2$
+-  $x-1=0 \Rightarrow x=1$
+-  $x+2=0 \Rightarrow x=-2$
 
-定义域：$x \neq 1$ 且 $x \neq -2$
+Domain: $x \neq 1$ and $x \neq -2$
 
-### 第二步：因式分解第三个分母
+### Step 2: Factor the third denominator
 
-注意到 $x^2+x-2$ 可因式分解：
+Notice that $x^2+x-2$ can be factored:
 
-$$
+ $$
 x^2 + x - 2 = (x-1)(x+2)
 $$
 
-验证：$(x-1)(x+2) = x^2 + 2x - x - 2 = x^2 + x - 2$ ✓
+Verification: $(x-1)(x+2) = x^2 + 2x - x - 2 = x^2 + x - 2$ ✓
 
-### 第三步：去分母
+### Step 3: Remove the denominator
 
-方程两边同乘 $(x-1)(x+2)$（在定义域内不为零）：
+Multiply both sides of the equation by $(x-1)(x+2)$ (not zero in the domain):
 
-$$
+ $$
 \frac{x}{x-1} \cdot (x-1)(x+2) + \frac{1}{x+2} \cdot (x-1)(x+2) = \frac{3}{(x-1)(x+2)} \cdot (x-1)(x+2)
 $$
 
-各项化简：
+Various simplifications:
 
-- 第一项：$x \cdot (x+2) = x(x+2)$
-- 第二项：$1 \cdot (x-1) = x-1$
-- 右边：$3$
+- First item: $x \cdot (x+2) = x(x+2)$
+- Second item: $1 \cdot (x-1) = x-1$
+- Right: $3$
 
-得到整式方程：
+Get the integral equation:
 
-$$
+ $$
 x(x+2) + (x-1) = 3
 $$
 
-### 第四步：化简并求解
+### Step 4: Simplify and solve
 
-展开：
+Expand:
 
-$$
+ $$
 x^2 + 2x + x - 1 = 3
 $$
 
-合并同类项：
+Merge similar items:
 
-$$
+ $$
 x^2 + 3x - 1 = 3
 $$
 
-移项：
+Move items:
 
-$$
+ $$
 x^2 + 3x - 4 = 0
 $$
 
-因式分解：
+Factoring:
 
-$$
+ $$
 (x+4)(x-1) = 0
 $$
 
-解得：
+Solution:
 
-$$
-x = -4 \quad \text{或} \quad x = 1
+ $$
+x = -4 \quad \text{or} \quad x = 1
 $$
 
-### 第五步：检验增根
+### Step 5: Check the root increase
 
-**检验 $x = 1$：**
+**Check $x = 1$:**
 
-代入原方程，分母 $x-1 = 0$，分式无意义。**$x=1$ 是增根，舍去。**
+Substituting into the original equation, the denominator is $x-1 = 0$, and the fraction is meaningless. ** $x=1$ is to increase the root and discard it. **
 
-**检验 $x = -4$：**
+**Check $x = -4$:**
 
-左式 $= \frac{-4}{-4-1} + \frac{1}{-4+2} = \frac{-4}{-5} + \frac{1}{-2} = \frac{4}{5} - \frac{1}{2}$
+Left style $= \frac{-4}{-4-1} + \frac{1}{-4+2} = \frac{-4}{-5} + \frac{1}{-2} = \frac{4}{5} - \frac{1}{2}$
 
-通分：$\frac{4}{5} = \frac{8}{10}$，$\frac{1}{2} = \frac{5}{10}$
+Common scores: $\frac{4}{5} = \frac{8}{10}$, $\frac{1}{2} = \frac{5}{10}$
 
-左式 $= \frac{8}{10} - \frac{5}{10} = \frac{3}{10}$
+Left style $= \frac{8}{10} - \frac{5}{10} = \frac{3}{10}$
 
-右式 $= \frac{3}{(-4)^2 + (-4) - 2} = \frac{3}{16 - 4 - 2} = \frac{3}{10}$
+Right style $= \frac{3}{(-4)^2 + (-4) - 2} = \frac{3}{16 - 4 - 2} = \frac{3}{10}$
 
-左式 $=$ 右式 $= \frac{3}{10}$ ✓
+Left style $=$ Right style $= \frac{3}{10}$ ✓
 
-## 验算
+## Check calculation
 
-### 验算方法一：代入原方程（增根检验）
+### Calculation method one: Substitute into the original equation (increased root test)
 
-已在上步完成，$x=-4$ 满足原方程。
+Completed in the previous step, $x=-4$ satisfies the original equation.
 
-### 验算方法二：将原方程通分验证
+### Calculation method two: Verify the original equation by dividing it into two parts
 
-对 $x = -4$，验证原方程等价于恒等式：
+For $x = -4$ , verify that the original equation is equivalent to the identity:
 
-左式通分：$\frac{x}{x-1} + \frac{1}{x+2} = \frac{x(x+2) + (x-1)}{(x-1)(x+2)} = \frac{x^2+3x-1}{(x-1)(x+2)}$
+Common participle of left-hand formula: $\frac{x}{x-1} + \frac{1}{x+2} = \frac{x(x+2) + (x-1)}{(x-1)(x+2)} = \frac{x^2+3x-1}{(x-1)(x+2)}$
 
-右式：$\frac{3}{x^2+x-2} = \frac{3}{(x-1)(x+2)}$
+Right form: $\frac{3}{x^2+x-2} = \frac{3}{(x-1)(x+2)}$
 
-原方程等价于：$\frac{x^2+3x-1}{(x-1)(x+2)} = \frac{3}{(x-1)(x+2)}$
+The original equation is equivalent to: $\frac{x^2+3x-1}{(x-1)(x+2)} = \frac{3}{(x-1)(x+2)}$
 
-即 $x^2+3x-1 = 3$（在 $x \neq 1, -2$ 时），即 $x^2+3x-4=0$。
+That is $x^2+3x-1 = 3$ (at $x \neq 1, -2$ ), which is $x^2+3x-4=0$ .
 
-当 $x=-4$ 时：$(-4)^2 + 3(-4) - 4 = 16 - 12 - 4 = 0$，满足。✓
+When $x=-4$: $(-4)^2 + 3(-4) - 4 = 16 - 12 - 4 = 0$, satisfied. ✓
 
-### 验算方法三：逆向推理
+### Calculation method three: reverse reasoning
 
-从 $x=-4$ 出发反推：
+Starting from $x=-4$ and working backwards:
 
-若 $x=-4$，则方程等价形式 $x^2+3x-4=0$ 成立。
+If $x=-4$ , then the equivalent form of the equation $x^2+3x-4=0$ holds.
 
-原方程等价于 $\frac{x^2+3x-1}{(x-1)(x+2)} = \frac{3}{(x-1)(x+2)}$
+The original equation is equivalent to $\frac{x^2+3x-1}{(x-1)(x+2)} = \frac{3}{(x-1)(x+2)}$
 
-代入 $x=-4$，两边分母均为 $(-5)(-2)=10$，分子左边为 $16-12-1=3$，右边为 $3$，两边相等。
+Substituting $x=-4$, the denominator on both sides is $(-5)(-2)=10$, the left side of the numerator is $16-12-1=3$, and the right side is $3$, both sides are equal.
 
-综上，$x=-4$ 确为原方程的解。
+In summary, $x=-4$ is indeed the solution to the original equation.
 
-## 最终答案
+## Final answer
 
-$$
+ $$
 \boxed{x = -4}
 $$
 
-（$x=1$ 为增根，舍去。）
+($x=1$ is to increase the root, discard it.)
 
-## 易错点
-1. **忘记检验增根：** 去分母后解得两个根 $x=-4$ 和 $x=1$，若不做检验直接将 $x=1$ 也当作答案，会丢分。**所有分式方程都必须检验增根。**
-2. **定义域分析缺失：** 未在解题开始时写下 $x \neq 1$ 且 $x \neq -2$，后续检验容易遗漏。
-3. **去分母操作错误：** 忘了每项都要乘，尤其是右边整式 $3$ 也要乘 $(x-1)(x+2)$。
-4. **因式分解错误：** $x^2+x-2$ 容易误分解为 $(x-1)(x-2)$，正确的交叉验算可以避免此错误。
-5. **增根类型混淆：** 增根 $x=1$ 恰好与定义域排除值重合，不是碰巧——去分母时乘的因子恰为 $(x-1)(x+2)$，所以 $x=1$ 和 $x=-2$ 都是潜在增根（虽然本题中 $x=-2$ 不是方程的根）。
+## Easy to make mistakes
+1. **Forgot to check for root addition:** After removing the denominator, we get two roots $x=-4$ and $x=1$. If we don’t do the test and directly use $x=1$ as the answer, we will lose points. **All fractional equations must be tested for increasing roots. **
+2. **Loss of domain analysis:** If $x \neq 1$ and $x \neq -2$ are not written down at the beginning of solving the problem, it is easy to miss them in subsequent tests.
+3. **Error in removing the denominator:** Forgot to multiply each item, especially the integer $3$ on the right must also be multiplied by $(x-1)(x+2)$.
+4. **Factorization error:** $x^2+x-2$ is easily decomposed into $(x-1)(x-2)$ by mistake. Correct cross-checking can avoid this error.
+5. **Confusion of augmented root types:** The augmented root $x=1$ happens to coincide with the domain exclusion value. It is not a coincidence - the factor multiplied when removing the denominator is exactly $(x-1)(x+2)$, so $x=1$ and $x=-2$ are both potential augmented roots (although $x=-2$ in this question is not a root of the equation).
diff --git a/examples/functions.md b/examples/functions.md
index fc5e785..ef37216 100644
--- a/examples/functions.md
+++ b/examples/functions.md
@@ -1,108 +1,108 @@
-# 函数的单调性与极值
+# Monotonicity and extreme values ​​of functions
 
-## 用户输入
-求函数 $f(x) = x^3 - 3x^2 - 9x + 5$ 在区间 $[-2, 4]$ 上的单调区间与极值（包括极大值、极小值、最大值、最小值）。
+## User input
+Find the monotonic interval and extreme value (including maximum value, minimum value, maximum value and minimum value) of function $f(x) = x^3 - 3x^2 - 9x + 5$ on the interval $[-2, 4]$.
 
-## Skill 分类
-函数（Functions）— 导数应用：单调性分析与极值问题
+## Skill Category
+Functions—derivative applications: monotonicity analysis and extreme value problems
 
-## 题意解析
+## Question meaning analysis
 
-**已知条件：**
-- 函数：$f(x) = x^3 - 3x^2 - 9x + 5$
-- 闭区间：$x \in [-2, 4]$
+**Known conditions:**
+- Function: $f(x) = x^3 - 3x^2 - 9x + 5$
+- Closed interval: $x \in [-2, 4]$
 
-**求解目标：**
-- 函数的单调递增区间和递减区间
-- 函数的极大值点与极大值、极小值点与极小值
-- 闭区间上的最大值与最小值
+**Solution goal:**
+- Monotonically increasing and decreasing intervals of the function
+- Maximum points and maxima, minimum points and minima of functions
+- Maximum and minimum values ​​on a closed interval
 
-**变量与定义域：**
-- 函数 $f(x)$ 是三次多项式，在 $\mathbb{R}$ 上连续可导，给定闭区间 $[-2, 4]$。
+**Variables and Domains:**
+- The function $f(x)$ is a cubic polynomial that is continuously differentiable on $\mathbb{R}$, given the closed interval $[-2, 4]$.
 
-**隐式条件：**
-- 闭区间上的最值可能出现在临界点（导数为零的点）或区间端点，需要全部比较。
-- 导数是二次函数，判别式 $> 0$，存在两个不同的实根，函数有两个驻点。
+**Implicit condition:**
+- The maximum value on a closed interval may appear at the critical point (the point where the derivative is zero) or at the end of the interval, requiring all comparisons.
+- The derivative is a quadratic function, the discriminant $> 0$, there are two different real roots, and the function has two stationary points.
 
-## 方法选择
+## Method selection
 
-**选择方法：** 导数法（求导 → 求驻点 → 符号分析 → 单调性判断）+ 端点比较
+**Selection method:** Derivative method (derivative → stationary point → symbolic analysis → monotonicity judgment) + endpoint comparison
 
-**理由：**
-- 多项式函数求导简单直接。
-- 通过一阶导数的符号判断单调性是最标准的方法。
-- 二阶导数辅助判断极值类型。
+**reason:**
+- Derivation of polynomial functions is simple and straightforward.
+- The most standard way to judge monotonicity is by the sign of the first derivative.
+- The second-order derivative assists in determining the type of extreme value.
 
-**备选方法：**
-- 三次函数可以通过 $f(x)$ 的增减性间接判断，但不系统。
-- 数值试探法（列表计算）——不够精确，只能作为验算手段。
+**Alternative method:**
+- Cubic functions can be judged indirectly through the increase or decrease of $f(x)$, but it is not systematic.
+- Numerical heuristics (list calculations) - not accurate enough and can only be used as a verification method.
 
-## 解题过程
+## Problem solving process
 
-### 第一步：求导
+### Step 1: Derivation
 
-$$
+ $$
 f'(x) = 3x^2 - 6x - 9
 $$
 
-提取公因式 $3$：
+Extract the common factor $3$:
 
-$$
+ $$
 f'(x) = 3(x^2 - 2x - 3)
 $$
 
-因式分解：
+Factoring:
 
-$$
+ $$
 f'(x) = 3(x-3)(x+1)
 $$
 
-### 第二步：求驻点
+### Step 2: Find the stationary point
 
-令 $f'(x) = 0$：
+Let $f'(x) = 0$ :
 
-$$
-3(x-3)(x+1) = 0 \quad\Rightarrow\quad x = -1 \ \text{或}\ x = 3
+ $$
+3(x-3)(x+1) = 0 \quad\Rightarrow\quad x = -1 \ \text{or}\ x = 3
 $$
 
-两个驻点都在区间 $[-2, 4]$ 内。
+Both stationary points are within the interval $[-2, 4]$.
 
-### 第三步：单调性分析
+### Step 3: Monotonicity Analysis
 
-根据 $f'(x) = 3(x-3)(x+1)$ 的符号判断原函数的单调性。
+Judge the monotonicity of the original function based on the sign of $f'(x) = 3(x-3)(x+1)$.
 
-| 区间 | $x$ 范围 | $x+1$ | $x-3$ | $f'(x)$ | $f(x)$ 单调性 |
+| interval | $x$ range | $x+1$ | $x-3$ | $f'(x)$ | $f(x)$ monotonicity |
 |------|----------|-------|-------|---------|--------------|
-| $[-2, -1)$ | $-2 \leq x < -1$ | $\leq 0$/$< 0$ | $< 0$ | $> 0$ | 严格递增 ↗ |
-| $(-1, 3)$ | $-1 < x < 3$ | $> 0$ | $< 0$ | $< 0$ | 严格递减 ↘ |
-| $(3, 4]$ | $3 < x \leq 4$ | $> 0$ | $> 0$ | $> 0$ | 严格递增 ↗ |
+| $[-2, -1)$ | $-2 \leq x < -1$ | $\leq 0$ / $< 0$ | $< 0$ | $> 0$ | Strictly increasing ↗ |
+| $(-1, 3)$ | $-1 < x < 3$ | $> 0$ | $< 0$ | $< 0$ | Strictly Decreasing ↘ |
+| $(3, 4]$ | $3 < x \leq 4$ | $> 0$ | $> 0$ | $> 0$ | Strictly increasing ↗ |
 
-详细推导：
+Detailed derivation:
 
-- 当 $x \in [-2, -1)$：$x+1 < 0$，$x-3 < 0$，两负相乘得正，乘以 $3$ 仍为正，$f'(x) > 0$。函数递增。
-- 当 $x \in (-1, 3)$：$x+1 > 0$，$x-3 < 0$，一正一负相乘得负，$f'(x) < 0$。函数递减。
-- 当 $x \in (3, 4]$：$x+1 > 0$，$x-3 > 0$，两正相乘得正，$f'(x) > 0$。函数递增。
+- When $x \in [-2, -1)$ : $x+1 < 0$ , $x-3 < 0$ , the two negatives are multiplied together to get a positive, multiplied by $3$ is still positive, $f'(x) > 0$ . The function is incremented.
+- When $x \in (-1, 3)$ : $x+1 > 0$ , $x-3 < 0$ , one positive and one negative are multiplied together to get negative, $f'(x) < 0$ . function decreases.
+- When $x \in (3, 4]$ : $x+1 > 0$ , $x-3 > 0$ , the two positives are multiplied to get a positive value, $f'(x) > 0$ . The function is incremented.
 
-因此：
-- **单调递增区间：** $[-2, -1]$ 和 $[3, 4]$
-- **单调递减区间：** $[-1, 3]$
+therefore:
+- **Monotonically increasing intervals:** $[-2, -1]$ and $[3, 4]$
+- **Monotonically decreasing interval:** $[-1, 3]$
 
-### 第四步：判断极值类型
+### Step 4: Determine the extreme value type
 
-求二阶导数：
+Find the second derivative:
 
-$$
+ $$
 f''(x) = 6x - 6 = 6(x-1)
 $$
 
-- $f''(-1) = 6(-1-1) = -12 < 0$ → $x = -1$ 处为**极大值点**
-- $f''(3) = 6(3-1) = 12 > 0$ → $x = 3$ 处为**极小值点**
+- $f''(-1) = 6(-1-1) = -12 < 0$ → $x = -1$ is the **maximum value point**
+- $f''(3) = 6(3-1) = 12 > 0$ → $x = 3$ is the **minimum point**
 
-### 第五步：计算各关键点的函数值
+### Step 5: Calculate the function value of each key point
 
-计算驻点和端点的函数值：
+Compute function values ​​for stationary points and endpoints:
 
-$$
+ $$
 \begin{aligned}
 f(-2) &= (-2)^3 - 3(-2)^2 - 9(-2) + 5 = -8 - 12 + 18 + 5 = 3 \\[6pt]
 f(-1) &= (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10 \\[6pt]
@@ -111,67 +111,67 @@ f(4) &= 4^3 - 3 \times 4^2 - 9 \times 4 + 5 = 64 - 48 - 36 + 5 = -15
 \end{aligned}
 $$
 
-### 第六步：确定极值与最值
+### Step 6: Determine the extreme value and maximum value
 
-| 位置 | $x$ | $f(x)$ | 性质 |
+| Location | $x$ | $f(x)$ | Properties |
 |------|-----|--------|------|
-| 左端点 | $-2$ | $3$ | — |
-| 驻点 | $-1$ | $10$ | 极大值，最大值 |
-| 驻点 | $3$ | $-22$ | 极小值，最小值 |
-| 右端点 | $4$ | $-15$ | — |
+| Left endpoint | $-2$ | $3$ | — |
+| Stationary point | $-1$ | $10$ | Maximum value, maximum value |
+| Stationary point | $3$ | $-22$ | Minimum value, minimum value |
+| Right endpoint | $4$ | $-15$ | — |
 
-比较四个值：
+Compare four values:
 
-- 最大值：$\max\{3, 10, -22, -15\} = 10$，在 $x = -1$ 处取得
-- 最小值：$\min\{3, 10, -22, -15\} = -22$，在 $x = 3$ 处取得
-- 极大值：$10$（$x=-1$）
-- 极小值：$-22$（$x=3$）
+- Maximum value: $\max\{3, 10, -22, -15\} = 10$, obtained at $x = -1$
+- Minimum value: $\min\{3, 10, -22, -15\} = -22$, taken at $x = 3$
+- Maximum value: $10$ ( $x=-1$ )
+- Minimum value: $-22$ ( $x=3$ )
 
-## 验算
+## Check calculation
 
-### 验算方法一：二阶导数验证极值类型
+### Verification method one: Second-order derivative verification extreme value type
 
-上步已用二阶导数验证，在 $x=-1$ 处 $f''(-1) = -12 < 0$（极大值），在 $x=3$ 处 $f''(3) = 12 > 0$（极小值）。判断正确。
+The previous step has been verified with the second derivative, at $x=-1$ $f''(-1) = -12 < 0$ (maximum value), at $x=3$ $f''(3) = 12 > 0$ (minimum value). The judgment is correct.
 
-### 验算方法二：取点验证单调性
+### Calculation method two: take points to verify monotonicity
 
-在单调递增区间取两点，验证 $f(x_1) < f(x_2)$；在递减区间取两点，验证 $f(x_1) > f(x_2)$。
+Pick two points in the monotonically increasing interval and verify $f(x_1) < f(x_2)$; pick two points in the decreasing interval and verify $f(x_1) > f(x_2)$.
 
-- 递增区间 $[-2, -1]$：$f(-2) = 3$，$f(-1.5) = (-1.5)^3 - 3(2.25) + 13.5 + 5 = -3.375 - 6.75 + 13.5 + 5 = 8.375$。$3 < 8.375 < 10$，递增 ✓
-- 递减区间 $[-1, 3]$：$f(-1) = 10$，$f(0) = 5$，$f(1) = 1-3-9+5 = -6$，$f(2) = 8-12-18+5 = -17$。$10 > 5 > -6 > -17 > -22$，递减 ✓
-- 递增区间 $[3, 4]$：$f(3) = -22$，$f(3.5) = 42.875 - 36.75 - 31.5 + 5 = -20.375$，$f(4) = -15$。$-22 < -20.375 < -15$，递增 ✓
+- Increasing intervals $[-2, -1]$: $f(-2) = 3$, $f(-1.5) = (-1.5)^3 - 3(2.25) + 13.5 + 5 = -3.375 - 6.75 + 13.5 + 5 = 8.375$. $3 < 8.375 < 10$ , increment ✓
+- Decreasing interval $[-1, 3]$: $f(-1) = 10$, $f(0) = 5$, $f(1) = 1-3-9+5 = -6$, $f(2) = 8-12-18+5 = -17$. $10 > 5 > -6 > -17 > -22$ , decreasing ✓
+- Increasing intervals $[3, 4]$: $f(3) = -22$, $f(3.5) = 42.875 - 36.75 - 31.5 + 5 = -20.375$, $f(4) = -15$. $-22 < -20.375 < -15$ , increment ✓
 
-### 验算方法三：导数符号数值验证
+### Verification method three: Numerical verification of derivative symbols
 
-在关键区间取点验证 $f'(x)$ 的符号：
+Take points in the key interval to verify the sign of $f'(x)$:
 
-- $x = -1.5$（在 $[-2, -1)$）：$f'(-1.5) = 3(2.25) - 6(-1.5) - 9 = 6.75 + 9 - 9 = 6.75 > 0$ ✓
-- $x = 0$（在 $(-1, 3)$）：$f'(0) = 0 - 0 - 9 = -9 < 0$ ✓
-- $x = 3.5$（在 $(3, 4]$）：$f'(3.5) = 3(12.25) - 6(3.5) - 9 = 36.75 - 21 - 9 = 6.75 > 0$ ✓
+- $x = -1.5$ (in $[-2, -1)$ ): $f'(-1.5) = 3(2.25) - 6(-1.5) - 9 = 6.75 + 9 - 9 = 6.75 > 0$ ✓
+- $x = 0$ (in $(-1, 3)$ ): $f'(0) = 0 - 0 - 9 = -9 < 0$ ✓
+- $x = 3.5$ (in $(3, 4]$ ): $f'(3.5) = 3(12.25) - 6(3.5) - 9 = 36.75 - 21 - 9 = 6.75 > 0$ ✓
 
-## 最终答案
+## Final answer
 
-**单调性：**
+**Monotonicity:**
 
-$$
-\boxed{\text{递增区间：}[-2, -1] \cup [3, 4],\quad \text{递减区间：}[-1, 3]}
+ $$
+\boxed{\text{Increasing interval:}[-2, -1] \cup [3, 4],\quad \text{Decreasing interval:}[-1, 3]}
 $$
 
-**极值与最值：**
+**Extreme value and maximum value:**
 
-$$
+ $$
 \boxed{
 \begin{aligned}
-&\text{极大值：} f(-1) = 10 \quad (\text{也是最大值}) \\[4pt]
-&\text{极小值：} f(3) = -22 \quad (\text{也是最小值})
+&\text{Maximum value:} f(-1) = 10 \quad (\text{Also the maximum value}) \\[4pt]
+&\text{Minimum value:} f(3) = -22 \quad (\text{Also the minimum value})
 \end{aligned}
 }
 $$
 
-## 易错点
-1. **忘记比较端点值：** 在闭区间上求最值时，驻点的极值不一定是全局最值，必须与端点值比较。本例题中极值恰好是最值，但并非总是如此。
-2. **驻点不在区间内：** 如果某个驻点不在给定区间内，则不应纳入比较。本题中两个驻点 $x=-1$ 和 $x=3$ 都在 $[-2,4]$ 内。
-3. **单调区间书写不规范：** 在 $x=-1$ 处函数不可导（本题中可导），但单调区间仍可以包含端点。实际上 $f(x)$ 在 $x=-1$ 处是极大值点，左侧递增右侧递减，单调区间可以写成 $[-2, -1]$ 和 $[-1, 3]$。
-4. **二阶导数为零的情况：** 若 $f''(c)=0$，二阶导数判别法失效，需用一阶导数符号变化判断。本题中 $f''(1)=0$ 但 $x=1$ 不是驻点，不构成问题。
-5. **导数求错：** 常数项 $5$ 求导后为零，容易被遗留。各项系数要注意——特别是 $-9x$ 求导后为 $-9$。
-6. **极值与最值的区别：** 极大值/极小值是局部概念（导数判断），最大值/最小值是全局概念（比较所有可能点）。不要混淆。
+## Easy to make mistakes
+1. **Forgot to compare endpoint values:** When finding the maximum value on a closed interval, the extreme value of the stationary point is not necessarily the global maximum value and must be compared with the endpoint value. In this example the extreme value happens to be the maximum value, but this is not always the case.
+2. **Station point not within the interval:** If a stationary point is not within the given interval, it should not be included in the comparison. In this question, the two stationary points $x=-1$ and $x=3$ are both within $[-2,4]$.
+3. **Irregular writing of monotonic intervals:** The function is not differentiable at $x=-1$ (it is differentiable in this question), but the monotonic interval can still include endpoints. In fact, $f(x)$ is the maximum value point at $x=-1$. The left side increases and the right side decreases. The monotonic interval can be written as $[-2, -1]$ and $[-1, 3]$.
+4. **The case where the second-order derivative is zero:** If $f''(c)=0$, the second-order derivative discrimination method fails, and the sign change of the first-order derivative needs to be used to judge. In this question, $f''(1)=0$ but $x=1$ are not stationary points and do not pose a problem.
+5. **Error in derivation:** The constant term $5$ is zero after derivation and is easily left behind. Pay attention to the coefficients - especially $-9x$ which is $-9$ after derivation.
+6. **The difference between extreme value and maximum value:** Maximum value/minimum value is a local concept (derivative judgment), and maximum value/minimum value is a global concept (compare all possible points). Don't get confused.
diff --git a/examples/geometry.md b/examples/geometry.md
index 583adf7..669432a 100644
--- a/examples/geometry.md
+++ b/examples/geometry.md
@@ -1,22 +1,22 @@
-# 等腰三角形的角度计算
+# Calculation of angles of isosceles triangle
 
-## 用户输入
-在 $\triangle ABC$ 中，$AB = AC$，点 $D$ 在边 $BC$ 上。已知 $\angle BAD = 30^\circ$，$\angle CAD = 20^\circ$，求 $\angle B$。
+## User input
+In $\triangle ABC$ , $AB = AC$ , point $D$ is on edge $BC$ . Given $\angle BAD = 30^\circ$, $\angle CAD = 20^\circ$, find $\angle B$.
 
-## Skill 分类
-几何（Geometry）— 三角形角度计算
+## Skill Category
+Geometry—triangle angle calculation
 
-## 题意解析
+## Question meaning analysis
 
-**已知条件：**
-- $\triangle ABC$ 是等腰三角形，$AB = AC$（两腰相等）
-- 点 $D$ 在边 $BC$ 上
-- $\angle BAD = 30^\circ$（射线 $AD$ 与 $AB$ 的夹角）
-- $\angle CAD = 20^\circ$（射线 $AD$ 与 $AC$ 的夹角）
+**Known conditions:**
+- $\triangle ABC$ is an isosceles triangle, $AB = AC$ (both sides are equal)
+- point $D$ on edge $BC$
+- $\angle BAD = 30^\circ$ (the angle between rays $AD$ and $AB$)
+- $\angle CAD = 20^\circ$ (the angle between rays $AD$ and $AC$)
 
-**图形结构（文字描述）：**
+**Graphic structure (text description):**
 
-```
+ ```
         A
        /|\
       / | \
@@ -28,194 +28,194 @@
 B-------+-------C
 ```
 
-射线 $AD$ 从顶点 $A$ 出发，经过 $BC$ 边上的点 $D$，将 $\angle A$ 分成 $\angle BAD = 30^\circ$ 和 $\angle CAD = 20^\circ$。
+Ray $AD$ starts from vertex $A$ and passes through point $D$ on the edge of $BC$, dividing $\angle A$ into $\angle BAD = 30^\circ$ and $\angle CAD = 20^\circ$.
 
-**求解目标：**
-- $\angle B$ 的度数。
+**Solution goal:**
+- Degree of $\angle B$.
 
-**隐式条件：**
-- 点 $D$ 在边 $BC$ 上，意味着 $AD$ 是 $\angle A$ 内部的一条线段，将 $\angle A$ 分为两个部分。因此 $\angle A = \angle BAD + \angle CAD = 30^\circ + 20^\circ = 50^\circ$。
-- $AB = AC$ 意味着 $\angle B = \angle C$（等腰三角形的两个底角相等）。
-- 三角形内角和为 $180^\circ$。
+**Implicit condition:**
+- Point $D$ is on edge $BC$, meaning $AD$ is a line segment inside $\angle A$, dividing $\angle A$ into two parts. Hence $\angle A = \angle BAD + \angle CAD = 30^\circ + 20^\circ = 50^\circ$ .
+- $AB = AC$ means $\angle B = \angle C$ (the two base angles of an isosceles triangle are equal).
+- The sum of the interior angles of a triangle is $180^\circ$.
 
-**解的个数：**
-- 角度值唯一确定。
+**Number of solutions:**
+- The angle value is uniquely determined.
 
-## 方法选择
+## Method selection
 
-**选择方法：** 等腰三角形性质 + 三角形内角和定理
+**Selection method:** Isosceles triangle properties + triangle interior angle sum theorem
 
-**理由：**
-- 解题关键是利用 $AB = AC$ 得出 $\angle B = \angle C$ 的性质。
-- 通过 $D$ 在 $BC$ 上一条件，将 $\angle A$ 表示为两个已知角度之和。
-- 然后用内角和定理求 $\angle B$。
+**reason:**
+- The key to solving the problem is to use $AB = AC$ to get the properties of $\angle B = \angle C$.
+- Express $\angle A$ as the sum of two known angles by passing $D$ on $BC$ .
+- Then use the sum of interior angles theorem to find $\angle B$.
 
-**备选方法：**
-- 正弦定理法：设定一边长度，用已知角度解三角形——虽然可行，但对纯角度计算来说过于复杂。
-- 坐标几何法：建立坐标系计算——适合作为验算手段，不适合作为主解法。
+**Alternative method:**
+- Sine theorem method: set the length of one side and solve the triangle using known angles - although feasible, it is too complex for pure angle calculations.
+- Coordinate geometry method: establish coordinate system calculation - suitable as a verification method, not suitable as the main solution method.
 
-## 解题过程
+## Problem solving process
 
-### 第一步：求 $\angle A$
+### Step one: Ask for $\angle A$
 
-因为 $D$ 在边 $BC$ 上，射线 $AD$ 在 $\angle A$ 内部，将 $\angle A$ 分为 $\angle BAD$ 和 $\angle CAD$：
+Since $D$ is on edge $BC$ and ray $AD$ is inside $\angle A$ , split $\angle A$ into $\angle BAD$ and $\angle CAD$ :
 
-$$
+ $$
 \angle A = \angle BAD + \angle CAD = 30^\circ + 20^\circ = 50^\circ
 $$
 
-### 第二步：利用等腰三角形性质
+### Step 2: Use the properties of isosceles triangles
 
-因为 $AB = AC$，根据"等边对等角"性质：
+Because $AB = AC$ , according to the "equilateral opposite angles" property:
 
-$$
+ $$
 \angle B = \angle C
 $$
 
-### 第三步：应用三角形内角和定理
+### Step 3: Apply the triangle interior angle sum theorem
 
-三角形三个内角之和为 $180^\circ$：
+The sum of the three interior angles of a triangle is $180^\circ$:
 
-$$
+ $$
 \angle A + \angle B + \angle C = 180^\circ
 $$
 
-代入 $\angle C = \angle B$ 和 $\angle A = 50^\circ$：
+Substitute $\angle C = \angle B$ and $\angle A = 50^\circ$ :
 
-$$
+ $$
 50^\circ + \angle B + \angle B = 180^\circ
 $$
 
-$$
+ $$
 50^\circ + 2\angle B = 180^\circ
 $$
 
-$$
+ $$
 2\angle B = 180^\circ - 50^\circ = 130^\circ
 $$
 
-$$
+ $$
 \angle B = 65^\circ
 $$
 
-因此 $\angle B = \angle C = 65^\circ$。
+Hence $\angle B = \angle C = 65^\circ$ .
 
-### 第四步：验证几何合理性
+### Step 4: Verify geometric rationality
 
-检查所有已知条件是否满足：
+Check that all known conditions are met:
 
-- $\angle A + \angle B + \angle C = 50^\circ + 65^\circ + 65^\circ = 180^\circ$ ✓
-- $\angle BAD + \angle CAD = 30^\circ + 20^\circ = 50^\circ = \angle A$ ✓
-- $D$ 在 $BC$ 上：在 $\triangle ABD$ 中，$\angle B = 65^\circ$，$\angle BAD = 30^\circ$，则 $\angle ADB = 180^\circ - 65^\circ - 30^\circ = 85^\circ$。在 $\triangle ADC$ 中，$\angle C = 65^\circ$，$\angle CAD = 20^\circ$，则 $\angle ADC = 180^\circ - 65^\circ - 20^\circ = 95^\circ$。$\angle ADB + \angle ADC = 85^\circ + 95^\circ = 180^\circ$，说明 $B$、$D$、$C$ 三点共线 ✓
+-  $\angle A + \angle B + \angle C = 50^\circ + 65^\circ + 65^\circ = 180^\circ$  ✓
+-  $\angle BAD + \angle CAD = 30^\circ + 20^\circ = 50^\circ = \angle A$  ✓
+- $D$ on $BC$ : on $\triangle ABD$ , $\angle B = 65^\circ$ , $\angle BAD = 30^\circ$ , then $\angle ADB = 180^\circ - 65^\circ - 30^\circ = 85^\circ$ . In $\triangle ADC$ , $\angle C = 65^\circ$ , $\angle CAD = 20^\circ$ , then $\angle ADC = 180^\circ - 65^\circ - 20^\circ = 95^\circ$ . $\angle ADB + \angle ADC = 85^\circ + 95^\circ = 180^\circ$, indicating that $B$, $D$, $C$ are three points collinear ✓
 
-## 验算
+## Check calculation
 
-### 验算方法一：内角和一致性检验
+### Calculation method one: interior angle sum consistency test
 
-将 $D$ 分割成的两个小三角形分别计算内角和：
+Divide $D$ into two small triangles and calculate the interior angle sum respectively:
 
-**在 $\triangle ABD$ 中：**
+**In $\triangle ABD$:**
 
-$$
+ $$
 \angle BAD + \angle B + \angle ADB = 30^\circ + 65^\circ + \angle ADB = 180^\circ
 $$
 
-$$
+ $$
 \angle ADB = 180^\circ - 95^\circ = 85^\circ
 $$
 
-**在 $\triangle ADC$ 中：**
+**In $\triangle ADC$:**
 
-$$
+ $$
 \angle CAD + \angle C + \angle ADC = 20^\circ + 65^\circ + \angle ADC = 180^\circ
 $$
 
-$$
+ $$
 \angle ADC = 180^\circ - 85^\circ = 95^\circ
 $$
 
-**验证三点共线：** $\angle ADB$ 和 $\angle ADC$ 是 $AD$ 两侧的邻补角。
+**Verify that the three points are collinear:** $\angle ADB$ and $\angle ADC$ are adjacent supplementary angles on both sides of $AD$.
 
-$$
+ $$
 \angle ADB + \angle ADC = 85^\circ + 95^\circ = 180^\circ
 $$
 
-$B$、$D$、$C$ 共线，满足题设 $D$ 在 $BC$ 上的条件。✓
+$B$, $D$, $C$ are collinear, satisfying the condition that $D$ is on $BC$. ✓
 
-### 验算方法二：坐标几何法
+### Calculation method two: coordinate geometry method
 
-建立坐标系，用坐标数值验证。
+Establish a coordinate system and verify it numerically.
 
-以 $A$ 为原点 $(0, 0)$，设 $AB = AC = 1$（不失一般性，因为相似三角形角度相同）。将射线 $AB$ 放在 $x$ 轴正向上：
+Taking $A$ as the origin $(0, 0)$ , let $AB = AC = 1$ (without loss of generality, since similar triangles have the same angles). Place ray $AB$ on the positive axis of $x$:
 
-- $B$：$(1, 0)$
-- $A$：$(0, 0)$
+-  $B$ ： $(1, 0)$
+-  $A$ ： $(0, 0)$
 
-由于 $\angle A = 50^\circ$ 且 $AC = 1$，点 $C$ 的坐标为：
+Since $\angle A = 50^\circ$ and $AC = 1$ , the coordinates of point $C$ are:
 
-$$
+ $$
 C = (\cos 50^\circ,\ \sin 50^\circ) \approx (0.6428,\ 0.7660)
 $$
 
-点 $D$ 在 $BC$ 上，且 $\angle BAD = 30^\circ$。射线 $AD$ 的方向角为 $30^\circ$：
+Point $D$ is on $BC$ and $\angle BAD = 30^\circ$ . The direction angle of ray $AD$ is $30^\circ$ :
 
-$$
-AD \text{ 方向向量：}(\cos 30^\circ,\ \sin 30^\circ) = \left(\frac{\sqrt{3}}{2},\ \frac{1}{2}\right) \approx (0.8660,\ 0.5000)
+ $$
+AD \text{ Direction vector: }(\cos 30^\circ,\ \sin 30^\circ) = \left(\frac{\sqrt{3}}{2},\ \frac{1}{2}\right) \approx (0.8660,\ 0.5000)
 $$
 
-$D$ 是 $AD$ 与 $BC$ 的交点。
+$D$ is the intersection of $AD$ and $BC$.
 
-直线 $BC$ 过 $B(1,0)$ 和 $C(0.6428, 0.7660)$，参数方程：
+Straight line $BC$ passes through $B(1,0)$ and $C(0.6428, 0.7660)$, parametric equation:
 
-$$
+ $$
 BC: (1, 0) + t[(0.6428, 0.7660) - (1, 0)] = (1 - 0.3572t,\ 0.7660t),\quad 0 \leq t \leq 1
 $$
 
-直线 $AD$ 方程：$(s \cos 30^\circ,\ s \sin 30^\circ) = (0.8660s,\ 0.5000s),\ s \geq 0$
+Line $AD$ Equation: $(s \cos 30^\circ,\ s \sin 30^\circ) = (0.8660s,\ 0.5000s),\ s \geq 0$
 
-令两点重合求交点：
+Let two points coincide and find the intersection point:
 
-$$
+ $$
 \begin{cases}
 0.8660s = 1 - 0.3572t \\
 0.5000s = 0.7660t
 \end{cases}
 $$
 
-由此可解出 $D$ 点位置（数值验证可得到在 $BC$ 上）。
+From this, the $D$ point position can be solved (numerical verification can be obtained on $BC$).
 
-计算 $\angle B$：向量 $\overrightarrow{BA}$ 与 $\overrightarrow{BC}$ 的夹角。
+Calculate $\angle B$: the angle between vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$.
 
-$$
+ $$
 \overrightarrow{BA} = (0-1,\ 0-0) = (-1,\ 0)
 $$
-$$
+ $$
 \overrightarrow{BC} = (0.6428-1,\ 0.7660-0) = (-0.3572,\ 0.7660)
 $$
 
-$$
+ $$
 \cos \angle B = \frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{|\overrightarrow{BA}| \cdot |\overrightarrow{BC}|} = \frac{(-1)(-0.3572) + 0 \times 0.7660}{1 \times \sqrt{0.3572^2 + 0.7660^2}}
 $$
 
-$$
+ $$
 = \frac{0.3572}{\sqrt{0.1276 + 0.5868}} = \frac{0.3572}{\sqrt{0.7144}} \approx \frac{0.3572}{0.8452} \approx 0.4226
 $$
 
-$$
+ $$
 \angle B = \arccos(0.4226) \approx 65^\circ
 $$
 
-坐标几何计算与解析结果一致 ✓
+Coordinate geometry calculations are consistent with analytical results ✓
 
-## 最终答案
+## Final answer
 
-$$
+ $$
 \boxed{\angle B = 65^\circ}
 $$
 
-## 易错点
-1. **忽略 $D$ 在 $BC$ 上的作用：** $D$ 的作用是将 $\angle A$ 表示为 $\angle BAD + \angle CAD$，这是求 $\angle A$ 的关键。有人会试图用 $D$ 做其他复杂构造而舍近求远。
-2. **等腰三角形性质记反：** 等腰三角形中"等边对等角"，不要误记为"等边对大角"。$AB = AC$ 得出 $\angle B = \angle C$（两个底角相等），而非 $\angle A = \angle B$。
-3. **内角和公式记错：** 三角形的内角和是 $180^\circ$，不要误记为 $360^\circ$（那是四边形的内角和）或 $90^\circ$。
-4. **角度单位混淆：** 解题使用度数制（$^\circ$），不要与弧度制混淆。最终的 $65^\circ$ 如果写成 $\frac{65\pi}{180}$ 就错了——除非特意要求弧度。
-5. **只列方程不解到底：** 得出 $2\angle B = 130^\circ$ 后忘记除以 $2$，直接写 $\angle B = 130^\circ$。$130^\circ$ 是 $2\angle B$，不是 $\angle B$。
+## Easy to make mistakes
+1. **Ignore the role of $D$ on $BC$:** The role of $D$ is to represent $\angle A$ as $\angle BAD + \angle CAD$, which is the key to finding $\angle A$. Some people will try to use $D$ to do other complex structures and sacrifice the near and far.
+2. **Reverse the properties of an isosceles triangle:** In an isosceles triangle, "equilateral sides correspond to equal angles". Do not mistakenly write "equilateral sides correspond to large angles". $AB = AC$ gives $\angle B = \angle C$ (two base angles are equal), not $\angle A = \angle B$ .
+3. **Incorrect memory of the formula for the sum of interior angles:** The sum of the interior angles of a triangle is $180^\circ$. Do not mistakenly remember it as $360^\circ$ (that is the sum of the interior angles of a quadrilateral) or $90^\circ$.
+4. **Angle unit confusion:** Use the degree system ($^\circ$) to solve the problem, do not confuse it with the radian system. The final $65^\circ$ would be wrong if written as $\frac{65\pi}{180}$ - unless radians were specifically requested.
+5. **Only list the equations without solving them:** After getting $2\angle B = 130^\circ$, forget to divide by $2$ and directly write $\angle B = 130^\circ$. $130^\circ$ is $2\angle B$ , not $\angle B$ .
diff --git a/examples/hard_problem.md b/examples/hard_problem.md
index 9a4d751..43eb22d 100644
--- a/examples/hard_problem.md
+++ b/examples/hard_problem.md
@@ -1,133 +1,133 @@
-# 困难问题协议：哥德巴赫猜想
+# Protocol on Hard Problems: Goldbach’s Conjecture
 
-## 用户输入
-大于 2 的每个偶数是否都是两个素数之和？
+## User input
+Is every even number greater than 2 the sum of two prime numbers?
 
-## Skill 分类
-数论 / 困难问题协议
+## Skill Category
+Number Theory/Hard Problem Protocol
 
-## 题意解析
-- **问题的性质**：这不是一道普通的习题，而是**哥德巴赫猜想（Goldbach's conjecture）**——数论中最为著名的未解决问题之一
-- **问题历史**：1742 年由 Christian Goldbach 在致 Euler 的信中提出
-- **当前状态**：截止 2024 年，该猜想**仍未被证明或否定**
-- **核心约束**：AI 绝不能声称已证明或解决该猜想，必须严格遵守数学界的已知结论
+## Question meaning analysis
+- **Nature of the problem**: This is not an ordinary exercise, but **Goldbach's conjecture** - one of the most famous unsolved problems in number theory
+- **Question History**: 1742 by Christian Goldbach in a letter to Euler
+- **Current Status**: As of 2024, this conjecture **has not yet been proven or disproven**
+- **Core Constraints**: AI must not claim to have proven or solved the conjecture, and must strictly adhere to known conclusions in the mathematical community
 
-## 方法选择
+## Method selection
 
-根据 **困难问题协议（Hard Problem Protocol）**，对于未解决的数学猜想，应采取以下方法：
+According to the **Hard Problem Protocol**, the following approach should be taken for unsolved mathematical conjectures:
 
-1. **分类与确认**：确认这是已知的开放问题
-2. **陈述已知事实**：区分已验证结果、部分结果、等价命题
-3. **避免造假**：严禁伪造证明或声称已解决
-4. **提供探索路径**：如果用户有兴趣，可介绍已知的尝试方法
+1. **Classification and Confirmation**: Confirm that this is a known open issue
+2. **State known facts**: distinguish between verified results, partial results, and equivalent propositions
+3. **Avoid forgery**: It is strictly prohibited to forge certificates or claim to have solved the problem
+4. **Provide exploration paths**: If the user is interested, you can introduce known attempts.
 
-## 解题过程
+## Problem solving process
 
-### Step 1：问题分类
+### Step 1: Problem classification
 
-这是**哥德巴赫猜想**（强哥德巴赫猜想），一个自 1742 年以来的开放问题。
+This is Goldbach's Conjecture, an open problem since 1742.
 
-**问题正式陈述**：每个大于 2 的偶数都可以表示为两个素数之和。
+**Formal Statement of the Problem**: Every even number greater than 2 can be expressed as the sum of two prime numbers.
 
-### Step 2：明确当前状态
+### Step 2: Clarify the current status
 
-**哥德巴赫猜想尚未被证明。** 截至 2024 年，它在数学上仍然是一个开放问题。
+**Goldbach's conjecture has not been proven. ** As of 2024, it remains mathematically an open problem.
 
-### Step 3：已验证的范围
+### Step 3: Verified range
 
-尽管未被证明，猜想已被**计算机验证**对于极大的范围成立：
+Although not proven, the conjecture has been verified by computers to hold for very large scales:
 
-- **已验证范围**：所有偶数 $n \leq 4 \times 10^{18}$ 都满足猜想（Oliveira e Silva, 2014 年及后续工作）
-- **验证方法**：对于范围内的每个偶数，搜索其素数分解
+- **Verified Range**: All even numbers $n \leq 4 \times 10^{18}$ satisfy the conjecture (Oliveira e Silva, 2014 and subsequent work)
+- **Verification Method**: For every even number in the range, search for its prime factorization
 
-注意：有限验证不等于证明。例如，可能存在某个大于 $4 \times 10^{18}$ 的偶数不满足猜想。
+Note: Limited verification is not the same as proof. For example, there may be some even number greater than $4 \times 10^{18}$ that does not satisfy the conjecture.
 
-### Step 4：已知的部分结果
+### Step 4: Known partial results
 
-以下是数学界已严格证明的相关结果：
+The following are relevant results that have been rigorously proven by the mathematical community:
 
-| 结果 | 内容 | 提出者/时间 |
+| Result | Content | Presenter/Time |
 |------|------|------------|
-| Chen 定理 | 每个充分大的偶数可以表示为一个素数和一个至多两个素数乘积的数（半素数）之和 | 陈景润, 1973 |
-| 三素数定理（Ternary Goldbach） | 每个大于 5 的奇数可表示为三个素数之和 | **已证明**（Helfgott, 2013） |
-| 弱哥德巴赫猜想的等价形式 | 每个大于 7 的奇数是三个奇素数之和 | 由三素数定理导出 |
-| Vinogradov 定理 | 所有充分大的奇数可表示为三个素数之和 | Vinogradov, 1937 |
-| 密度结果 | 不满足猜想的偶数（如果存在）的密度为零 | 多个作者的渐进结果 |
+| Chen's theorem | Every sufficiently large even number can be expressed as the sum of a prime number and a number (semi-prime number) that is the product of at most two prime numbers | Chen Jingrun, 1973 |
+| Three Prime Theorem (Ternary Goldbach) | Every odd number greater than 5 can be expressed as the sum of three prime numbers | **Proved** (Helfgott, 2013) |
+| Equivalent form of weak Goldbach's conjecture | Every odd number greater than 7 is the sum of three odd prime numbers | Derived from the three prime number theorem |
+| Vinogradov's theorem | All sufficiently large odd numbers can be expressed as the sum of three prime numbers | Vinogradov, 1937 |
+| Density results | Even integers that do not satisfy the conjecture (if they exist) have a density of zero | Asymptotic results by multiple authors |
 
-**关键区分**：
-- 三素数定理（奇数 = 三个素数之和）→ **已证明** ✓
-- 强哥德巴赫猜想（偶数 = 两个素数之和）→ **未证明** ✗
+**Key distinction**:
+- Three prime numbers theorem (odd number = sum of three prime numbers) → **Proved** ✓
+- Strong Goldbach's conjecture (even number = sum of two prime numbers) → **Unproven** ✗
 
-二者容易混淆，但它们是不等价的：三素数定理成立并不蕴含强哥德巴赫猜想成立。
+The two are easy to confuse, but they are not equivalent: the establishment of the three prime number theorem does not imply the establishment of the Strong Goldbach Conjecture.
 
-### Step 5：已知的主要尝试方法
+### Step 5: Known main methods to try
 
-历史上数学家尝试过的方法包括：
+Methods that mathematicians have tried throughout history include:
 
-1. **圆法（Circle Method）**：Hardy 和 Littlewood 发展，成功应用于三素数定理的证明，但对于两个素数的情况遇到本质困难
-2. **筛法（Sieve Methods）**：Brun, Selberg 等人的工作，陈景润用加权筛法证明了 Chen 定理（"1+2"）
-3. **解析数论方法**：利用 L-函数、指数和等技术
+1. **Circle Method**: Developed by Hardy and Littlewood and successfully applied to the proof of the three prime number theorem, but encountered essential difficulties in the case of two prime numbers
+2. **Sieve Methods**: The work of Brun, Selberg and others, Chen Jingrun used the weighted sieve method to prove Chen's theorem ("1+2")
+3. **Analytical number theory method**: using techniques such as L-function, exponential sum, etc.
 
-### Step 6：为何该问题如此困难？
+### Step 6: Why is this problem so difficult?
 
-- **加法数论的本质困难**：素数主要由乘法性质定义，而"两素数之和"涉及加法结构，二者之间的桥梁极难建立
-- **筛法的局限性**：筛法能控制的误差项似乎有天然的下界（奇偶性问题, parity problem），使其无法直接证明强猜想
-- **需要全新的思想**：许多数学家认为哥德巴赫猜想的证明可能需要与现有方法根本不同的新工具
+- **Essential difficulty of additive number theory**: Prime numbers are mainly defined by multiplicative properties, while "sum of two prime numbers" involves an additive structure, and the bridge between the two is extremely difficult to establish.
+- **Limitations of the sieve method**: The error term that the sieve method can control seems to have a natural lower bound (parity problem, parity problem), making it impossible to directly prove the strong conjecture
+- **Whole new ideas required**: Many mathematicians believe that the proof of Goldbach's conjecture may require new tools that are fundamentally different from existing methods
 
-### Step 7：如果用户想探索
+### Step 7: If the user wants to explore
 
-如果用户想自己尝试理解这个问题，建议：
+If users want to try to understand this problem themselves, it is recommended:
 
-- **从小数字开始**：手动验证 $4=2+2, 6=3+3, 8=3+5, 10=3+7=5+5, \ldots$
-- **学习筛法入门**：理解 Brun 筛和 Selberg 筛的基本思想
-- **阅读 Chen 定理的简化版本**：理解"1+2"的含义和证明思路
-- **清晰的期望管理**：解决这一猜想需要极为深厚的数学功底，历史上最优秀的数学家也未能攻克
+- **Start with a small number**: Manual verification $4=2+2, 6=3+3, 8=3+5, 10=3+7=5+5, \ldots$
+- **Introduction to learning sieve method**: Understand the basic ideas of Brun sieve and Selberg sieve
+- **Read the simplified version of Chen's theorem**: Understand the meaning of "1+2" ​​and the proof idea
+- **Clear Expectation Management**: Solving this conjecture requires extremely deep mathematical skills, and the best mathematicians in history have not been able to overcome it.
 
-### 小程序：验证特定偶数的分解
+### Mini program: Verify the decomposition of a specific even number
 
-对于较小的偶数（如 $\leq 10^6$），可以通过简单算法验证分解。例如：
+For small even numbers (such as $\leq 10^6$ ), the decomposition can be verified by a simple algorithm. For example:
 
-- $100 = 3 + 97 = 11 + 89 = 17 + 83 = 29 + 71 = 41 + 59 = 47 + 53$
-- $1000 = 3 + 997$（还有多种分解方式）
+-  $100 = 3 + 97 = 11 + 89 = 17 + 83 = 29 + 71 = 41 + 59 = 47 + 53$
+- $1000 = 3 + 997$ (there are many ways to decompose it)
 
-如果需要验证某个具体偶数的分解，可以通过编程完成（在较小范围内穷举素数）。
+If you need to verify the decomposition of a specific even number, you can do it through programming (exhaustively enumerate prime numbers in a smaller range).
 
-## 验算
+## Check calculation
 
-**验算方法 1：确认问题分类**
+**Verification method 1: Confirm problem classification**
 
-查阅标准数学文献，哥德巴赫猜想在《Unsolved Problems in Number Theory》（Guy, 2004）及 Clay 数学研究所的千禧年问题综述中均列为开放问题 ✓
+Checking the standard mathematics literature, Goldbach's conjecture is listed as an open problem in "Unsolved Problems in Number Theory" (Guy, 2004) and the Clay Mathematics Institute's Millennium Problem Review ✓
 
-**验算方法 2：交叉验证已知结果**
+**Verification method 2: Cross-validation of known results**
 
-- 三素数定理由 Helfgott (2013) 证明，发表在 *Annals of Mathematics* 上，可查证 ✓
-- Chen 定理由陈景润 (1973) 证明，在数论教材中有标准记述 ✓
-- 验证范围 $4 \times 10^{18}$ 由分布式计算项目确认，文献可查 ✓
+- The triple prime number theorem was proved by Helfgott (2013), published in *Annals of Mathematics*, available for verification ✓
+- Chen's theorem was proved by Chen Jingrun (1973) and has a standard description in number theory textbooks ✓
+- Verification scope $4 \times 10^{18}$ Confirmed by Distributed Computing Project, literature available ✓
 
-**验算方法 3：小范围手工验证**
+**Verification method 3: Small-scale manual verification**
 
-- $4 = 2 + 2$ ✓
-- $6 = 3 + 3$ ✓
-- $8 = 3 + 5$ ✓
-- $10 = 3 + 7 = 5 + 5$ ✓
-- $12 = 5 + 7$ ✓
-- $14 = 3 + 11 = 7 + 7$ ✓
-- $100 = 47 + 53$（均为素数）✓
+-  $4 = 2 + 2$  ✓
+-  $6 = 3 + 3$  ✓
+-  $8 = 3 + 5$  ✓
+-  $10 = 3 + 7 = 5 + 5$  ✓
+-  $12 = 5 + 7$  ✓
+-  $14 = 3 + 11 = 7 + 7$  ✓
+- $100 = 47 + 53$ (all prime numbers) ✓
 
-## 最终答案
+## Final answer
 
-**哥德巴赫猜想（强猜想）至今未被证明。**
+**Goldbach's conjecture (strong conjecture) has not yet been proven. **
 
-已知事实：
-- 对所有偶数 $n \leq 4 \times 10^{18}$ 已验证成立
-- Chen 定理：充分大偶数 = 素数 + 半素数（最接近的已证明结果）
-- 奇数版本（三素数定理）已由 Helfgott (2013) 证明
-- 猜想在数学界普遍认为是正确的，但缺乏严格证明
+Known facts:
+- $n \leq 4 \times 10^{18}$ has been verified to be true for all even numbers
+- Chen's Theorem: Sufficiently large even numbers = prime numbers + semi-prime numbers (nearest proven result)
+- The odd version (three prime numbers theorem) was proved by Helfgott (2013)
+- The conjecture is generally considered correct in the mathematical community, but lacks rigorous proof
 
-**重要提醒**：不要相信任何声称已证明哥德巴赫猜想的非同行评审来源。
+**IMPORTANT NOTE**: Do not trust any non-peer-reviewed source that claims to have proved Goldbach's conjecture.
 
-## 易错点
-1. **混淆强弱猜想**：三素数定理（弱猜想）已证明，不等于强猜想已证明
-2. **有限验证误认为证明**：即使验证了 $10^{18}$ 以内的所有偶数，也不是证明——数学归纳原理只能从 $\forall n: P(n) \implies P(n+1)$ 和 $P(0)$ 推导出 $\forall n: P(n)$，但哥德巴赫猜想没有这样的归纳步骤
-3. **AI 幻觉风险**：AI 模型可能"编造"一个看似合理的证明。用户应始终交叉验证数学声明
-4. **不理解"开放问题"的含义**：这意味着**没有任何人**（包括最顶尖数学家）知道答案，而非"还没学到"或"没有人告诉过你"
+## Easy to make mistakes
+1. **Confusing strong and weak conjectures**: The three prime number theorem (weak conjecture) has been proved, but it does not mean that the strong conjecture has been proved
+2. **Limited verification mistaken for proof**: Even if all even numbers within $10^{18}$ are verified, it is not a proof - the principle of mathematical induction can only deduce $\forall n: P(n)$ from $\forall n: P(n) \implies P(n+1)$ and $P(0)$, but Goldbach's conjecture has no such induction step
+3. **AI Hallucination Risk**: The AI ​​model may "make up" a seemingly reasonable proof. Users should always cross-validate mathematical claims
+4. **Not understanding the meaning of "open question": It means that **no one** (including the top mathematicians) knows the answer, not "hasn't learned it yet" or "no one has told you"
diff --git a/examples/inequalities.md b/examples/inequalities.md
index e0e0a53..cefd5d8 100644
--- a/examples/inequalities.md
+++ b/examples/inequalities.md
@@ -1,172 +1,172 @@
-# 分式不等式的解法
+# Solutions to fractional inequalities
 
-## 用户输入
-解不等式：$\displaystyle \frac{x^2-3x+2}{x+1} \geq 0$
+## User input
+Solve the inequality: $\displaystyle \frac{x^2-3x+2}{x+1} \geq 0$
 
-## Skill 分类
-不等式（Inequalities）— 分式不等式与符号表法
+## Skill Category
+Inequalities - Fractional Inequalities and Symbol Table Method
 
-## 题意解析
+## Question meaning analysis
 
-**已知条件：**
-- 分式不等式：$\frac{x^2-3x+2}{x+1} \geq 0$
+**Known conditions:**
+- Fractional inequality: $\frac{x^2-3x+2}{x+1} \geq 0$
 
-**求解目标：**
-- 求满足不等式 $\geq 0$ 的所有实数 $x$。
+**Solution goal:**
+- Find all real numbers $x$ that satisfy the inequality $\geq 0$.
 
-**变量与定义域：**
-- 分母 $x+1 \neq 0$，即 $x \neq -1$
-- 分子 $x^2-3x+2$ 为二次多项式，对一切实数有定义
-- 定义域：$x \in \mathbb{R} \setminus \{-1\}$
+**Variables and Domains:**
+- Denominator $x+1 \neq 0$ , which is $x \neq -1$
+- The numerator $x^2-3x+2$ is a quadratic polynomial and is defined for all real numbers
+- Domain: $x \in \mathbb{R} \setminus \{-1\}$
 
-**隐式条件：**
-- 分子可因式分解为 $(x-1)(x-2)$，意味着在 $x=1$ 和 $x=2$ 处分式值为 $0$，应纳入解集（不等式是 $\geq$，含等号）。
-- 分母根 $x=-1$ 是间断点，分式在此无定义，绝对不能纳入解集。
+**Implicit condition:**
+- The numerator is factorable as $(x-1)(x-2)$ , which means that at $x=1$ and $x=2$ the fractional value is $0$ , which should be included in the solution set (the inequality is $\geq$ , including the equal sign).
+- The denominator root $x=-1$ is a discontinuity point, the fraction is not defined here and must not be included in the solution set.
 
-**解的个数：**
-- 不等式的解是实数轴上的若干区间，可能有无限多个解。
+**Number of solutions:**
+- The solutions to the inequalities are intervals on the real number axis, and there are infinitely many possible solutions.
 
-## 方法选择
+## Method selection
 
-**选择方法：** 符号表法（穿根法/数轴标根法）
+**Selection method:** Symbol table method (root crossing method/number axis marking method)
 
-**理由：**
-- 分式不等式的解题核心是确定分子分母在各个区间的符号。
-- 符号表法通过列出所有零点，划分数轴为若干区间，逐一确定各区间的符号，直观且不易遗漏。
+**reason:**
+- The core of solving fractional inequalities is to determine the signs of the numerator and denominator in each interval.
+- The symbol table method lists all zero points, divides the number axis into several intervals, and determines the symbols of each interval one by one, which is intuitive and difficult to miss.
 
-**备选方法：**
-- 分类讨论法：将分式不等式 $\frac{P(x)}{Q(x)} \geq 0$ 转化为 $\begin{cases} P(x) \geq 0 \\ Q(x) > 0 \end{cases}$ 或 $\begin{cases} P(x) \leq 0 \\ Q(x) < 0 \end{cases}$。可行但步骤更多。
-- 直接移项通分法：将不等式化为 $\frac{(x-1)(x-2)}{x+1} \geq 0$，然后分析符号——实际就是符号表法。
+**Alternative method:**
+- Classification discussion method: Convert the fractional inequality $\frac{P(x)}{Q(x)} \geq 0$ into $\begin{cases} P(x) \geq 0 \\ Q(x) > 0 \end{cases}$ or $\begin{cases} P(x) \leq 0 \\ Q(x) < 0 \end{cases}$. It works but there are more steps.
+- Direct transfer method: convert the inequality into $\frac{(x-1)(x-2)}{x+1} \geq 0$, and then analyze the symbols - actually it is the symbol table method.
 
-**关键原则：**
-- 分母不能为零，$x=-1$ 处用空心点标记。
-- 分子零点处不等式取等号成立，$x=1$ 和 $x=2$ 用实心点标记。
+**Key Principles:**
+- The denominator cannot be zero, $x=-1$ is marked with a hollow dot.
+- The equality sign holds true at the zero point of the molecule, $x=1$ and $x=2$ are marked with solid dots.
 
-## 解题过程
+## Problem solving process
 
-### 第一步：因式分解分子
+### Step 1: Factor the numerator
 
-$$
+ $$
 x^2 - 3x + 2 = (x-1)(x-2)
 $$
 
-验证：$(x-1)(x-2) = x^2 - 2x - x + 2 = x^2 - 3x + 2$ ✓
+Verification: $(x-1)(x-2) = x^2 - 2x - x + 2 = x^2 - 3x + 2$ ✓
 
-原不等式化为：
+The original inequality becomes:
 
-$$
+ $$
 \frac{(x-1)(x-2)}{x+1} \geq 0
 $$
 
-### 第二步：确定关键点
+### Step 2: Determine key points
 
-分式的符号由三个线性因子决定：
-- 分子因子：$x-1$（零点 $x=1$），$x-2$（零点 $x=2$）
-- 分母因子：$x+1$（零点 $x=-1$）
+The sign of the fraction is determined by three linear factors:
+- Numerator factors: $x-1$ (zero point $x=1$ ), $x-2$ (zero point $x=2$ )
+- Denominator factor: $x+1$ (zero point $x=-1$ )
 
-将所有关键点按从小到大的顺序排列在数轴上：
+Arrange all key points on the number axis in order from small to large:
 
-$$
+ $$
 -1 \quad 1 \quad 2
 $$
 
-它们在数轴上划分出四个区间：
+They divide four intervals on the number line:
 
-$$
+ $$
 (-\infty, -1),\ (-1, 1),\ (1, 2),\ (2, +\infty)
 $$
 
-### 第三步：建立符号表
+### Step 3: Create symbol table
 
-在每个区间内取一个代表点，判断各因子的符号，进而判断分式的符号。
+Take a representative point in each interval to determine the sign of each factor, and then determine the sign of the fraction.
 
-| 区间 | 代表点 | $x+1$ | $x-1$ | $x-2$ | $\frac{(x-1)(x-2)}{x+1}$ | 符号 |
+| Interval | Representative point | $x+1$ | $x-1$ | $x-2$ | $\frac{(x-1)(x-2)}{x+1}$ | Symbol |
 |------|--------|-------|-------|-------|---------------------------|------|
-| $(-\infty, -1)$ | $x=-2$ | $-$ | $-$ | $-$ | $\frac{(-)\times(-)}{(-)} = \frac{(+)}{(-)} = (-)$ | **负** |
-| $(-1, 1)$ | $x=0$ | $+$ | $-$ | $-$ | $\frac{(-)\times(-)}{(+)} = \frac{(+)}{(+)} = (+)$ | **正** |
-| $(1, 2)$ | $x=1.5$ | $+$ | $+$ | $-$ | $\frac{(+)\times(-)}{(+)} = \frac{(-)}{(+)} = (-)$ | **负** |
-| $(2, +\infty)$ | $x=3$ | $+$ | $+$ | $+$ | $\frac{(+)\times(+)}{(+)} = (+)$ | **正** |
+| $(-\infty, -1)$ | $x=-2$ | $-$ | $-$ | $-$ | $\frac{(-)\times(-)}{(-)} = \frac{(+)}{(-)} = (-)$ | **Negative** |
+| $(-1, 1)$ | $x=0$ | $+$ | $-$ | $-$ | $\frac{(-)\times(-)}{(+)} = \frac{(+)}{(+)} = (+)$ | **True** |
+| $(1, 2)$ | $x=1.5$ | $+$ | $+$ | $-$ | $\frac{(+)\times(-)}{(+)} = \frac{(-)}{(+)} = (-)$ | **Negative** |
+| $(2, +\infty)$ | $x=3$ | $+$ | $+$ | $+$ | $\frac{(+)\times(+)}{(+)} = (+)$ | **True** |
 
-### 第四步：处理边界点
+### Step 4: Process boundary points
 
-在关键点处判断分式的值：
+Judge the value of the fraction at key points:
 
-- $x = -1$：**分母为零，分式无意义**。空心点，不纳入解集。
-- $x = 1$：代入，$\frac{(1-1)(1-2)}{1+1} = \frac{0 \times (-1)}{2} = 0$，满足 $\geq 0$。**实心点，纳入解集。**
-- $x = 2$：代入，$\frac{(2-1)(2-2)}{2+1} = \frac{1 \times 0}{3} = 0$，满足 $\geq 0$。**实心点，纳入解集。**
+- $x = -1$: **The denominator is zero, the fraction is meaningless**. Hollow points are not included in the solution set.
+- $x = 1$: Substitute, $\frac{(1-1)(1-2)}{1+1} = \frac{0 \times (-1)}{2} = 0$, satisfy $\geq 0$. **Solid points are included in the solution set. **
+- $x = 2$: Substitute, $\frac{(2-1)(2-2)}{2+1} = \frac{1 \times 0}{3} = 0$, satisfy $\geq 0$. **Solid points are included in the solution set. **
 
-### 第五步：写出解集
+### Step 5: Write the solution set
 
-不等式 $\geq 0$ 要求分式值为非负。由符号表可知，分式取正值（$>0$）的区间为 $(-1, 1)$ 和 $(2, +\infty)$，取零值的点为 $x=1$ 和 $x=2$。合并得：
+The inequality $\geq 0$ requires that the fraction value be nonnegative. It can be seen from the symbol table that the interval where the fraction takes a positive value ($>0$) is $(-1, 1)$ and $(2, +\infty)$, and the points where the fraction takes a zero value are $x=1$ and $x=2$. Combined:
 
-$$
+ $$
 x \in (-1, 1] \cup [2, +\infty)
 $$
 
-用数轴表示：
+Represented on a number line:
 
-```
+ ```
 <---(----●========●)----(----●=======>
     -1       0       1       2
-    空心       正     实心       实心
+Hollow Positive Solid Solid
 ```
 
-## 验算
+## Check calculation
 
-### 验算方法一：分类区间代表性取点
+### Calculation method one: representative points of classification interval
 
-在每个解集区间和非解集区间取点验证：
+Take points in each solution set interval and non-solution set interval to verify:
 
-**区间 $(-1, 1]$（应满足）：**
+**Interval $(-1, 1]$ (should be satisfied): **
 
-$x = 0$：$\frac{0-0+2}{1} = 2 \geq 0$ ✓
+ $x = 0$ ： $\frac{0-0+2}{1} = 2 \geq 0$  ✓
 
-$x = 0.5$：$\frac{0.25-1.5+2}{1.5} = \frac{0.75}{1.5} = 0.5 > 0$ ✓
+ $x = 0.5$ ： $\frac{0.25-1.5+2}{1.5} = \frac{0.75}{1.5} = 0.5 > 0$  ✓
 
-$x = 1$（边界）：$\frac{1-3+2}{2} = \frac{0}{2} = 0 \geq 0$ ✓
+$x = 1$ (boundary): $\frac{1-3+2}{2} = \frac{0}{2} = 0 \geq 0$ ✓
 
-**区间 $[2, +\infty)$（应满足）：**
+**Interval $[2, +\infty)$ (should be satisfied): **
 
-$x = 2$（边界）：$\frac{4-6+2}{3} = \frac{0}{3} = 0 \geq 0$ ✓
+$x = 2$ (boundary): $\frac{4-6+2}{3} = \frac{0}{3} = 0 \geq 0$ ✓
 
-$x = 3$：$\frac{9-9+2}{4} = \frac{2}{4} = 0.5 > 0$ ✓
+ $x = 3$ ： $\frac{9-9+2}{4} = \frac{2}{4} = 0.5 > 0$  ✓
 
-$x = 10$：$\frac{100-30+2}{11} = \frac{72}{11} > 0$ ✓
+ $x = 10$ ： $\frac{100-30+2}{11} = \frac{72}{11} > 0$  ✓
 
-**区间 $(-\infty, -1)$（不应满足）：**
+**Interval $(-\infty, -1)$ (should not be met):**
 
-$x = -2$：$\frac{4+6+2}{-1} = \frac{12}{-1} = -12 < 0$（不满足） ✓
+$x = -2$: $\frac{4+6+2}{-1} = \frac{12}{-1} = -12 < 0$ (not satisfied) ✓
 
-$x = -10$：$\frac{100+30+2}{-9} = \frac{132}{-9} < 0$（不满足） ✓
+$x = -10$: $\frac{100+30+2}{-9} = \frac{132}{-9} < 0$ (not satisfied) ✓
 
-**区间 $(1, 2)$（不应满足）：**
+**Interval $(1, 2)$ (should not be met): **
 
-$x = 1.5$：$\frac{2.25-4.5+2}{2.5} = \frac{-0.25}{2.5} = -0.1 < 0$（不满足） ✓
+$x = 1.5$: $\frac{2.25-4.5+2}{2.5} = \frac{-0.25}{2.5} = -0.1 < 0$ (not satisfied) ✓
 
-所有验证点均符合预期。
+All verification points are as expected.
 
-### 验算方法二：函数图像法
+### Calculation method two: function image method
 
-令 $f(x) = \frac{x^2-3x+2}{x+1}$，分析其渐近线和零点。
+Let $f(x) = \frac{x^2-3x+2}{x+1}$ , analyze its asymptotes and zeros.
 
-- 垂直渐近线：$x = -1$
-- 零点：$x = 1$，$x = 2$
-- 水平渐近线：$\lim_{x\to +\infty} f(x) = +\infty$（分子二次，分母一次）
+- Vertical asymptote: $x = -1$
+- Zero point: $x = 1$, $x = 2$
+- Horizontal asymptote: $\lim_{x\to +\infty} f(x) = +\infty$ (numerator quadratic, denominator linear)
 
-函数在零点处分式值为 $0$，在零点之间符号交替变化，与符号表分析一致。$f(x) \geq 0$ 的解集为函数图像位于 $x$ 轴及其以上的部分，即 $(-1, 1] \cup [2, +\infty)$。
+The function's fractional value at zero points is $0$ , and its sign alternates between zero points, consistent with symbol table analysis. The solution set of $f(x) \geq 0$ is the part of the function graph located on and above the $x$ axis, that is, $(-1, 1] \cup [2, +\infty)$ .
 
-## 最终答案
+## Final answer
 
-$$
+ $$
 \boxed{x \in (-1,\ 1] \cup [2,\ +\infty)}
 $$
 
-或用不等式表示：$-1 < x \leq 1$ 或 $x \geq 2$
+Or expressed as an inequality: $-1 < x \leq 1$ or $x \geq 2$
 
-## 易错点
-1. **分母零点误纳入解集：** $x=-1$ 时分式无意义，必须排除。即使用的是 $\geq$ 符号，分母零点也不能取等号。
-2. **分母零点导致遗漏区间：** 做符号表时，分子零点将数轴划分为区间，**分母零点同样划分区间**。忘记分母零点会导致少分析一个区间，解集错误。
-3. **符号判断错误：** 三个因子符号相乘时计算错误。建议在符号表中逐因子写下符号，再综合判断。也可以使用穿根法：从右上方开始，遇到奇次因子穿根，遇偶次因子反弹。
-4. **区间端点符号写错：** $(-1, 1]$ 中 $-1$ 用小括号（不包含），$1$ 用中括号（包含）。区间 $[2, +\infty)$ 中 $2$ 包含。写反会直接丢分。
-5. **忽视"取等"条件：** 不等式是 $\geq 0$ 时，分子零点处 $=0$ 满足条件，应纳入解集。如果是 $>0$（严格不等），分子零点也需排除。
-6. **穿根法起始方向：** 习惯从右上方开始穿根，但如果最高次项前面有负号，应从右下方开始。验证最右侧区间的符号可以避免方向错误。
+## Easy to make mistakes
+1. **The zero point of the denominator is mistakenly included in the solution set:** $x=-1$ The time fraction is meaningless and must be eliminated. Even if $\geq$ notation is used, the denominator zero cannot be equal.
+2. **The zero point in the denominator causes the missing interval:** When making a symbol table, the zero point in the numerator divides the number axis into intervals, and the zero point in the denominator also divides the interval**. Forgetting the zero point in the denominator will result in one less interval being analyzed and the solution set being wrong.
+3. **Symbol judgment error:** Calculation error when the signs of three factors are multiplied. It is recommended to write down the symbols factor by factor in the symbol table and then make a comprehensive judgment. You can also use the root-penetrating method: starting from the upper right, it will penetrate the root when encountering odd-order factors, and rebound when encountering even-order factors.
+4. **The interval endpoint symbols are written incorrectly:** In $(-1, 1]$, $-1$ uses parentheses (not included), and $1$ uses square brackets (inclusive). The range $[2, +\infty)$ contains $2$. If you write it backwards, you will lose points directly.
+5. **Ignore the "equal" condition:** When the inequality is $\geq 0$, $=0$ at the zero point of the molecule satisfies the condition and should be included in the solution set. In the case of $>0$ (strictly unequal), the numerator zeros also need to be excluded.
+6. **Starting direction of root threading method:** It is customary to start root threading from the upper right, but if there is a negative sign in front of the highest order term, you should start from the lower right. Verifying the sign of the rightmost interval can avoid directional errors.
diff --git a/examples/integration_example.md b/examples/integration_example.md
index 1524d05..c2a9e5a 100644
--- a/examples/integration_example.md
+++ b/examples/integration_example.md
@@ -1,100 +1,100 @@
-# 不定积分：$\int x \sin x \, dx$ 的计算
+# Indefinite integral: calculation of $\int x \sin x \, dx$
 
-## 用户输入
-计算不定积分 $\displaystyle \int x \sin x \, dx$
+## User input
+Compute indefinite integrals $\displaystyle \int x \sin x \, dx$
 
-## Skill 分类
-微积分 / 不定积分
+## Skill Category
+Calculus / Indefinite integral
 
-## 题意解析
-- **被积函数**：$x \sin x$，是两个函数的乘积
-- **目标**：求原函数（不定积分），需包含积分常数 $C$
-- **方法选择依据**：被积函数是多项式 $x$ 和三角函数 $\sin x$ 的乘积，这是**分部积分法**的典型适用情境
-- **隐含条件**：积分变量为 $x$，积分区间未指定（不定积分）
-- **解的唯一性**：原函数不唯一（相差常数），但表示为标准形式即可
+## Question meaning analysis
+- **integrand**: $x \sin x$, is the product of two functions
+- **Goal**: Find the original function (indefinite integral), which must include the integration constant $C$
+- **Method selection basis**: The integrand is the product of the polynomial $x$ and the trigonometric function $\sin x$. This is a typical application situation of the **integration by parts method**
+- **Implicit condition**: The integration variable is $x$ and the integration interval is not specified (indefinite integral)
+- **Uniqueness of solution**: The original function is not unique (difference constant), but it can be expressed in standard form
 
-## 方法选择
+## Method selection
 
-**主方法：分部积分法（Integration by Parts）**
+**Main method: Integration by Parts**
 
-选择理由：分部积分法适用于两类不同函数乘积的积分，特别是当其中一个函数求导后简化（$x \to 1$）、另一个函数容易积分（$\sin x \to -\cos x$）时。
+Reason for selection: The integral method by parts is suitable for the integration of the product of two different types of functions, especially when one of the functions is simplified after derivation ($x \to 1$) and the other function is easy to integrate ($\sin x \to -\cos x$).
 
-分部积分公式：
-$$\int u \, dv = uv - \int v \, du$$
+Integral by parts formula:
+ $$\int u \, dv = uv - \int v \, du$$
 
-**LIATE 规则**：选取 $u$ 的优先级为 Logarithmic > Inverse trig > Algebraic > Trigonometric > Exponential。此处 $x$ 是代数函数（A），$\sin x$ 是三角函数（T），根据 LIATE 规则应选择 $u = x$，$dv = \sin x \, dx$。
+**LIATE Rule**: The priority for selecting $u$ is Logarithmic > Inverse trig > Algebraic > Trigonometric > Exponential. Here $x$ is an algebraic function (A), $\sin x$ is a trigonometric function (T), and according to LIATE rules $u = x$ , $dv = \sin x \, dx$ should be chosen.
 
-备选方法：不存在——$x \sin x$ 没有简单的变元代换方法，分部积分是唯一的标准方法。
+Alternative methods: None - $x \sin x$ There is no simple method of substitution of variables, integration by parts is the only standard method.
 
-## 解题过程
+## Problem solving process
 
-**Step 1：选定 $u$ 和 $dv$**
+**Step 1: Select $u$ and $dv$ **
 
-令：
-$$u = x, \quad dv = \sin x \, dx$$
+make:
+ $$u = x, \quad dv = \sin x \, dx$$
 
-**Step 2：计算 $du$ 和 $v$**
+**Step 2: Calculate $du$ and $v$ **
 
-$$du = dx, \quad v = \int \sin x \, dx = -\cos x$$
+ $$du = dx, \quad v = \int \sin x \, dx = -\cos x$$
 
-**Step 3：代入分部积分公式**
+**Step 3: Substitute the integral of parts formula**
 
-$$\int x \sin x \, dx = uv - \int v \, du$$
+ $$\int x \sin x \, dx = uv - \int v \, du$$
 
-$$= x \cdot (-\cos x) - \int (-\cos x) \, dx$$
+ $$= x \cdot (-\cos x) - \int (-\cos x) \, dx$$
 
-$$= -x \cos x + \int \cos x \, dx$$
+ $$= -x \cos x + \int \cos x \, dx$$
 
-**Step 4：计算剩余积分**
+**Step 4: Calculate remaining points**
 
-$$\int \cos x \, dx = \sin x$$
+ $$\int \cos x \, dx = \sin x$$
 
-**Step 5：得出最终结果**
+**Step 5: Get the final result**
 
-$$\int x \sin x \, dx = -x \cos x + \sin x + C$$
+ $$\int x \sin x \, dx = -x \cos x + \sin x + C$$
 
-其中 $C$ 为积分常数。
+Where $C$ is the integration constant.
 
-## 验算
+## Check calculation
 
-**验算方法 1：对结果求导验证**
+**Verification method 1: Derivative verification of the results**
 
-对 $F(x) = -x \cos x + \sin x + C$ 求导：
+Derivative of $F(x) = -x \cos x + \sin x + C$:
 
-$$\frac{d}{dx}[-x \cos x] = -1 \cdot \cos x + (-x) \cdot (-\sin x) = -\cos x + x \sin x$$
-（使用了乘积法则：$(fg)' = f'g + fg'$）
+ $$\frac{d}{dx}[-x \cos x] = -1 \cdot \cos x + (-x) \cdot (-\sin x) = -\cos x + x \sin x$$
+(Used the product rule: $(fg)' = f'g + fg'$ )
 
-$$\frac{d}{dx}[\sin x] = \cos x$$
+ $$\frac{d}{dx}[\sin x] = \cos x$$
 
-$$\frac{d}{dx}[C] = 0$$
+ $$\frac{d}{dx}[C] = 0$$
 
-求和：
-$$F'(x) = (-\cos x + x \sin x) + \cos x + 0 = x \sin x$$
+Sum:
+ $$F'(x) = (-\cos x + x \sin x) + \cos x + 0 = x \sin x$$
 
-恰好等于被积函数 $x \sin x$ ✓
+Exactly equal to the integrand $x \sin x$ ✓
 
-**验算方法 2：分部积分的反向推导**
+**Check Calculation Method 2: Backward Derivation of Integral by Parts**
 
-从结果出发，分部积分应为：
-$$\int x \sin x \, dx = -x \cos x - \int (-\cos x) \cdot 1 \, dx = -x \cos x + \sin x + C$$
+Starting from the result, the integral of parts should be:
+ $$\int x \sin x \, dx = -x \cos x - \int (-\cos x) \cdot 1 \, dx = -x \cos x + \sin x + C$$
 
-与计算过程一致 ✓
+Consistent with the calculation process ✓
 
-**验算方法 3：数值验证（取 $C=0$）**
+**Verification method 3: Numerical verification (take $C=0$)**
 
-令 $F(x) = -x \cos x + \sin x$：
-- $F(0) = -0 \cdot 1 + 0 = 0$
-- $F(\pi/2) = -\frac{\pi}{2} \cdot 0 + 1 = 1$
+Let $F(x) = -x \cos x + \sin x$ :
+-  $F(0) = -0 \cdot 1 + 0 = 0$
+-  $F(\pi/2) = -\frac{\pi}{2} \cdot 0 + 1 = 1$
 
-而 $\int_0^{\pi/2} x \sin x \, dx = [-x \cos x + \sin x]_0^{\pi/2} = 1 - 0 = 1$ ✓
+And $\int_0^{\pi/2} x \sin x \, dx = [-x \cos x + \sin x]_0^{\pi/2} = 1 - 0 = 1$ ✓
 
-数值 $\int_0^{\pi/2} x \sin x \, dx$ 可近似：在 $[0, \pi/2]$ 上 $x \sin x$ 为正函数，积分 $1$ 合理 ✓
+Numerical $\int_0^{\pi/2} x \sin x \, dx$ can be approximated: $x \sin x$ is a positive function on $[0, \pi/2]$, and the integral $1$ is reasonable ✓
 
-## 最终答案
-$$\int x \sin x \, dx = -x \cos x + \sin x + C$$
+## Final answer
+ $$\int x \sin x \, dx = -x \cos x + \sin x + C$$
 
-## 易错点
-1. **$u$ 和 $dv$ 的选择错误**：若选 $u = \sin x$，$dv = x \, dx$，则 $du = \cos x \, dx$，$v = x^2/2$，积分变为 $\frac{1}{2}x^2\sin x - \frac{1}{2}\int x^2 \cos x \, dx$，反而更复杂——说明 LIATE 规则的重要性
-2. **$v$ 的符号**：$\int \sin x \, dx = -\cos x$（不是 $\cos x$），符号错误会导致验证失败
-3. **忘记积分常数 $C$**：不定积分必须包含积分常数，遗漏会扣分
-4. **乘积法则方向反了**：求导验证时注意 $(uv)' = u'v + uv'$，不要写错顺序
+## Easy to make mistakes
+1. **Wrong selection of $u$ and $dv$**: If you select $u = \sin x$, $dv = x \, dx$, then $du = \cos x \, dx$, $v = x^2/2$, the points become $\frac{1}{2}x^2\sin x - \frac{1}{2}\int x^2 \cos x \, dx$, which is more complicated - indicating the importance of the LIATE rule
+2. **Symbol of $v$**: $\int \sin x \, dx = -\cos x$ (not $\cos x$). Symbol error will cause verification to fail.
+3. **Forgot the integral constant $C$ **: Indefinite integral must contain the integral constant, points will be deducted if omitted
+4. **The direction of the product rule is reversed**: Pay attention to $(uv)' = u'v + uv'$ when deriving and verifying, and do not write in the wrong order.
diff --git a/examples/linear_algebra.md b/examples/linear_algebra.md
index 4caa1a4..8738507 100644
--- a/examples/linear_algebra.md
+++ b/examples/linear_algebra.md
@@ -1,112 +1,112 @@
-# 矩阵的特征值与对角化
+# Eigenvalues ​​and diagonalization of matrices
 
-## 用户输入
-求矩阵 $A = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$ 的特征值、特征向量，并求可逆矩阵 $P$ 和对角矩阵 $D$ 使得 $A = PDP^{-1}$。
+## User input
+Find the eigenvalues ​​and eigenvectors of matrix $A = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$, and find the invertible matrix $P$ and the diagonal matrix $D$ such that $A = PDP^{-1}$.
 
-## Skill 分类
-线性代数
+## Skill Category
+linear algebra
 
-## 题意解析
-- **已知条件**：$2 \times 2$ 实对称矩阵 $A = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$。
-- **求解目标**：(1) 特征值 $\lambda_1, \lambda_2$；(2) 对应的特征向量 $\mathbf{v}_1, \mathbf{v}_2$；(3) 对角化 $A = PDP^{-1}$。
-- **关键性质**：$A$ 是实对称矩阵，因此特征值均为实数，特征向量正交，且一定可对角化。
-- **隐含条件**：$P$ 的列就是特征向量，$D$ 的对角线是特征值（顺序与 $P$ 中的特征向量对应）。
+## Question meaning analysis
+- **Known conditions**: $2 \times 2$ real symmetric matrix $A = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$.
+- **Solution objectives**: (1) eigenvalue $\lambda_1, \lambda_2$; (2) corresponding eigenvector $\mathbf{v}_1, \mathbf{v}_2$; (3) diagonalization $A = PDP^{-1}$.
+- **Key Properties**: $A$ is a real symmetric matrix, so the eigenvalues ​​are all real numbers, the eigenvectors are orthogonal, and they must be diagonalizable.
+- **Implicit condition**: The columns of $P$ are the eigenvectors, and the diagonals of $D$ are the eigenvalues ​​(the order corresponds to the eigenvectors in $P$).
 
-## 方法选择
-**首选方法**：求解特征方程 $\det(A - \lambda I) = 0$ 得到特征值，再对每个 $\lambda$ 解齐次方程组 $(A - \lambda I)\mathbf{v} = \mathbf{0}$ 得到特征向量。
+## Method selection
+**Preferred method**: Solve the characteristic equation $\det(A - \lambda I) = 0$ to get the eigenvalues, and then solve the homogeneous system of equations $(A - \lambda I)\mathbf{v} = \mathbf{0}$ for each $\lambda$ to get the eigenvectors.
 
-**备选方法**：利用对称矩阵的谱定理进行正交对角化。对于 $2 \times 2$ 矩阵，也可以直接用迹和行列式求特征值（$\lambda_1 + \lambda_2 = \text{tr}(A) = 6$，$\lambda_1 \lambda_2 = \det(A) = 8$）。
+**Alternative method**: Use the spectral theorem for symmetric matrices to perform orthogonal diagonalization. For the $2 \times 2$ matrix, you can also directly use the trace and determinant to find the eigenvalues ​​($\lambda_1 + \lambda_2 = \text{tr}(A) = 6$, $\lambda_1 \lambda_2 = \det(A) = 8$).
 
-## 解题过程
+## Problem solving process
 
-### 第一步：求特征值
+### Step 1: Find eigenvalues
 
-特征多项式：
+Characteristic polynomial:
 
-$$\det(A - \lambda I) = \begin{vmatrix} 3 - \lambda & 1 \\ 1 & 3 - \lambda \end{vmatrix}$$
+ $$\det(A - \lambda I) = \begin{vmatrix} 3 - \lambda & 1 \\ 1 & 3 - \lambda \end{vmatrix}$$
 
-$$= (3 - \lambda)^2 - 1 = \lambda^2 - 6\lambda + 8 = (\lambda - 4)(\lambda - 2)$$
+ $$= (3 - \lambda)^2 - 1 = \lambda^2 - 6\lambda + 8 = (\lambda - 4)(\lambda - 2)$$
 
-$$(\lambda - 4)(\lambda - 2) = 0$$
+ $$(\lambda - 4)(\lambda - 2) = 0$$
 
-特征值：
+Eigenvalues:
 
-$$\lambda_1 = 2,\quad \lambda_2 = 4$$
+ $$\lambda_1 = 2,\quad \lambda_2 = 4$$
 
-### 第二步：求特征向量
+### Step 2: Find the feature vector
 
-**对 $\lambda_1 = 2$**，解 $(A - 2I)\mathbf{v} = \mathbf{0}$：
+**For $\lambda_1 = 2$ **, solve for $(A - 2I)\mathbf{v} = \mathbf{0}$:
 
-$$\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$
+ $$\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$
 
-即 $x + y = 0$，取 $x = 1$ 则 $y = -1$。
+That is $x + y = 0$, take $x = 1$ then $y = -1$.
 
-$$\mathbf{v}_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$$
+ $$\mathbf{v}_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$$
 
-**对 $\lambda_2 = 4$**，解 $(A - 4I)\mathbf{v} = \mathbf{0}$：
+**For $\lambda_2 = 4$**, solve $(A - 4I)\mathbf{v} = \mathbf{0}$:
 
-$$\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$
+ $$\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$
 
-即 $-x + y = 0$，取 $x = 1$ 则 $y = 1$。
+That is $-x + y = 0$, take $x = 1$ then $y = 1$.
 
-$$\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$
+ $$\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$
 
-### 第三步：构造 $P$ 和 $D$
+### Step 3: Construct $P$ and $D$
 
-$$P = \begin{pmatrix} \mathbf{v}_1 & \mathbf{v}_2 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$$
+ $$P = \begin{pmatrix} \mathbf{v}_1 & \mathbf{v}_2 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$$
 
-$$D = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}$$
+ $$D = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}$$
 
-### 第四步：验证 $A = PDP^{-1}$
+### Step 4: Verify $A = PDP^{-1}$
 
-求 $P^{-1}$：
+Asking for $P^{-1}$:
 
-$$\det(P) = 1 \cdot 1 - 1 \cdot (-1) = 2$$
+ $$\det(P) = 1 \cdot 1 - 1 \cdot (-1) = 2$$
 
-$$P^{-1} = \frac{1}{2} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$$
+ $$P^{-1} = \frac{1}{2} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$$
 
-验证：
+verify:
 
-$$PDP^{-1} = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$$
+ $$PDP^{-1} = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$$
 
-$$= \frac{1}{2} \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 2 & -2 \\ 4 & 4 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 6 & 2 \\ 2 & 6 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} = A$$
+ $$= \frac{1}{2} \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 2 & -2 \\ 4 & 4 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 6 & 2 \\ 2 & 6 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} = A$$
 
 ✓
 
-## 验算
+## Check calculation
 
-### 方法一：验证 $A\mathbf{v}_i = \lambda_i \mathbf{v}_i$
+### Method 1: Verify $A\mathbf{v}_i = \lambda_i \mathbf{v}_i$
 
-$$A\mathbf{v}_1 = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 - 1 \\ 1 - 3 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \lambda_1 \mathbf{v}_1$$
+ $$A\mathbf{v}_1 = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 - 1 \\ 1 - 3 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \lambda_1 \mathbf{v}_1$$
 
-$$A\mathbf{v}_2 = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 + 1 \\ 1 + 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \end{pmatrix} = 4 \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \lambda_2 \mathbf{v}_2$$
+ $$A\mathbf{v}_2 = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 + 1 \\ 1 + 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \end{pmatrix} = 4 \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \lambda_2 \mathbf{v}_2$$
 
 ✓
 
-### 方法二：迹与行列式验证
+### Method 2: Trace and determinant verification
 
-$$\text{tr}(A) = 3 + 3 = 6 = 2 + 4 = \lambda_1 + \lambda_2$$
+ $$\text{tr}(A) = 3 + 3 = 6 = 2 + 4 = \lambda_1 + \lambda_2$$
 
-$$\det(A) = 3 \cdot 3 - 1 \cdot 1 = 8 = 2 \cdot 4 = \lambda_1 \lambda_2$$
+ $$\det(A) = 3 \cdot 3 - 1 \cdot 1 = 8 = 2 \cdot 4 = \lambda_1 \lambda_2$$
 
 ✓
 
-### 方法三：特征向量的正交性
+### Method 3: Orthogonality of eigenvectors
 
-$$\mathbf{v}_1 \cdot \mathbf{v}_2 = 1 \cdot 1 + (-1) \cdot 1 = 0$$
+ $$\mathbf{v}_1 \cdot \mathbf{v}_2 = 1 \cdot 1 + (-1) \cdot 1 = 0$$
 
-特征向量正交，符合实对称矩阵的性质。✓
+The eigenvectors are orthogonal and conform to the properties of real symmetric matrices. ✓
 
-## 最终答案
+## Final answer
 
-$$\boxed{\lambda_1 = 2,\; \mathbf{v}_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}}$$
+ $$\boxed{\lambda_1 = 2,\; \mathbf{v}_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}}$$
 
-$$\boxed{\lambda_2 = 4,\; \mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}}$$
+ $$\boxed{\lambda_2 = 4,\; \mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}}$$
 
-$$\boxed{P = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix},\quad D = \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix},\quad A = PDP^{-1}}$$
+ $$\boxed{P = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix},\quad D = \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix},\quad A = PDP^{-1}}$$
 
-## 易错点
-1. **特征方程的符号**：$\det(A - \lambda I) = 0$ 还是 $\det(\lambda I - A) = 0$？前者展开后特征多项式为 $(-1)^n \det(\lambda I - A)$，两者对求特征值等价，但注意符号细节。
-2. **特征向量的顺序**：$P$ 中特征向量的顺序必须与 $D$ 对角线上特征值的顺序一致。若交换 $\mathbf{v}_1$ 和 $\mathbf{v}_2$ 的位置，$D$ 的对角线也应变为 $\operatorname{diag}(4, 2)$。
-3. **特征向量缩放**：特征向量不唯一（可乘以任意非零标量），但 $PDP^{-1}$ 的结果不变。
-4. **检查可对角化前提**：不是所有矩阵都可对角化（如 Jordan 块）。实对称矩阵天然可对角化，但一般矩阵需要检查几何重数 = 代数重数。
+## Easy to make mistakes
+1. **Symbol of characteristic equation**: $\det(A - \lambda I) = 0$ or $\det(\lambda I - A) = 0$? After the expansion of the former, the characteristic polynomial is $(-1)^n \det(\lambda I - A)$. The two are equivalent to find the eigenvalues, but pay attention to the symbolic details.
+2. **Order of eigenvectors**: The order of eigenvectors in $P$ must be consistent with the order of eigenvalues ​​on the diagonal of $D$. If the positions of $\mathbf{v}_1$ and $\mathbf{v}_2$ are swapped, the diagonal of $D$ should also become $\operatorname{diag}(4, 2)$ .
+3. **Eigenvector Scaling**: The eigenvector is not unique (can be multiplied by any non-zero scalar), but the result of $PDP^{-1}$ remains unchanged.
+4. **Check diagonalizability premise**: Not all matrices are diagonalizable (such as Jordan blocks). Real symmetric matrices are naturally diagonalizable, but general matrices need to check geometric multiplicity = algebraic multiplicity.
diff --git a/examples/modeling.md b/examples/modeling.md
index 2c8efd2..49627e2 100644
--- a/examples/modeling.md
+++ b/examples/modeling.md
@@ -1,142 +1,142 @@
-# 数学建模：河边围栏最大面积问题
+#Mathematical modeling: Maximum area problem of riverside fence
 
-## 用户输入
-一位农民有 100 米篱笆，想靠河围一个矩形区域（靠河的一边不需要篱笆）。求使面积最大的矩形尺寸。
+## User input
+A farmer has a 100 meter fence and wants to enclose a rectangular area close to the river (no fence is needed on the river side). Find the size of the rectangle that maximizes the area.
 
-## Skill 分类
-数学建模 / 优化问题
+## Skill Category
+Mathematical modeling/optimization problems
 
-## 题意解析
+## Question meaning analysis
 
-**问题翻译**：
-- **资源**：100 米篱笆
-- **区域形状**：矩形
-- **约束**：矩形的一条边临河，该边不需篱笆（即只有三条边需要建篱笆）
-- **目标**：最大化矩形面积
+**Question translation**:
+- **RESOURCE**: 100m Fence
+- **Area Shape**: Rectangle
+- **Constraint**: One side of the rectangle faces the river, and no fence is needed on this side (that is, only three sides need to build fences)
+- **Goal**: Maximize the area of ​​the rectangle
 
-**变量定义**：
-- 设 $x$：垂直于河岸的边长（宽度，共两条）
-- 设 $y$：平行于河岸的边长（长度，一条，靠河不需篱笆）
+**Variable definition**:
+- Let $x$: the length of the side perpendicular to the river bank (width, two in total)
+- Let $y$: the length of the side parallel to the river bank (length, one piece, no fence is needed near the river)
 
-**约束条件**：
-- 篱笆总长：$2x + y = 100$
-- 物理意义：$x > 0$，$y > 0$（边长为正）
-- 隐含定义域：$x \in (0, 50)$（因为 $y = 100 - 2x > 0$ 要求 $x < 50$；$x > 0$ 是必要条件）
+**Constraints**:
+- Total fence length: $2x + y = 100$
+- Physical meaning: $x > 0$, $y > 0$ (side length is positive)
+- Implied domain: $x \in (0, 50)$ (because $y = 100 - 2x > 0$ requires $x < 50$; $x > 0$ is a necessary condition)
 
-**目标函数**：面积 $A = xy$
+**Objective function**: Area $A = xy$
 
-**解的约束**：最大面积时 $x$ 应为定义域的内点（非边界）
+**Solution constraints**: When the maximum area is, $x$ should be the interior point of the domain (not the boundary)
 
-## 方法选择
+## Method selection
 
-**主方法：单变量优化（消元 + 求导）**
+**Main method: single variable optimization (elimination + derivation)**
 
-利用约束 $y = 100 - 2x$，将二元优化问题化为单变量优化问题，然后使用导数求极值。
+Use the constraint $y = 100 - 2x$ to convert the binary optimization problem into a univariate optimization problem, and then use derivatives to find the extreme value.
 
-选择理由：该问题只有一个等式约束，消元后得到一元二次函数，求导即可。这是典型的建模-消元-求极值三步流程。
+Reason for selection: This problem has only one equality constraint. After elimination, a quadratic function of one variable can be obtained and just need to be differentiated. This is a typical three-step process of modeling-elimination-finding extreme values.
 
-**备选方法**：
-1. **配方法（完全平方）**：将 $A(x) = 100x - 2x^2$ 配方，直接读出最大值
-2. **不等式法（AM-GM）**：利用 $2x + y = 100$ 约束下的 $A = xy$ 最大化
-3. **Lagrange 乘数法**：二元约束优化的一般方法（此处稍显过度）
+**Alternative Method**:
+1. **Preparation method (complete square)**: Use the $A(x) = 100x - 2x^2$ formula and directly read the maximum value
+2. **Inequality method (AM-GM)**: Maximize $A = xy$ using $2x + y = 100$ constraints
+3. **Lagrange multiplier method**: a general method for binary constrained optimization (a little excessive here)
 
-本文展示消元求导法（主方法）和配方法（验算），并进行全面验证。
+This article demonstrates the elimination derivation method (main method) and combination method (check calculation), and conducts comprehensive verification.
 
-## 解题过程
+## Problem solving process
 
-### Step 1：建立数学模型
+### Step 1: Establish mathematical model
 
-**模型要素**：
+**Model elements**:
 
-| 要素 | 内容 |
+| Elements | Content |
 |------|------|
-| 决策变量 | $x$（宽度），$y$（长度） |
-| 目标函数 | $A = x \cdot y$（最大化） |
-| 约束条件 | $2x + y = 100$，$x > 0$，$y > 0$ |
+| Decision variables | $x$ (width), $y$ (length) |
+| Objective function | $A = x \cdot y$ (maximize) |
+| Constraints | $2x + y = 100$, $x > 0$, $y > 0$ |
 
-**模型类型**：线性约束下的二次目标函数最大化问题（连续优化）。
+**Model type**: Quadratic objective function maximization problem under linear constraints (continuous optimization).
 
-### Step 2：消元化为一元问题
+### Step 2: Eliminate into a univariate problem
 
-由约束 $2x + y = 100$ 得 $y = 100 - 2x$。
+From the constraint $2x + y = 100$, we get $y = 100 - 2x$.
 
-代入目标函数：
-$$A(x) = x \cdot (100 - 2x) = 100x - 2x^2$$
+Substitute into the objective function:
+ $$A(x) = x \cdot (100 - 2x) = 100x - 2x^2$$
 
-定义域：$x \in (0, 50)$（因为 $y = 100 - 2x > 0$）。
+Domain: $x \in (0, 50)$ (because $y = 100 - 2x > 0$ ).
 
-问题转化为：求一元函数 $A(x) = 100x - 2x^2$ 在 $(0, 50)$ 上的最大值。
+The problem is transformed into: find the maximum value of the unary function $A(x) = 100x - 2x^2$ on $(0, 50)$.
 
-### Step 3：求导找临界点
+### Step 3: Find the critical point by derivation
 
-$$A'(x) = 100 - 4x$$
+ $$A'(x) = 100 - 4x$$
 
-令 $A'(x) = 0$：
-$$100 - 4x = 0 \implies x = 25$$
+Let $A'(x) = 0$ :
+ $$100 - 4x = 0 \implies x = 25$$
 
-在定义域 $(0, 50)$ 内，只有一个临界点 $x = 25$。
+Within the domain $(0, 50)$, there is only one critical point $x = 25$.
 
-### Step 4：判定极值类型
+### Step 4: Determine the extreme value type
 
-**二阶导数检验**：
-$$A''(x) = -4 < 0$$
+**Second Derivative Test**:
+ $$A''(x) = -4 < 0$$
 
-$A''(25) = -4 < 0$，故 $x = 25$ 是极大值点。由于函数在 $(0, 50)$ 上只有一个临界点且二阶导数恒为负，这是全局最大值。
+$A''(25) = -4 < 0$, so $x = 25$ is the maximum value point. Since the function has only one critical point on $(0, 50)$ and the second derivative is always negative, this is the global maximum.
 
-### Step 5：计算最优解
+### Step 5: Calculate the optimal solution
 
-$$x = 25 \text{ m}$$
-$$y = 100 - 2 \times 25 = 50 \text{ m}$$
-$$A_{\max} = 25 \times 50 = 1250 \text{ m}^2$$
+ $$x = 25 \text{ m}$$
+ $$y = 100 - 2 \times 25 = 50 \text{ m}$$
+ $$A_{\max} = 25 \times 50 = 1250 \text{ m}^2$$
 
-### Step 6：边界行为确认
+### Step 6: Boundary behavior confirmation
 
-- 当 $x \to 0^+$：$y \to 100$，$A \to 0$（退化为一维线段，面积为零）
-- 当 $x \to 50^-$：$y \to 0$，$A \to 0$（退化为二维线段，面积为零）
+- When $x \to 0^+$ : $y \to 100$ , $A \to 0$ (degenerate into a one-dimensional line segment with zero area)
+- When $x \to 50^-$ : $y \to 0$ , $A \to 0$ (degenerated into a two-dimensional line segment with zero area)
 
-边界值均为 0，小于内点 $x=25$ 处的 $1250$，确认这是全局最大值。
+The boundary values ​​are all 0, which is less than $1250$ at the interior point $x=25$, confirming that this is the global maximum.
 
-## 验算
+## Check calculation
 
-### 验算方法 1：配方法（代数验证）
+### Verification method 1: Matching method (algebraic verification)
 
-$$A(x) = 100x - 2x^2 = -2(x^2 - 50x)$$
-$$= -2\left[(x^2 - 50x + 625) - 625\right] = -2(x - 25)^2 + 1250$$
+ $$A(x) = 100x - 2x^2 = -2(x^2 - 50x)$$
+ $$= -2\left[(x^2 - 50x + 625) - 625\right] = -2(x - 25)^2 + 1250$$
 
-由于 $-2(x - 25)^2 \leq 0$ 且当 $x = 25$ 时取 0，故：
-$$A_{\max} = 1250, \quad \text{在 } x = 25 \text{ 时取得}$$
+Since $-2(x - 25)^2 \leq 0$ takes 0 when $x = 25$, therefore:
+$$A_{\max} = 1250, \quad \text{obtained}$$ when } x = 25 \text{
 
-与求导结果一致 ✓
+Consistent with the derivation result ✓
 
-### 验算方法 2：邻域验证
+### Verification method 2: Neighborhood verification
 
-| $x$ | $y = 100 - 2x$ | $A = xy$ | 说明 |
+| $x$ | $y = 100 - 2x$ | $A = xy$ | Description |
 |-----|----------------|----------|------|
-| 24 | 52 | 1248 | 小于 1250 |
-| 25 | 50 | 1250 | 最大值 |
-| 26 | 48 | 1248 | 小于 1250 |
+| 24 | 52 | 1248 | Less than 1250 |
+| 25 | 50 | 1250 | Maximum value |
+| 26 | 48 | 1248 | Less than 1250 |
 
-验证了 $x=25$ 确实比邻近点给出更大的面积 ✓
+Verified that $x=25$ indeed gives a larger area than neighboring points ✓
 
-### 验算方法 3：量纲检查
+### Calculation method 3: Dimension check
 
-- 篱笆长度：$2x + y = 2 \times 25 + 50 = 100$ m ✓
-- 面积单位：$A = 25 \text{ m} \times 50 \text{ m} = 1250 \text{ m}^2$ ✓
-- 变量量纲一致：$x$ 和 $y$ 的单位均为米，$A$ 的单位为平方米 ✓
+- Fence length: $2x + y = 2 \times 25 + 50 = 100$ m ✓
+- Area unit: $A = 25 \text{ m} \times 50 \text{ m} = 1250 \text{ m}^2$ ✓
+- The variables have the same dimensions: $x$ and $y$ are both in meters, and $A$ is in square meters ✓
 
-### 验算方法 4：等周问题的类比验证
+### Verification method 4: Analogous verification of isoperiodic problems
 
-对于固定周长的矩形面积最大化：
-- 封闭矩形（四面篱笆）：周长 $P = 2x + 2y$ 固定时，$x = y$ 时面积最大（正方形）
-- 半封闭矩形（三面篱笆，一面靠河）：$2x + y = 100$，得出 $y = 2x$ 时面积最大
+Maximizing the area of ​​a rectangle with fixed perimeter:
+- Closed rectangle (four-sided fence): When the perimeter is fixed at $P = 2x + 2y$, the area is the largest (square) when $x = y$
+- Semi-closed rectangle (with fences on three sides and one side by the river): $2x + y = 100$, the largest area is obtained when $y = 2x$ is obtained
 
-这与直觉一致：靠河面"免费"，所以应该让靠河面（$y$）更宽——两边各建一半宽度的篱笆用于形成两翼。结果 $y:x = 50:25 = 2:1$ 符合此直觉 ✓
+This is consistent with intuition: the river front is "free", so the river front ( $y$ ) should be made wider - with half-width fences on each side to form the wings. The result $y:x = 50:25 = 2:1$ fits this intuition ✓
 
-### 验算方法 5：枚举法粗略验证
+### Verification method 5: Rough verification by enumeration method
 
-用步长 $\Delta x = 5$ 枚举几个值：
+Enumerate several values ​​with step size $\Delta x = 5$:
 
-| $x$ | $y$ | $A$ |
+|  $x$  |  $y$  |  $A$  |
 |-----|-----|-----|
 | 5 | 90 | 450 |
 | 10 | 80 | 800 |
@@ -148,19 +148,19 @@ $$A_{\max} = 1250, \quad \text{在 } x = 25 \text{ 时取得}$$
 | 40 | 20 | 800 |
 | 45 | 10 | 450 |
 
-$A$ 在 $x=25$ 处最大 ✓
+$A$ is maximum at $x=25$ ✓
 
-## 最终答案
+## Final answer
 
-最优方案：垂直于河岸的宽度 $x = 25$ 米，平行于河岸的长度 $y = 50$ 米，最大面积为 $1250$ 平方米。
+Optimal solution: width perpendicular to the river bank $x = 25$ meters, length parallel to the river bank $y = 50$ meters, maximum area $1250$ square meters.
 
-数学表达：
-$$x^* = 25 \text{ m}, \quad y^* = 50 \text{ m}, \quad A_{\max} = 1250 \text{ m}^2$$
+Mathematical expression:
+ $$x^* = 25 \text{ m}, \quad y^* = 50 \text{ m}, \quad A_{\max} = 1250 \text{ m}^2$$
 
-## 易错点
-1. **约束等式写错**：靠河面不需篱笆，故篱笆总长为 $2x + y = 100$（不是 $2x + 2y = 100$），这是最容易掉入的陷阱
-2. **忽略物理约束**：$x$ 和 $y$ 必须为正，导致定义域为 $(0, 50)$ 而非 $\mathbb{R}$。$x=0$ 和 $x=50$ 虽然在数学上对应退化情况，但物理意义上不可接受
-3. **极值与最值的区分**：一阶导数为零给出的是极值点，需要二阶导数检验 + 边界检查才能确认是全局最大值
-4. **求导错误**：$A(x) = 100x - 2x^2$ 的导数是 $100 - 4x$（注意 $2x^2$ 的导数是 $4x$，不是 $2x$）
-5. **题目条件的过度推广**：该结论（$y = 2x$）仅在矩形且一面靠河时成立。如果形状不是矩形，或靠河面不止一面，结论会完全不同
-6. **优化问题的完整性**：除了给出最优解，还应给出最优值（最大面积），这是数学建模题目的常见扣分点
+## Easy to make mistakes
+1. **The constraint equation is written incorrectly**: There is no need for a fence near the river, so the total length of the fence is $2x + y = 100$ (not $2x + 2y = 100$). This is the easiest trap to fall into.
+2. **Ignore physical constraints**: $x$ and $y$ must be positive, causing the domain to be $(0, 50)$ instead of $\mathbb{R}$. Although $x=0$ and $x=50$ correspond to degenerate situations mathematically, they are not acceptable in a physical sense
+3. **Distinguishing between extreme values ​​and maximum values**: The first-order derivative of zero gives the extreme point, and second-order derivative test + boundary check is required to confirm that it is the global maximum.
+4. **derivative error**: The derivative of $A(x) = 100x - 2x^2$ is $100 - 4x$ (note that the derivative of $2x^2$ is $4x$, not $2x$)
+5. **Excessive generalization of question conditions**: This conclusion ($y = 2x$) is only true when it is a rectangle and one side is close to the river. If the shape is not rectangular, or if it is close to the river on more than one side, the conclusion will be completely different.
+6. **Completeness of the optimization problem**: In addition to giving the optimal solution, the optimal value (maximum area) should also be given. This is a common deduction point for mathematical modeling questions.
diff --git a/examples/multivariable_calculus.md b/examples/multivariable_calculus.md
index 9127416..67593e5 100644
--- a/examples/multivariable_calculus.md
+++ b/examples/multivariable_calculus.md
@@ -1,41 +1,41 @@
-# 多元函数临界点与分类
+# Critical points and classification of multivariate functions
 
-## 用户输入
-求函数 $f(x, y) = x^3 + y^3 - 3xy$ 的所有临界点，并判断每个临界点的类型（极大值点、极小值点或鞍点）。
+## User input
+Find all critical points of function $f(x, y) = x^3 + y^3 - 3xy$ and determine the type of each critical point (maximum point, minimum point or saddle point).
 
-## Skill 分类
-多元微积分
+## Skill Category
+multivariable calculus
 
-## 题意解析
-- **函数形式**：$f(x, y) = x^3 + y^3 - 3xy$，定义域为 $\mathbb{R}^2$。
-- **求解目标**：(1) 所有临界点（即 $\nabla f = \mathbf{0}$ 的点）；(2) 分类各临界点。
-- **关键定理**：二阶导数检验（Hessian 行列式判别法）。对临界点 $(x_0, y_0)$，计算
+## Question meaning analysis
+- **Function form**: $f(x, y) = x^3 + y^3 - 3xy$, the domain is $\mathbb{R}^2$.
+- **Solution objectives**: (1) All critical points (that is, the points of $\nabla f = \mathbf{0}$); (2) Classify each critical point.
+- **Key theorem**: Second derivative test (Hessian determinant method). For the critical point $(x_0, y_0)$, calculate
 
-  $$D = f_{xx}(x_0, y_0) \cdot f_{yy}(x_0, y_0) - [f_{xy}(x_0, y_0)]^2$$
+   $$D = f_{xx}(x_0, y_0) \cdot f_{yy}(x_0, y_0) - [f_{xy}(x_0, y_0)]^2$$
 
-  - 若 $D > 0$ 且 $f_{xx} > 0$：极小值点
-  - 若 $D > 0$ 且 $f_{xx} < 0$：极大值点
-  - 若 $D < 0$：鞍点
-  - 若 $D = 0$：无法判定
+- If $D > 0$ and $f_{xx} > 0$ : Minimum point
+- If $D > 0$ and $f_{xx} < 0$: maximum value point
+- if $D < 0$ : saddle point
+- If $D = 0$: Unable to determine
 
-- **隐含条件**：$f$ 是 $C^\infty$ 函数（无限阶可导），所有偏导数存在且连续，二阶导数检验条件满足。
+- **Implicit condition**: $f$ is a $C^\infty$ function (infinitely differentiable), all partial derivatives exist and are continuous, and the second-order derivative test conditions are met.
 
-## 方法选择
-**首选方法**：梯度为零求临界点 + Hessian 矩阵行列式分类。这是多元函数极值问题的标准流程。
+## Method selection
+**Preferred method**: Find the critical point with zero gradient + Hessian matrix determinant classification. This is the standard procedure for the extreme value problem of multivariate functions.
 
-**备选方法**：Taylor 展开到二阶项判断二次型的正定性。本质与 Hessian 判别等价。
+**Alternative method**: Taylor expansion to second-order terms determines the positive definiteness of the quadratic form. Essentially equivalent to the Hessian discriminant.
 
-## 解题过程
+## Problem solving process
 
-### 第一步：求一阶偏导数并令其为零
+### Step 1: Find the first-order partial derivative and make it zero
 
-$$f_x = \frac{\partial f}{\partial x} = 3x^2 - 3y$$
+ $$f_x = \frac{\partial f}{\partial x} = 3x^2 - 3y$$
 
-$$f_y = \frac{\partial f}{\partial y} = 3y^2 - 3x$$
+ $$f_y = \frac{\partial f}{\partial y} = 3y^2 - 3x$$
 
-令 $\nabla f = \mathbf{0}$：
+Let $\nabla f = \mathbf{0}$ :
 
-$$\begin{cases}
+ $$\begin{cases}
 3x^2 - 3y = 0 \\
 3y^2 - 3x = 0
 \end{cases}
@@ -45,82 +45,82 @@ x^2 = y \\
 y^2 = x
 \end{cases}$$
 
-### 第二步：求解方程组
+### Step 2: Solve the system of equations
 
-将 $y = x^2$ 代入 $y^2 = x$：
+Substitute $y = x^2$ into $y^2 = x$ :
 
-$$(x^2)^2 = x \quad \Rightarrow \quad x^4 = x \quad \Rightarrow \quad x^4 - x = 0 \quad \Rightarrow \quad x(x^3 - 1) = 0$$
+ $$(x^2)^2 = x \quad \Rightarrow \quad x^4 = x \quad \Rightarrow \quad x^4 - x = 0 \quad \Rightarrow \quad x(x^3 - 1) = 0$$
 
-$$x(x - 1)(x^2 + x + 1) = 0$$
+ $$x(x - 1)(x^2 + x + 1) = 0$$
 
-实数解：$x = 0$ 或 $x = 1$（$x^2 + x + 1 = 0$ 无实根，判别式 $1 - 4 = -3 < 0$）。
+Real solution: $x = 0$ or $x = 1$ ($x^2 + x + 1 = 0$ has no real roots, discriminant $1 - 4 = -3 < 0$).
 
-- 当 $x = 0$：$y = x^2 = 0$，临界点为 $(0, 0)$。
-- 当 $x = 1$：$y = x^2 = 1$，临界点为 $(1, 1)$。
+- When $x = 0$ : $y = x^2 = 0$ , the critical point is $(0, 0)$ .
+- When $x = 1$ : $y = x^2 = 1$ , the critical point is $(1, 1)$ .
 
-### 第三步：求二阶偏导数
+### Step 3: Find the second-order partial derivative
 
-$$f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6x$$
+ $$f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6x$$
 
-$$f_{yy} = \frac{\partial^2 f}{\partial y^2} = 6y$$
+ $$f_{yy} = \frac{\partial^2 f}{\partial y^2} = 6y$$
 
-$$f_{xy} = f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = -3$$
+ $$f_{xy} = f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = -3$$
 
-### 第四步：对每个临界点计算 Hessian 行列式
+### Step 4: Calculate the Hessian determinant for each critical point
 
-**临界点 $(0, 0)$**：
+**Critical point $(0, 0)$ **:
 
-$$f_{xx}(0,0) = 0, \quad f_{yy}(0,0) = 0, \quad f_{xy}(0,0) = -3$$
+ $$f_{xx}(0,0) = 0, \quad f_{yy}(0,0) = 0, \quad f_{xy}(0,0) = -3$$
 
-$$D(0,0) = 0 \cdot 0 - (-3)^2 = 0 - 9 = -9 < 0$$
+ $$D(0,0) = 0 \cdot 0 - (-3)^2 = 0 - 9 = -9 < 0$$
 
-$D < 0$，所以 $(0, 0)$ 是**鞍点**。
+$D < 0$ , so $(0, 0)$ is a saddle point.
 
-**临界点 $(1, 1)$**：
+**Critical point $(1, 1)$ **:
 
-$$f_{xx}(1,1) = 6, \quad f_{yy}(1,1) = 6, \quad f_{xy}(1,1) = -3$$
+ $$f_{xx}(1,1) = 6, \quad f_{yy}(1,1) = 6, \quad f_{xy}(1,1) = -3$$
 
-$$D(1,1) = 6 \cdot 6 - (-3)^2 = 36 - 9 = 27 > 0$$
+ $$D(1,1) = 6 \cdot 6 - (-3)^2 = 36 - 9 = 27 > 0$$
 
-$D > 0$ 且 $f_{xx}(1,1) = 6 > 0$，所以 $(1, 1)$ 是**极小值点**。
+$D > 0$ and $f_{xx}(1,1) = 6 > 0$ , so $(1, 1)$ is the minimum point.
 
-极小值为：
+The minimum value is:
 
-$$f(1, 1) = 1^3 + 1^3 - 3 \cdot 1 \cdot 1 = 1 + 1 - 3 = -1$$
+ $$f(1, 1) = 1^3 + 1^3 - 3 \cdot 1 \cdot 1 = 1 + 1 - 3 = -1$$
 
-## 验算
+## Check calculation
 
-### 方法一：验证梯度在临界点处为零
+### Method 1: Verify that the gradient is zero at the critical point
 
-$(0, 0)$ 点：$f_x(0,0) = 3(0)^2 - 3(0) = 0$，$f_y(0,0) = 3(0)^2 - 3(0) = 0$。✓
+$(0, 0)$ points: $f_x(0,0) = 3(0)^2 - 3(0) = 0$ , $f_y(0,0) = 3(0)^2 - 3(0) = 0$ . ✓
 
-$(1, 1)$ 点：$f_x(1,1) = 3(1)^2 - 3(1) = 0$，$f_y(1,1) = 3(1)^2 - 3(1) = 0$。✓
+$(1, 1)$ points: $f_x(1,1) = 3(1)^2 - 3(1) = 0$ , $f_y(1,1) = 3(1)^2 - 3(1) = 0$ . ✓
 
-### 方法二：函数值比较法
+### Method 2: Function value comparison method
 
-在 $(0, 0)$ 附近取测试点：
+Take a test point near $(0, 0)$:
 
-- 沿 $y = x$ 方向：$f(t, t) = 2t^3 - 3t^2$。当 $t$ 很小时，$f(t, t) \approx -3t^2 < 0 = f(0,0)$（$t \neq 0$）。
-- 沿 $y = -x$ 方向：$f(t, -t) = t^3 + (-t)^3 - 3t(-t) = 0 + 3t^2 = 3t^2 > 0 = f(0,0)$（$t \neq 0$）。
+- Along direction $y = x$: $f(t, t) = 2t^3 - 3t^2$. When $t$ is small, $f(t, t) \approx -3t^2 < 0 = f(0,0)$ ( $t \neq 0$ ).
+- Along the $y = -x$ direction: $f(t, -t) = t^3 + (-t)^3 - 3t(-t) = 0 + 3t^2 = 3t^2 > 0 = f(0,0)$ ( $t \neq 0$ ).
 
-函数值在 $(0,0)$ 附近既有大于 $f(0,0)$ 的也有小于 $f(0,0)$ 的，确认为鞍点。✓
+The function values ​​near $(0,0)$ are both greater than $f(0,0)$ and less than $f(0,0)$, which are confirmed as saddle points. ✓
 
-在 $(1, 1)$ 附近取测试点，二阶 Taylor 展开：
+Take a test point near $(1, 1)$ and perform second-order Taylor expansion:
 
-$$f(1+h,1+k) \approx f(1,1) + \frac{1}{2}(6h^2 + 6k^2 - 6hk) = -1 + \frac{3}{2}((h-k)^2 + h^2 + k^2) > -1$$
+ $$f(1+h,1+k) \approx f(1,1) + \frac{1}{2}(6h^2 + 6k^2 - 6hk) = -1 + \frac{3}{2}((h-k)^2 + h^2 + k^2) > -1$$
 
-对于 $(h,k) \neq (0,0)$，函数值大于 $-1$，确认为极小值点。✓
+For $(h,k) \neq (0,0)$, the function value is greater than $-1$, which is confirmed as a minimum value point. ✓
 
-## 最终答案
+## Final answer
 
-$$\boxed{\text{临界点：}(0, 0) \text{ 和 } (1, 1)}$$
+$$\boxed{\text{Critical points: }(0, 0) \text{ and } (1, 1)}$$
 
-$$\boxed{(0, 0) \text{ 是鞍点}}$$
+$$\boxed{(0, 0) \text{ is a saddle point}}$$
 
-$$\boxed{(1, 1) \text{ 是极小值点，极小值 } f_{\min} = -1}$$
+$$\boxed{(1, 1) \text{ is the minimum value point, minimum value } f_{\min} = -1}$$
 
-## 易错点
-1. **遗漏 $x = 0$ 解**：解 $x^4 = x$ 时不要把 $x = 0$ 约掉（即不能两边同时除以 $x$），应移项写成 $x^4 - x = 0$，因式分解求得全部解。
-2. **Hessian 行列式的符号判断**：$D > 0$ 且 $f_{xx} > 0$ 是极小值；$D > 0$ 且 $f_{xx} < 0$ 是极大值。有些教材使用 $-D$ 的判别，注意正负号方向。
-3. **仅有两个临界点**：别被方程组 $x^2 = y$ 和 $y^2 = x$ 的对称性误导以为有无穷多解。通过代入消元可知只有两个实数解。
-4. **$D = 0$ 的情况**：若 Hessian 行列式为零，二阶导数检验失效，需使用更高阶展开或其他方法判定。本题不涉及此情况。
+## Easy to make mistakes
+1. **Omission of $x = 0$ Solution**: When solving $x^4 = x$, do not eliminate $x = 0$ (that is, you cannot divide both sides by $x$ at the same time). You should move the terms and write it as $x^4 - x = 0$, and factor it to get all the solutions.
+2. **Sign judgment of Hessian determinant**: $D > 0$ and $f_{xx} > 0$ are minimum values; $D > 0$ and $f_{xx} < 0$ are maximum values. Some textbooks use the judgment of $-D$, pay attention to the direction of the positive and negative signs.
+3. **Only Two Critical Points**: Don’t let the symmetry of the system of equations $x^2 = y$ and $y^2 = x$ fool you into thinking there are infinitely many solutions. By substituting elimination, we can see that there are only two real solutions.
+4. ** Case of $D = 0$**: If the Hessian determinant is zero, the second-order derivative test fails, and higher-order expansion or other methods need to be used to determine. This question does not cover this situation.
diff --git a/examples/ode.md b/examples/ode.md
index 75842c6..9c9f2d6 100644
--- a/examples/ode.md
+++ b/examples/ode.md
@@ -1,139 +1,139 @@
-# 常系数线性微分方程的初值问题
+# Initial value problem of linear differential equations with constant coefficients
 
-## 用户输入
-求解微分方程初值问题：
+## User input
+Solve the initial value problem of differential equations:
 
-$$y'' - 3y' + 2y = e^x,\quad y(0) = 0,\; y'(0) = 1$$
+ $$y'' - 3y' + 2y = e^x,\quad y(0) = 0,\; y'(0) = 1$$
 
-## Skill 分类
-常微分方程
+## Skill Category
+ordinary differential equations
 
-## 题意解析
-- **方程类型**：二阶常系数线性非齐次 ODE，形如 $y'' + py' + qy = g(x)$，其中 $p = -3$，$q = 2$，$g(x) = e^x$。
-- **求解目标**：满足初始条件 $y(0) = 0$ 和 $y'(0) = 1$ 的特解。
-- **解的结构**：通解 = 齐次解 $y_h$ + 特解 $y_p$。齐次解含 2 个待定常数，由初始条件确定。
-- **关键观察**：非齐次项 $g(x) = e^x$。需检查 $e^x$ 是否出现在齐次解中（即 $r = 1$ 是否为特征根）。
-- **定义域**：$x \in \mathbb{R}$，整个实数轴。
+## Question meaning analysis
+- **Equation type**: Second-order constant coefficient linear non-homogeneous ODE, in the form of $y'' + py' + qy = g(x)$, where $p = -3$, $q = 2$, $g(x) = e^x$.
+- **Solution Goal**: A special solution that satisfies the initial conditions $y(0) = 0$ and $y'(0) = 1$.
+- **Structure of solution**: General solution = homogeneous solution $y_h$ + special solution $y_p$. The homogeneous solution contains 2 undetermined constants, determined by the initial conditions.
+- **Key Observation**: The non-homogeneous term $g(x) = e^x$ . It is necessary to check whether $e^x$ appears in the homogeneous solution (that is, whether $r = 1$ is a characteristic root).
+- **Domain**: $x \in \mathbb{R}$, the entire real axis.
 
-## 方法选择
-**首选方法**：先求齐次解（特征方程法），再求非齐次特解（待定系数法）。这是处理常系数线性 ODE 的标准化流程。
+## Method selection
+**Preferred method**: First find the homogeneous solution (characteristic equation method), and then find the non-homogeneous special solution (undetermined coefficient method). This is a standardized procedure for handling linear ODEs with constant coefficients.
 
-**备选方法**：变参数法（参数变易法）、拉普拉斯变换法。待定系数法在右端函数为指数、多项式、正弦余弦时最简洁。
+**Alternative methods**: variable parameter method (parameter change method), Laplace transform method. The undetermined coefficient method is the most concise when the right-hand function is exponential, polynomial, sine and cosine.
 
-## 解题过程
+## Problem solving process
 
-### 第一步：求齐次解 $y_h$
+### Step 1: Find the homogeneous solution $y_h$
 
-特征方程：
+Characteristic equation:
 
-$$r^2 - 3r + 2 = 0$$
+ $$r^2 - 3r + 2 = 0$$
 
-$$(r - 1)(r - 2) = 0$$
+ $$(r - 1)(r - 2) = 0$$
 
-特征根：$r_1 = 1$，$r_2 = 2$。
+Characteristic roots: $r_1 = 1$, $r_2 = 2$.
 
-齐次解：
+Homogeneous solution:
 
-$$y_h(x) = C_1 e^{x} + C_2 e^{2x}$$
+ $$y_h(x) = C_1 e^{x} + C_2 e^{2x}$$
 
-### 第二步：求非齐次特解 $y_p$
+### Step 2: Find non-homogeneous special solution $y_p$
 
-非齐次项 $g(x) = e^x$。注意 $r = 1$ 是特征根，因此标准尝试形式 $y_p = Ae^x$ 已经在齐次解中。
+Non-homogeneous term $g(x) = e^x$ . Note that $r = 1$ is a characteristic root, so the standard attempt form $y_p = Ae^x$ is already in the homogeneous solution.
 
-根据待定系数法的修正规则，$e^x$ 对应于 $m = 1$ 重根，尝试形式应乘以 $x^m = x$：
+According to the modified rules of the method of undetermined coefficients, $e^x$ corresponds to the $m = 1$ multiple root, and the tried form should be multiplied by $x^m = x$ :
 
-$$y_p = A x e^x$$
+ $$y_p = A x e^x$$
 
-求导：
+Find the derivative:
 
-$$y_p' = A e^x + A x e^x = A e^x(1 + x)$$
+ $$y_p' = A e^x + A x e^x = A e^x(1 + x)$$
 
-$$y_p'' = A e^x(1 + x) + A e^x = A e^x(2 + x)$$
+ $$y_p'' = A e^x(1 + x) + A e^x = A e^x(2 + x)$$
 
-代入原方程 $y_p'' - 3y_p' + 2y_p = e^x$：
+Substitute into the original equation $y_p'' - 3y_p' + 2y_p = e^x$:
 
-$$A e^x(2 + x) - 3A e^x(1 + x) + 2A x e^x = e^x$$
+ $$A e^x(2 + x) - 3A e^x(1 + x) + 2A x e^x = e^x$$
 
-除以 $e^x$（恒正）：
+Divide by $e^x$ (constant):
 
-$$A(2 + x) - 3A(1 + x) + 2Ax = A(2 + x - 3 - 3x + 2x) = A(-1) = 1$$
+ $$A(2 + x) - 3A(1 + x) + 2Ax = A(2 + x - 3 - 3x + 2x) = A(-1) = 1$$
 
-$$A = -1$$
+ $$A = -1$$
 
-因此：
+therefore:
 
-$$y_p(x) = -x e^x$$
+ $$y_p(x) = -x e^x$$
 
-### 第三步：写出通解并应用初始条件
+### Step 3: Write the general solution and apply initial conditions
 
-通解：
+General explanation:
 
-$$y(x) = y_h(x) + y_p(x) = C_1 e^x + C_2 e^{2x} - x e^x$$
+ $$y(x) = y_h(x) + y_p(x) = C_1 e^x + C_2 e^{2x} - x e^x$$
 
-应用 $y(0) = 0$：
+Apply $y(0) = 0$:
 
-$$C_1 e^0 + C_2 e^0 - 0 \cdot e^0 = C_1 + C_2 = 0 \quad \Rightarrow \quad C_2 = -C_1$$
+ $$C_1 e^0 + C_2 e^0 - 0 \cdot e^0 = C_1 + C_2 = 0 \quad \Rightarrow \quad C_2 = -C_1$$
 
-求导：
+Find the derivative:
 
-$$y'(x) = C_1 e^x + 2C_2 e^{2x} - e^x - x e^x$$
+ $$y'(x) = C_1 e^x + 2C_2 e^{2x} - e^x - x e^x$$
 
-应用 $y'(0) = 1$：
+Apply $y'(0) = 1$:
 
-$$C_1 + 2C_2 - 1 - 0 = 1 \quad \Rightarrow \quad C_1 + 2C_2 = 2$$
+ $$C_1 + 2C_2 - 1 - 0 = 1 \quad \Rightarrow \quad C_1 + 2C_2 = 2$$
 
-代入 $C_2 = -C_1$：
+Substitute $C_2 = -C_1$:
 
-$$C_1 + 2(-C_1) = -C_1 = 2 \quad \Rightarrow \quad C_1 = -2,\; C_2 = 2$$
+ $$C_1 + 2(-C_1) = -C_1 = 2 \quad \Rightarrow \quad C_1 = -2,\; C_2 = 2$$
 
-### 第四步：写出最终特解
+### Step 4: Write the final special solution
 
-$$y(x) = -2e^x + 2e^{2x} - x e^x = 2e^{2x} - (x + 2)e^x$$
+ $$y(x) = -2e^x + 2e^{2x} - x e^x = 2e^{2x} - (x + 2)e^x$$
 
-## 验算
+## Check calculation
 
-### 方法一：代入原方程验证
+### Method 1: Substitute into the original equation to verify
 
-计算 $y$、$y'$、$y''$ 并代入 $y'' - 3y' + 2y$：
+Calculate $y$ , $y'$ , $y''$ and substitute $y'' - 3y' + 2y$ :
 
-$$y = 2e^{2x} - (x + 2)e^x$$
+ $$y = 2e^{2x} - (x + 2)e^x$$
 
-$$y' = 4e^{2x} - e^x - (x + 2)e^x = 4e^{2x} - (x + 3)e^x$$
+ $$y' = 4e^{2x} - e^x - (x + 2)e^x = 4e^{2x} - (x + 3)e^x$$
 
-$$y'' = 8e^{2x} - e^x - (x + 3)e^x = 8e^{2x} - (x + 4)e^x$$
+ $$y'' = 8e^{2x} - e^x - (x + 3)e^x = 8e^{2x} - (x + 4)e^x$$
 
-代入左端：
+Substitute into the left end:
 
-$$y'' - 3y' + 2y = [8e^{2x} - (x + 4)e^x] - 3[4e^{2x} - (x + 3)e^x] + 2[2e^{2x} - (x + 2)e^x]$$
+ $$y'' - 3y' + 2y = [8e^{2x} - (x + 4)e^x] - 3[4e^{2x} - (x + 3)e^x] + 2[2e^{2x} - (x + 2)e^x]$$
 
-$$= 8e^{2x} - (x + 4)e^x - 12e^{2x} + 3(x + 3)e^x + 4e^{2x} - 2(x + 2)e^x$$
+ $$= 8e^{2x} - (x + 4)e^x - 12e^{2x} + 3(x + 3)e^x + 4e^{2x} - 2(x + 2)e^x$$
 
-合并 $e^{2x}$ 项：$(8 - 12 + 4)e^{2x} = 0 \cdot e^{2x} = 0$
+Merge $e^{2x}$ items: $(8 - 12 + 4)e^{2x} = 0 \cdot e^{2x} = 0$
 
-合并 $e^x$ 项：$[-(x + 4) + 3(x + 3) - 2(x + 2)]e^x = [-x - 4 + 3x + 9 - 2x - 4]e^x$
+Merge $e^x$ items: $[-(x + 4) + 3(x + 3) - 2(x + 2)]e^x = [-x - 4 + 3x + 9 - 2x - 4]e^x$
 
-$$= [(-x + 3x - 2x) + (-4 + 9 - 4)]e^x = [0x + 1]e^x = e^x$$
+ $$= [(-x + 3x - 2x) + (-4 + 9 - 4)]e^x = [0x + 1]e^x = e^x$$
 
-左端 = $e^x$ = 右端。✓
+Left end = $e^x$ = Right end. ✓
 
-### 方法二：验证初始条件
+### Method 2: Verify initial conditions
 
-$$y(0) = 2e^0 - (0 + 2)e^0 = 2 - 2 = 0$$
+ $$y(0) = 2e^0 - (0 + 2)e^0 = 2 - 2 = 0$$
 
-$$y'(0) = 4e^0 - (0 + 3)e^0 = 4 - 3 = 1$$
+ $$y'(0) = 4e^0 - (0 + 3)e^0 = 4 - 3 = 1$$
 
 ✓
 
-### 方法三：通解含 2 个常数验证
+### Method 3: General solution contains 2 constants verification
 
-齐次解 $y_h$ 含两个线性无关的基函数 $e^x$ 和 $e^{2x}$（Wronskian 非零），通解的自由度为 2，匹配方程阶数。✓
+The homogeneous solution $y_h$ contains two linearly independent basis functions $e^x$ and $e^{2x}$ (Wronskian non-zero), and the degree of freedom of the general solution is 2, matching the order of the equation. ✓
 
-## 最终答案
+## Final answer
 
-$$\boxed{y(x) = 2e^{2x} - (x + 2)e^x}$$
+ $$\boxed{y(x) = 2e^{2x} - (x + 2)e^x}$$
 
-## 易错点
-1. **待定系数法中 $Axe^x$ 的阶数**：由于 $r = 1$ 是特征根，特解形式是 $A x e^x$ 而非 $A e^x$。若错误地设为 $A e^x$，代入后左端为零，无法匹配右端的 $e^x$。
-2. **求导 $A x e^x$**：$y_p = Axe^x$ 的导数为 $Ae^x(1+x)$，二导为 $Ae^x(2+x)$。忘记乘积法则（漏掉 $x e^x$ 中 $x$ 的导数贡献）是常见错误。
-3. **特征根符号**：$r^2 - 3r + 2 = 0$ 的解是 $r = 1$ 和 $r = 2$（均为正），不是 $-1$ 和 $-2$。注意因式分解 $(\lambda - 1)(\lambda - 2)$ 的符号。
-4. **初始条件的应用顺序**：先求 $C_1, C_2$ 的通解表达式，再代入 $y(0)$ 和 $y'(0)$；不要在使用待定系数法前就代入初始条件。
+## Easy to make mistakes
+1. **Order of $Axe^x$ in the undetermined coefficient method**: Since $r = 1$ is a characteristic root, the special solution form is $A x e^x$ instead of $A e^x$. If it is set to $A e^x$ by mistake, the left end will be zero after substitution and cannot match $e^x$ on the right end.
+2. **Derivative $A x e^x$ **: The derivative of $y_p = Axe^x$ is $Ae^x(1+x)$ and the second derivative is $Ae^x(2+x)$. Forgetting the product rule (missing the derivative contribution of $x$ in $x e^x$) is a common mistake.
+3. **Characteristic root symbol**: The solutions of $r^2 - 3r + 2 = 0$ are $r = 1$ and $r = 2$ (both positive), not $-1$ and $-2$. Note the sign of factorization $(\lambda - 1)(\lambda - 2)$.
+4. **Sequence of application of initial conditions**: First find the general solution expression of $C_1, C_2$, and then substitute $y(0)$ and $y'(0)$; do not substitute initial conditions before using the undetermined coefficient method.
diff --git a/examples/probability.md b/examples/probability.md
index e943df2..f988201 100644
--- a/examples/probability.md
+++ b/examples/probability.md
@@ -1,106 +1,106 @@
-# 超几何分布的概率与期望
+# Probability and expectation of hypergeometric distribution
 
-## 用户输入
-一个袋子里有 5 个红球和 3 个蓝球。从中不放回地抽取 3 个球。求：(1) 恰好抽到 2 个红球的概率；(2) 抽到红球个数的期望。
+## User input
+There are 5 red balls and 3 blue balls in a bag. Draw 3 balls from it without replacement. Find: (1) the probability of drawing exactly 2 red balls; (2) the expectation of the number of red balls drawn.
 
-## Skill 分类
-概率论
+## Skill Category
+probability theory
 
-## 题意解析
-- **已知条件**：总体 $N = 8$ 个球，其中 $K = 5$ 个红球（成功类）、$N - K = 3$ 个蓝球；样本量 $n = 3$，不放回抽样。
-- **求解目标**：(1) $P(X = 2)$，其中 $X$ 为红球个数；(2) $E[X]$。
-- **分布类型识别**：从有限总体中不放回抽样，红球个数 $X$ 服从**超几何分布** $X \sim \text{Hypergeometric}(N=8, K=5, n=3)$。
-- **$X$ 的取值范围**：$\max\{0,\, n - (N - K)\} \leq k \leq \min\{n, K\}$，即 $k \in \{0, 1, 2, 3\}$。
-- **关键前提**：抽样是完全随机的，每个球被抽中的概率相等。
+## Question meaning analysis
+- **Known conditions**: There are $N = 8$ balls in total, including $K = 5$ red balls (successful type) and $N - K = 3$ blue balls; sample size $n = 3$, sampling without replacement.
+- **Solution target**: (1) $P(X = 2)$, where $X$ is the number of red balls; (2) $E[X]$.
+- **Distribution type identification**: Sampling from a finite population without replacement, the number of red balls $X$ obeys **Hypergeometric distribution** $X \sim \text{Hypergeometric}(N=8, K=5, n=3)$.
+- **The value range of $X$**: $\max\{0,\, n - (N - K)\} \leq k \leq \min\{n, K\}$, that is, $k \in \{0, 1, 2, 3\}$.
+- **Key Premise**: Sampling is completely random, and each ball has an equal probability of being drawn.
 
-## 方法选择
-**首选方法**：超几何分布公式。超几何分布的概率质量函数为
+## Method selection
+**Preferred method**: Hypergeometric distribution formula. The probability mass function of the hypergeometric distribution is
 
-$$P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}$$
+ $$P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}$$
 
-期望公式为
+The expected formula is
 
-$$E[X] = n \cdot \frac{K}{N}$$
+ $$E[X] = n \cdot \frac{K}{N}$$
 
-**备选方法**：古典概率直接列举；指示变量法分解期望（利用线性性）。
+**Alternative methods**: Direct enumeration of classical probabilities; decomposition of expectations by indicator variable method (using linearity).
 
-## 解题过程
+## Problem solving process
 
-### (1) 恰好抽到 2 个红球的概率
+### (1) The probability of drawing exactly 2 red balls
 
-代入超几何分布公式，$k = 2$：
+Substitute into the hypergeometric distribution formula, $k = 2$:
 
-$$P(X = 2) = \frac{\binom{5}{2} \binom{3}{1}}{\binom{8}{3}}$$
+ $$P(X = 2) = \frac{\binom{5}{2} \binom{3}{1}}{\binom{8}{3}}$$
 
-分别计算组合数：
+Calculate the number of combinations separately:
 
-$$\binom{5}{2} = \frac{5 \times 4}{2} = 10$$
+ $$\binom{5}{2} = \frac{5 \times 4}{2} = 10$$
 
-$$\binom{3}{1} = 3$$
+ $$\binom{3}{1} = 3$$
 
-$$\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$$
+ $$\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$$
 
-代入：
+Substitute:
 
-$$P(X = 2) = \frac{10 \times 3}{56} = \frac{30}{56} = \frac{15}{28}$$
+ $$P(X = 2) = \frac{10 \times 3}{56} = \frac{30}{56} = \frac{15}{28}$$
 
-### (2) 红球个数的期望
+### (2) Expectation of the number of red balls
 
-法一：直接使用超几何分布期望公式
+Method 1: Directly use the expectation formula of hypergeometric distribution
 
-$$E[X] = n \cdot \frac{K}{N} = 3 \cdot \frac{5}{8} = \frac{15}{8} = 1.875$$
+ $$E[X] = n \cdot \frac{K}{N} = 3 \cdot \frac{5}{8} = \frac{15}{8} = 1.875$$
 
-法二：通过分布律计算（用于验证）
+Method 2: Calculation through distribution law (for verification)
 
-先计算完整的分布律：
+First calculate the complete distribution law:
 
-| $k$ | $\binom{5}{k}$ | $\binom{3}{3-k}$ | $P(X=k)$ |
+|  $k$  |  $\binom{5}{k}$  |  $\binom{3}{3-k}$  |  $P(X=k)$  |
 |-----|----------------|-------------------|----------|
-| 0   | 1              | 1                 | $1/56$   |
-| 1   | 5              | 3                 | $15/56$  |
-| 2   | 10             | 3                 | $30/56$  |
-| 3   | 10             | 1                 | $10/56$  |
+| 0   | 1              | 1                 |  $1/56$    |
+| 1   | 5              | 3                 |  $15/56$   |
+| 2   | 10             | 3                 |  $30/56$   |
+| 3   | 10             | 1                 |  $10/56$   |
 
-$$E[X] = 0 \cdot \frac{1}{56} + 1 \cdot \frac{15}{56} + 2 \cdot \frac{30}{56} + 3 \cdot \frac{10}{56} = \frac{15 + 60 + 30}{56} = \frac{105}{56} = \frac{15}{8}$$
+ $$E[X] = 0 \cdot \frac{1}{56} + 1 \cdot \frac{15}{56} + 2 \cdot \frac{30}{56} + 3 \cdot \frac{10}{56} = \frac{15 + 60 + 30}{56} = \frac{105}{56} = \frac{15}{8}$$
 
-两种方法结果一致。
+The results of both methods are consistent.
 
-## 验算
+## Check calculation
 
-### 方法一：概率总和验证
+### Method 1: Probability sum verification
 
-所有可能的 $k$ 值概率之和应为 $1$：
+The sum of the probabilities of all possible $k$ values ​​should be $1$ :
 
-$$P(X=0) + P(X=1) + P(X=2) + P(X=3) = \frac{1 + 15 + 30 + 10}{56} = \frac{56}{56} = 1$$
+ $$P(X=0) + P(X=1) + P(X=2) + P(X=3) = \frac{1 + 15 + 30 + 10}{56} = \frac{56}{56} = 1$$
 
-✓ 通过。
+✓ Passed.
 
-### 方法二：期望的指示变量分解
+### Method 2: Decomposition of expected indicator variables
 
-定义指示变量 $Y_i = \mathbb{1}_{\{\text{第 }i\text{ 个球是红色}\}}$，$i = 1,2,3$。则 $X = Y_1 + Y_2 + Y_3$。
+Define indicator variables $Y_i = \mathbb{1}_{\{\text{The }i\text{th ball is red}\}}$, $i = 1,2,3$. Then $X = Y_1 + Y_2 + Y_3$ .
 
-由对称性（尽管是不放回抽样，每个位置上的球是红色的概率相同）：
+Due to symmetry (despite sampling without replacement, the ball at each position has the same probability of being red):
 
-$$E[Y_i] = P(\text{第 $i$ 个球是红色}) = \frac{5}{8}$$
+$$E[Y_i] = P(\text{The $i$th ball is red}) = \frac{5}{8}$$
 
-利用期望的线性性：
+Exploiting linearity of expectations:
 
-$$E[X] = E[Y_1] + E[Y_2] + E[Y_3] = 3 \cdot \frac{5}{8} = \frac{15}{8}$$
+ $$E[X] = E[Y_1] + E[Y_2] + E[Y_3] = 3 \cdot \frac{5}{8} = \frac{15}{8}$$
 
-与公式计算结果完全一致。
+The calculation results are completely consistent with the formula.
 
-### 方法三：每个概率在 $[0, 1]$ 内
+### Method 3: Each probability is within $[0, 1]$
 
-$\frac{1}{56} \approx 0.018$，$\frac{15}{56} \approx 0.268$，$\frac{30}{56} \approx 0.536$，$\frac{10}{56} \approx 0.179$，均在 $[0, 1]$ 内。
+$\frac{1}{56} \approx 0.018$, $\frac{15}{56} \approx 0.268$, $\frac{30}{56} \approx 0.536$, $\frac{10}{56} \approx 0.179$, all within $[0, 1]$.
 
-## 最终答案
+## Final answer
 
-$$\boxed{P(X = 2) = \frac{15}{28}}$$
+ $$\boxed{P(X = 2) = \frac{15}{28}}$$
 
-$$\boxed{E[X] = \frac{15}{8} = 1.875}$$
+ $$\boxed{E[X] = \frac{15}{8} = 1.875}$$
 
-## 易错点
-1. **放回与不放回的混淆**：本题是不放回抽样，应用超几何分布；若为放回抽样，$X \sim \text{Binomial}(n=3, p=5/8)$，两者期望相同但方差不同。
-2. **组合数计算**：$\binom{8}{3} = 56$ 而非 $8 \times 7 \times 6 = 336$（忘记除以 $3!$）。
-3. **$k$ 取值范围遗漏**：$k = 0$ 是可能的（3 个都是蓝球），不要遗漏。同时 $k=3$ 也是可能的。
-4. **期望公式记忆**：$E[X] = n \cdot \frac{K}{N}$，分子是成功类总数 $K$，不是 $N-K$。
+## Easy to make mistakes
+1. **Confusion between replacement and non-replacement**: This question is about sampling without replacement and applying hypergeometric distribution; if it is sampling with replacement, $X \sim \text{Binomial}(n=3, p=5/8)$, the two have the same expectation but different variances.
+2. **Combination number calculation**: $\binom{8}{3} = 56$ instead of $8 \times 7 \times 6 = 336$ (forgot to divide by $3!$).
+3. ** $k$ value range is missing**: $k = 0$ is possible (all 3 are blue balls), don’t miss it. Also $k=3$ is possible.
+4. **Expected formula memory**: $E[X] = n \cdot \frac{K}{N}$, the numerator is the total number of successful classes $K$, not $N-K$.
diff --git a/examples/problem_generation.md b/examples/problem_generation.md
index 2400229..f860387 100644
--- a/examples/problem_generation.md
+++ b/examples/problem_generation.md
@@ -1,204 +1,204 @@
-# 二次方程练习题生成（含答案与评分标准）
+# Generation of quadratic equation practice questions (including answers and scoring standards)
 
-## 用户输入
-生成 3 道二次方程题目，难度递增，包含完整解答和评分标准。
+## User input
+Generate 3 quadratic equation questions with increasing difficulty, including complete solutions and scoring criteria.
 
-## Skill 分类
-代数 / 题目生成与评估
+## Skill Category
+Algebra/Question Generation and Assessment
 
-## 题意解析
-- **目标**：生成 3 道难度递增的二次方程（或二次方程相关）题目
-- **要求**：每道题需包含题目陈述、完整解答、常见错误、评分标准
-- **难度递增路径**：
-  - Easy：直接因式分解
-  - Medium：需用求根公式，系数非 1
-  - Hard：含参数讨论，需分析判别式
+## Question meaning analysis
+- **Goal**: Generate 3 quadratic equation (or quadratic equation related) questions of increasing difficulty
+- **Requirements**: Each question must include a problem statement, complete answer, common mistakes, and scoring criteria.
+- **Ascending Difficulty Paths**:
+- Easy: direct factorization
+- Medium: The root formula needs to be used, and the coefficient is not 1
+- Hard: Contains parameter discussion and needs to analyze the discriminant
 
 ---
 
-## Problem 1（简单）— 因式分解法
+## Problem 1 (Simple) — Factoring
 
-### 题目
-解方程：$x^2 - 7x + 12 = 0$
+### Title
+Solve the equation: $x^2 - 7x + 12 = 0$
 
-### 解题过程
+### Problem solving process
 
-**Step 1：因式分解**
+**Step 1: Factorization**
 
-找两个数，和为 $-7$，积为 $12$。这两个数是 $-3$ 和 $-4$。
+Find two numbers whose sum is $-7$ and their product is $12$. The two numbers are $-3$ and $-4$ .
 
-$$x^2 - 7x + 12 = (x - 3)(x - 4) = 0$$
+ $$x^2 - 7x + 12 = (x - 3)(x - 4) = 0$$
 
-验证：$(x-3)(x-4) = x^2 - 4x - 3x + 12 = x^2 - 7x + 12$ ✓
+Verification: $(x-3)(x-4) = x^2 - 4x - 3x + 12 = x^2 - 7x + 12$ ✓
 
-**Step 2：零因子性质**
+**Step 2: Zero factor property**
 
-$$(x - 3)(x - 4) = 0 \implies x - 3 = 0 \text{ 或 } x - 4 = 0$$
+$$(x - 3)(x - 4) = 0 \implies x - 3 = 0 \text{ or } x - 4 = 0$$
 
-$$x = 3 \quad \text{或} \quad x = 4$$
+$$x = 3 \quad \text{or} \quad x = 4$$
 
-**验算**：
-- $x = 3$：$3^2 - 7 \times 3 + 12 = 9 - 21 + 12 = 0$ ✓
-- $x = 4$：$4^2 - 7 \times 4 + 12 = 16 - 28 + 12 = 0$ ✓
+**Check**:
+-  $x = 3$ ： $3^2 - 7 \times 3 + 12 = 9 - 21 + 12 = 0$  ✓
+-  $x = 4$ ： $4^2 - 7 \times 4 + 12 = 16 - 28 + 12 = 0$  ✓
 
-### 最终答案
-$$x = 3 \quad \text{或} \quad x = 4$$
+### Final answer
+$$x = 3 \quad \text{or} \quad x = 4$$
 
-### 常见错误
-1. 找因数时符号错误，写成 $(x+3)(x+4)$，和变成了 $+7$ 而非 $-7$
-2. 因式分解后忘记设每个因子为 0
-3. 只给出一个根
+### Common mistakes
+1. The sign is wrong when finding the factors, it is written as $(x+3)(x+4)$, and the sum becomes $+7$ instead of $-7$
+2. Forgetting to set each factor to 0 after factoring
+3. Only one root is given
 
-### 评分标准（共 10 分）
+### Scoring criteria (total 10 points)
 
-| 评分项 | 分值 | 说明 |
+| Rating items | Points | Description |
 |--------|------|------|
-| 正确因式分解 | 4 分 | 写出 $(x-3)(x-4)=0$ |
-| 应用零因子性质 | 3 分 | 分别设 $x-3=0$ 和 $x-4=0$ |
-| 得出正确解 | 2 分 | $x=3$ 和 $x=4$ |
-| 验算 | 1 分 | 将解代入原方程验证 |
+| Correct factorization | 4 points | Write $(x-3)(x-4)=0$ |
+| Apply the zero factor property | 3 points | Let $x-3=0$ and $x-4=0$ respectively |
+| Get the correct solution | 2 points | $x=3$ and $x=4$ |
+| Check calculation | 1 point | Substitute the solution into the original equation to verify |
 
 ---
 
-## Problem 2（中等）— 求根公式法
+## Problem 2 (medium) - Root formula method
 
-### 题目
-解方程：$2x^2 + 3x - 2 = 0$
+### Title
+Solve the equation: $2x^2 + 3x - 2 = 0$
 
-### 解题过程
+### Problem solving process
 
-**分析**：系数 $a=2$，直接因式分解不易，使用求根公式。
+**Analysis**: Coefficient $a=2$, direct factorization is not easy, use the root formula.
 
-**Step 1：计算判别式**
+**Step 1: Calculate the discriminant**
 
-$$\Delta = b^2 - 4ac = 3^2 - 4 \times 2 \times (-2) = 9 + 16 = 25$$
+ $$\Delta = b^2 - 4ac = 3^2 - 4 \times 2 \times (-2) = 9 + 16 = 25$$
 
-$\Delta = 25 > 0$，有两个不等实根。
+$\Delta = 25 > 0$ , has two unequal real roots.
 
-**Step 2：应用求根公式**
+**Step 2: Apply the root formula**
 
-$$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-3 \pm \sqrt{25}}{2 \times 2} = \frac{-3 \pm 5}{4}$$
+ $$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-3 \pm \sqrt{25}}{2 \times 2} = \frac{-3 \pm 5}{4}$$
 
-$$x_1 = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}$$
+ $$x_1 = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}$$
 
-$$x_2 = \frac{-3 - 5}{4} = \frac{-8}{4} = -2$$
+ $$x_2 = \frac{-3 - 5}{4} = \frac{-8}{4} = -2$$
 
-**备选方法（因式分解）**：
+**Alternative method (factorization)**:
 
-$$2x^2 + 3x - 2 = (2x - 1)(x + 2) = 0$$
+ $$2x^2 + 3x - 2 = (2x - 1)(x + 2) = 0$$
 
-验证展开：$2x^2 + 4x - x - 2 = 2x^2 + 3x - 2$ ✓
+Verify expansion: $2x^2 + 4x - x - 2 = 2x^2 + 3x - 2$ ✓
 
-得 $2x - 1 = 0$ 或 $x + 2 = 0$，即 $x = \frac{1}{2}$ 或 $x = -2$，与求根公式一致。
+Get $2x - 1 = 0$ or $x + 2 = 0$, that is, $x = \frac{1}{2}$ or $x = -2$, which is consistent with the root formula.
 
-**验算**：
-- $x = \frac{1}{2}$：$2 \times \frac{1}{4} + 3 \times \frac{1}{2} - 2 = \frac{1}{2} + \frac{3}{2} - 2 = 0$ ✓
-- $x = -2$：$2 \times 4 + 3 \times (-2) - 2 = 8 - 6 - 2 = 0$ ✓
+**Check**:
+-  $x = \frac{1}{2}$ ： $2 \times \frac{1}{4} + 3 \times \frac{1}{2} - 2 = \frac{1}{2} + \frac{3}{2} - 2 = 0$  ✓
+-  $x = -2$ ： $2 \times 4 + 3 \times (-2) - 2 = 8 - 6 - 2 = 0$  ✓
 
-### 最终答案
-$$x = \frac{1}{2} \quad \text{或} \quad x = -2$$
+### Final answer
+$$x = \frac{1}{2} \quad \text{or} \quad x = -2$$
 
-### 常见错误
-1. 求根公式中分母写成 $a$ 而非 $2a$，即 $x = (-b \pm \sqrt{\Delta})/a$
-2. 计算 $\Delta$ 时 $4ac$ 的符号搞错（$c=-2$，$4 \times 2 \times (-2) = -16$，减去负数为加）
-3. 忘记化简分数
+### Common mistakes
+1. The denominator in the root formula is written as $a$ instead of $2a$, that is, $x = (-b \pm \sqrt{\Delta})/a$
+2. The sign of $4ac$ is wrong when calculating $\Delta$ ($c=-2$, $4 \times 2 \times (-2) = -16$, subtracting negative numbers is adding)
+3. Forgetting to simplify fractions
 
-### 评分标准（共 10 分）
+### Scoring criteria (total 10 points)
 
-| 评分项 | 分值 | 说明 |
+| Rating items | Points | Description |
 |--------|------|------|
-| 正确识别系数 $a, b, c$ | 2 分 | $a=2, b=3, c=-2$ |
-| 计算 $\Delta$ 正确 | 2 分 | $\Delta=25$ |
-| 正确代入求根公式 | 3 分 | $x = \frac{-3\pm 5}{4}$ |
-| 得出正确解 | 2 分 | $x=1/2$ 和 $x=-2$ |
-| 验算 | 1 分 | 代入验证 |
+| Correct identification coefficient $a, b, c$ | 2 points | $a=2, b=3, c=-2$ |
+| Calculate $\Delta$ Correct | 2 points | $\Delta=25$ |
+| Correctly substitute the root formula | 3 points | $x = \frac{-3\pm 5}{4}$ |
+| Get the correct solution | 2 points | $x=1/2$ and $x=-2$ |
+| Verification | 1 point | Substitution verification |
 
 ---
 
-## Problem 3（困难）— 含参数与判别式分析
+## Problem 3 (difficult)—including parameter and discriminant analysis
 
-### 题目
-求所有实数 $k$，使方程 $x^2 + 2kx + k + 2 = 0$ 有实根。
+### Title
+Find all real numbers $k$ such that the equation $x^2 + 2kx + k + 2 = 0$ has real roots.
 
-### 解题过程
+### Problem solving process
 
-**分析**：这是一个含参数 $k$ 的二次方程（首项系数 $1 \neq 0$，确为二次）。有实根的条件是判别式非负。
+**Analysis**: This is a quadratic equation with parameter $k$ (the first coefficient is $1 \neq 0$, which is indeed quadratic). The condition for having real roots is that the discriminant is non-negative.
 
-**Step 1：识别系数**
+**Step 1: Identification coefficient**
 
-$$a = 1, \quad b = 2k, \quad c = k + 2$$
+ $$a = 1, \quad b = 2k, \quad c = k + 2$$
 
-**Step 2：写出判别式并化简**
+**Step 2: Write the discriminant and simplify it**
 
-$$\Delta = b^2 - 4ac = (2k)^2 - 4 \times 1 \times (k + 2)$$
+ $$\Delta = b^2 - 4ac = (2k)^2 - 4 \times 1 \times (k + 2)$$
 
-$$\Delta = 4k^2 - 4(k + 2) = 4k^2 - 4k - 8$$
+ $$\Delta = 4k^2 - 4(k + 2) = 4k^2 - 4k - 8$$
 
-$$\Delta = 4(k^2 - k - 2)$$
+ $$\Delta = 4(k^2 - k - 2)$$
 
-**Step 3：因式分解判别式表达式**
+**Step 3: Factor the discriminant expression**
 
-$$k^2 - k - 2 = (k - 2)(k + 1)$$
+ $$k^2 - k - 2 = (k - 2)(k + 1)$$
 
-验证：$(k-2)(k+1) = k^2 + k - 2k - 2 = k^2 - k - 2$ ✓
+Verification: $(k-2)(k+1) = k^2 + k - 2k - 2 = k^2 - k - 2$ ✓
 
-$$\Delta = 4(k - 2)(k + 1)$$
+ $$\Delta = 4(k - 2)(k + 1)$$
 
-**Step 4：解不等式 $\Delta \geq 0$**
+**Step 4: Solve the inequality $\Delta \geq 0$ **
 
-需要有实根，即 $\Delta \geq 0$：
+Requires a real root, i.e. $\Delta \geq 0$ :
 
-$$4(k - 2)(k + 1) \geq 0$$
+ $$4(k - 2)(k + 1) \geq 0$$
 
-由于 $4 > 0$，不等式等价于：
+Because of $4 > 0$ , the inequality is equivalent to:
 
-$$(k - 2)(k + 1) \geq 0$$
+ $$(k - 2)(k + 1) \geq 0$$
 
-**Step 5：解二次不等式**
+**Step 5: Solve the quadratic inequality**
 
-零点：$k = -1$ 和 $k = 2$。
+Zero points: $k = -1$ and $k = 2$ .
 
-二次项系数为正，抛物线开口向上。因子 $(k-2)(k+1) \geq 0$ 的解为：
+The coefficient of the quadratic term is positive, and the parabola opens upward. The solution for factor $(k-2)(k+1) \geq 0$ is:
 
-$$k \leq -1 \quad \text{或} \quad k \geq 2$$
+$$k \leq -1 \quad \text{or} \quad k \geq 2$$
 
-即 $k \in (-\infty, -1] \cup [2, +\infty)$。
+That is $k \in (-\infty, -1] \cup [2, +\infty)$ .
 
-**Step 6：边界验证**
+**Step 6: Boundary verification**
 
-- $k = -1$：方程变为 $x^2 - 2x + 1 = 0 \implies (x-1)^2 = 0 \implies x = 1$（重根，$\Delta = 0$）✓
-- $k = 2$：方程变为 $x^2 + 4x + 4 = 0 \implies (x+2)^2 = 0 \implies x = -2$（重根，$\Delta = 0$）✓
-- $k = 0$（中间值，不应有实根）：方程变为 $x^2 + 0x + 2 = 0 \implies x^2 = -2$（无实根，$\Delta = -8$）✓
-- $k = -2$（小于 $-1$）：$\Delta = 4 \times (-4) \times (-1) = 16 > 0$，有实根 ✓
-- $k = 3$（大于 $2$）：$\Delta = 4 \times 1 \times 4 = 16 > 0$，有实根 ✓
+- $k = -1$ : the equation becomes $x^2 - 2x + 1 = 0 \implies (x-1)^2 = 0 \implies x = 1$ (double root, $\Delta = 0$ ) ✓
+- $k = 2$ : the equation becomes $x^2 + 4x + 4 = 0 \implies (x+2)^2 = 0 \implies x = -2$ (double root, $\Delta = 0$ ) ✓
+- $k = 0$ (intermediate value, should not have real roots): equation becomes $x^2 + 0x + 2 = 0 \implies x^2 = -2$ (no real roots, $\Delta = -8$ )✓
+- $k = -2$ (less than $-1$ ): $\Delta = 4 \times (-4) \times (-1) = 16 > 0$ , has real roots ✓
+- $k = 3$ (greater than $2$ ): $\Delta = 4 \times 1 \times 4 = 16 > 0$ , has real roots ✓
 
-### 最终答案
-$$k \in (-\infty, -1] \cup [2, +\infty)$$
+### Final answer
+ $$k \in (-\infty, -1] \cup [2, +\infty)$$
 
-即 $k \leq -1$ 或 $k \geq 2$。
+That is $k \leq -1$ or $k \geq 2$ .
 
-### 常见错误
-1. 忘记判别式是 "非负"（$\geq 0$），只考虑严格大于（遗漏边界值 $k=-1$ 和 $k=2$，此时有重根）
-2. 解不等式 $(k-2)(k+1) \geq 0$ 时写反方向（写为 $-1 \leq k \leq 2$），混淆了两侧和中间的关系
-3. 化简判别式时忘记提取公因子，导致不等式难解
-4. 误以为 $a=0$ 也要讨论（此处 $a=1$ 恒不为零，无需讨论退化为一次方程的情况）
+### Common mistakes
+1. Forget that the discriminant is "non-negative" ( $\geq 0$ ), only consider strictly greater than (missing boundary values ​​$k=-1$ and $k=2$ , in which case there are multiple roots)
+2. When solving the inequality $(k-2)(k+1) \geq 0$, write the opposite direction (written as $-1 \leq k \leq 2$), confusing the relationship between the two sides and the middle
+3. Forgetting to extract common factors when simplifying the discriminant, making the inequality difficult to solve
+4. Mistakenly thinking that $a=0$ also needs to be discussed (here $a=1$ is always not zero, so there is no need to discuss the situation of degenerating into a linear equation)
 
-### 评分标准（共 10 分）
+### Scoring criteria (total 10 points)
 
-| 评分项 | 分值 | 说明 |
+| Rating items | Points | Description |
 |--------|------|------|
-| 正确识别 $a,b,c$ | 1 分 | $a=1, b=2k, c=k+2$ |
-| 写出判别式并化简 | 3 分 | $\Delta = 4k^2 - 4k - 8 = 4(k^2 - k - 2)$ |
-| 因式分解判别式 | 2 分 | $\Delta = 4(k-2)(k+1)$ |
-| 解不等式 $\Delta \geq 0$ | 3 分 | 正确得出 $k \leq -1$ 或 $k \geq 2$ |
-| 边界验证 | 1 分 | 验证 $k=-1$ 和 $k=2$ 时有重根 |
+| Correct identification of $a,b,c$ | 1 point | $a=1, b=2k, c=k+2$ |
+| Write the discriminant and simplify it | 3 points | $\Delta = 4k^2 - 4k - 8 = 4(k^2 - k - 2)$ |
+| Factoring discriminant | 2 points | $\Delta = 4(k-2)(k+1)$ |
+| Solve the inequality $\Delta \geq 0$ | 3 points | Correctly derive $k \leq -1$ or $k \geq 2$ |
+| Bounds verification | 1 point | Verify that $k=-1$ and $k=2$ have duplicate roots |
 
 ---
 
-## 整体难度递进说明
+## Overall difficulty progression description
 
-| 难度 | 核心方法 | 认知要求 |
+| Difficulty | Core Methods | Cognitive Requirements |
 |------|----------|----------|
-| Easy | 因式分解 | 找出两个和为 $-7$ 积为 $12$ 的数 |
-| Medium | 求根公式 | 识别系数、计算判别式、代入公式、化简分数 |
-| Hard | 参数分析 | 含参判别式、因式分解、二次不等式求解、边界检查 |
+| Easy | Factorization | Find two numbers whose sum is $-7$ and whose product is $12$ |
+| Medium | Root formula | Identify coefficients, calculate discriminants, substitute formulas, simplify fractions |
+| Hard | Parametric analysis | Parametric discriminant, factorization, quadratic inequality solution, boundary checking |
diff --git a/examples/proof.md b/examples/proof.md
index 337f961..f6380c7 100644
--- a/examples/proof.md
+++ b/examples/proof.md
@@ -1,127 +1,127 @@
-# $\sqrt{2}$ 是无理数的证明
+# $\sqrt{2}$ is a proof of irrational numbers
 
-## 用户输入
-证明 $\sqrt{2}$ 是无理数。
+## User input
+Show that $\sqrt{2}$ is an irrational number.
 
-## Skill 分类
-数论 / 反证法
+## Skill Category
+Number theory / proof by contradiction
 
-## 题意解析
-- **条件**：定义 $\sqrt{2}$ 为满足 $x^2 = 2$ 的正实数
-- **目标**：证明不存在整数 $p, q$（$q \neq 0$）使得 $\sqrt{2} = p/q$
-- **隐含条件**：$p, q$ 可假设互质（即分数已约分到最简形式）
-- **方法约束**：经典反证法，假设存在有理数表达式然后导出矛盾
-- **边界情况**：$p = 0$ 时 $\sqrt{2} = 0$ 显然不成立，可排除
+## Question meaning analysis
+- **Condition**: Define $\sqrt{2}$ as a positive real number that satisfies $x^2 = 2$
+- **Goal**: Prove that there is no integer $p, q$ ( $q \neq 0$ ) such that $\sqrt{2} = p/q$
+- **Implicit condition**: $p, q$ can be assumed to be relatively prime (that is, the fraction has been reduced to its simplest form)
+- **Method Constraints**: Classical proof by contradiction, assuming there is a rational number expression and then deducing a contradiction
+- **Boundary case**: When $p = 0$, $\sqrt{2} = 0$ is obviously not true and can be excluded
 
-## 方法选择
+## Method selection
 
-**主方法：反证法（standard proof by contradiction）**
+**Main method: standard proof by contradiction**
 
-这是证明无理数最经典的方法。假设 $\sqrt{2}$ 是有理数，写成最简分数形式，然后利用奇偶性导出矛盾。
+This is the most classic way to prove irrational numbers. Assume that $\sqrt{2}$ is a rational number, write it in the simplest fractional form, and then use parity to derive a contradiction.
 
-**备选方法 1：有理根定理**
+**Alternative 1: Rational Root Theorem**
 
-利用有理根定理：若 $x = p/q$（最简形式）是有理数且满足 $x^2 - 2 = 0$，则 $p \mid 2$ 且 $q \mid 1$。故 $p \in \{\pm 1, \pm 2\}$，$q \in \{\pm 1\}$。验证 $(\pm 1)^2 = 1 \neq 2$，$(\pm 2)^2 = 4 \neq 2$，说明 $\sqrt{2}$ 不是有理数。
+Use the rational roots theorem: If $x = p/q$ (in its simplest form) is a rational number and satisfies $x^2 - 2 = 0$ , then $p \mid 2$ and $q \mid 1$ . So $p \in \{\pm 1, \pm 2\}$ , $q \in \{\pm 1\}$ . Verify $(\pm 1)^2 = 1 \neq 2$, $(\pm 2)^2 = 4 \neq 2$, indicating that $\sqrt{2}$ is not a rational number.
 
-**备选方法 2：无穷递降法（几何证明）**
+**Alternative 2: Infinite descent method (geometric proof)**
 
-假设 $\sqrt{2} = a/b$（$a > b > 0$），则 $a^2 = 2b^2$。考虑等腰直角三角形，可构造更小的正整数对 $(2b-a, a-b)$ 也满足类似关系，导出无穷递降矛盾。
+Assuming $\sqrt{2} = a/b$ ( $a > b > 0$ ), then $a^2 = 2b^2$ . Considering an isosceles right triangle, a smaller positive integer pair $(2b-a, a-b)$ can be constructed that also satisfies a similar relationship and derives an infinite descending contradiction.
 
-本文展示主方法和有理根定理两种证明。
+This article shows two proofs of the main method and the rational root theorem.
 
-## 解题过程
+## Problem solving process
 
-### 方法一：经典反证法
+### Method 1: Classical proof by contradiction
 
-**Step 1：假设与化简**
+**Step 1: Assumption and Simplification**
 
-假设 $\sqrt{2}$ 是有理数，即存在互质的正整数 $p, q$（$q \neq 0$）使得
-$$\sqrt{2} = \frac{p}{q}, \quad \gcd(p, q) = 1$$
+Assume that $\sqrt{2}$ is a rational number, that is, there is a relatively prime positive integer $p, q$ ( $q \neq 0$ ) such that
+ $$\sqrt{2} = \frac{p}{q}, \quad \gcd(p, q) = 1$$
 
-两边平方：
-$$2 = \frac{p^2}{q^2} \implies p^2 = 2q^2$$
+Square both sides:
+ $$2 = \frac{p^2}{q^2} \implies p^2 = 2q^2$$
 
-**Step 2：证明 $p$ 是偶数**
+**Step 2: Prove that $p$ is an even number**
 
-由 $p^2 = 2q^2$ 知 $p^2$ 是偶数。
+From $p^2 = 2q^2$, we know that $p^2$ is an even number.
 
-**论断**：若 $p^2$ 是偶数，则 $p$ 必为偶数。
+**Conclusion**: If $p^2$ is an even number, then $p$ must be an even number.
 
-**论证**：假设 $p$ 是奇数，即存在整数 $k$ 使得 $p = 2k + 1$，则
-$$p^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$$
+**Argument**: Suppose $p$ is an odd number, that is, there is an integer $k$ such that $p = 2k + 1$ , then
+ $$p^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$$
 
-$p^2$ 为奇数，与 $p^2$ 是偶数矛盾。故 $p$ 必须是偶数。
+$p^2$ is an odd number, which is inconsistent with $p^2$ being an even number. Therefore $p$ must be an even number.
 
-因此存在整数 $k$ 使得 $p = 2k$。
+Therefore there is an integer $k$ such that $p = 2k$ .
 
-**Step 3：证明 $q$ 也是偶数**
+**Step 3: Prove that $q$ is also an even number**
 
-代入 $p = 2k$ 到 $p^2 = 2q^2$：
-$$(2k)^2 = 2q^2 \implies 4k^2 = 2q^2 \implies 2k^2 = q^2$$
+Substitute $p = 2k$ into $p^2 = 2q^2$ :
+ $$(2k)^2 = 2q^2 \implies 4k^2 = 2q^2 \implies 2k^2 = q^2$$
 
-所以 $q^2$ 是偶数。同理，$q$ 必为偶数。
+So $q^2$ is an even number. In the same way, $q$ must be an even number.
 
-**Step 4：导出矛盾**
+**Step 4: Export contradictions**
 
-$p$ 和 $q$ 都是偶数，则 $\gcd(p, q) \geq 2$，与假设 $\gcd(p, q) = 1$ 矛盾。
+If $p$ and $q$ are both even numbers, then $\gcd(p, q) \geq 2$ is in conflict with the assumption $\gcd(p, q) = 1$.
 
-故假设不成立，$\sqrt{2}$ 不是有理数，即 $\sqrt{2}$ 是无理数。$\square$
+Therefore, the assumption does not hold, $\sqrt{2}$ is not a rational number, that is, $\sqrt{2}$ is an irrational number. $\square$
 
-### 方法二：有理根定理
+### Method 2: Rational Root Theorem
 
-**Step 1：构造多项式**
+**Step 1: Construct polynomial**
 
-$\sqrt{2}$ 是方程 $x^2 - 2 = 0$ 的正根。
+$\sqrt{2}$ is the positive root of the equation $x^2 - 2 = 0$.
 
-**Step 2：应用有理根定理**
+**Step 2: Apply the rational root theorem**
 
-设 $x = p/q$（$p, q$ 互质，$q > 0$）是 $x^2 - 2 = 0$ 的有理根，则 $p \mid (-2)$ 且 $q \mid 1$。
+Assume $x = p/q$ ( $p, q$ is relatively prime, $q > 0$ ) is a rational root of $x^2 - 2 = 0$, then $p \mid (-2)$ and $q \mid 1$ .
 
-故 $p \in \{\pm 1, \pm 2\}$，$q \in \{1\}$，$x \in \{\pm 1, \pm 2\}$。
+So $p \in \{\pm 1, \pm 2\}$ , $q \in \{1\}$ , $x \in \{\pm 1, \pm 2\}$ .
 
-**Step 3：逐一验证**
+**Step 3: Verify one by one**
 
-- $x = 1$：$1^2 - 2 = -1 \neq 0$ ✗
-- $x = -1$：$(-1)^2 - 2 = -1 \neq 0$ ✗
-- $x = 2$：$2^2 - 2 = 2 \neq 0$ ✗
-- $x = -2$：$(-2)^2 - 2 = 2 \neq 0$ ✗
+-  $x = 1$ ： $1^2 - 2 = -1 \neq 0$  ✗
+-  $x = -1$ ： $(-1)^2 - 2 = -1 \neq 0$  ✗
+-  $x = 2$ ： $2^2 - 2 = 2 \neq 0$  ✗
+-  $x = -2$ ： $(-2)^2 - 2 = 2 \neq 0$  ✗
 
-无一满足，故 $x^2 - 2 = 0$ 无有理根，$\sqrt{2}$ 是无理数。$\square$
+None of them are satisfied, so $x^2 - 2 = 0$ has no rational root and $\sqrt{2}$ is an irrational number. $\square$
 
-## 验算
+## Check calculation
 
-**验算方法 1：检查反证法每一步逻辑**
+**Check calculation method 1: Check the logic of each step of proof by contradiction**
 
-- 假设 $\sqrt{2} = p/q$，$\gcd(p,q)=1$ — 合理，任何有理数可写为最简分数 ✓
-- $p^2 = 2q^2$ — 平方运算正确 ✓
-- "$p^2$ 偶 $\implies$ $p$ 偶" — 由逆否命题 "$p$ 奇 $\implies$ $p^2$ 奇" 证明，计算 $(2k+1)^2 = 4k^2+4k+1 = 2(2k^2+2k)+1$ 正确 ✓
-- $q^2 = 2k^2$ — 代入 $p=2k$ 得 $4k^2=2q^2$，即 $q^2=2k^2$ ✓
-- "$q^2$ 偶 $\implies$ $q$ 偶" — 同理正确 ✓
-- $\gcd(p,q) \geq 2$ 矛盾 — 两个偶数必有公因子 2 ✓
+- Assumptions $\sqrt{2} = p/q$, $\gcd(p,q)=1$ — reasonable, any rational number can be written as the simplest fraction ✓
+- $p^2 = 2q^2$ — Square operation is correct ✓
+- "$p^2$ even $\implies$ $p$ even" — proved by the inverse proposition "$p$ odd $\implies$ $p^2$ odd", calculate $(2k+1)^2 = 4k^2+4k+1 = 2(2k^2+2k)+1$ correct ✓
+- $q^2 = 2k^2$ — Substitute $p=2k$ to get $4k^2=2q^2$, which is $q^2=2k^2$ ✓
+- "$q^2$ even $\implies$ $q$ even" — The same logic is correct ✓
+- $\gcd(p,q) \geq 2$ Contradiction — Two even numbers must have a common factor 2 ✓
 
-**验算方法 2：检查有理根定理的应用**
+**Check method 2: Check the application of the rational root theorem**
 
-- 多项式的首项系数为 $1$，常数项为 $-2$ ✓
-- 有理根定理要求 $p \mid \text{常数项}$，$q \mid \text{首项系数}$ ✓
-- 候选值 $\pm 1, \pm 2$ 完备 ✓
-- 代入验证无误 ✓
+- The leading coefficient of the polynomial is $1$ and the constant term is $-2$ ✓
+- The rational root theorem requires $p \mid \text{constant term}$, $q \mid \text{leading term coefficient}$ ✓
+- Candidate value $\pm 1, \pm 2$ Complete ✓
+- Substitution verification is correct ✓
 
-**验算方法 3：边界检查**
+**Check Calculation Method 3: Boundary Check**
 
-$p = 0$ 时 $\sqrt{2} = 0/q = 0$，但 $0^2 = 0 \neq 2$，故 $\sqrt{2} \neq 0$，$p = 0$ 不可能是解。排除该边界情况。
+$p = 0$ is $\sqrt{2} = 0/q = 0$ , but $0^2 = 0 \neq 2$ , so $\sqrt{2} \neq 0$ , $p = 0$ cannot be solutions. Exclude this edge case.
 
-**验算方法 4：数值验证**
+**Verification method 4: Numerical verification**
 
-$\sqrt{2} \approx 1.4142135623730951$，该数值的小数部分无限不循环（无理数的特征），与结论一致 ✓
+$\sqrt{2} \approx 1.4142135623730951$, the decimal part of this value is infinite and non-cyclic (characteristic of irrational numbers), consistent with the conclusion ✓
 
-## 最终答案
+## Final answer
 
-$$\sqrt{2} \notin \mathbb{Q}, \quad \text{即 } \sqrt{2} \text{ 是无理数}$$
+$$\sqrt{2} \notin \mathbb{Q}, \quad \text{i.e. } \sqrt{2} \text{ is an irrational number}$$
 
-两种证明方法均得出相同结论，且验证确认每一步推理正确无误。
+Both proof methods lead to the same conclusion, and verification confirms that each step of reasoning is correct.
 
-## 易错点
-1. **忘记设定 $\gcd(p, q) = 1$**：不假设互质则无法导出矛盾（$p, q$ 都可能是偶数但不矛盾，因为可以同时除以 2 继续约分）
-2. **混淆逆命题与逆否命题**：证明 "$p^2$ 偶 $\implies$ $p$ 偶" 用的是逆否命题 "$p$ 奇 $\implies$ $p^2$ 奇"，不是逆命题 "$p$ 偶 $\implies$ $p^2$ 偶"（后者也是对的但不是这里需要的）
-3. **错误推广**：不要以为用类似方法就能证明所有平方根都是无理数。例如 $\sqrt{4} = 2$ 是有理数，用同样的推理会在 "$p^2 = 4q^2$" 步骤后无法导出奇偶矛盾
-4. **有理根定理使用条件**：有理根定理要求多项式系数为整数，且首项系数不为零，此处 $x^2 - 2$ 满足条件
+## Easy to make mistakes
+1. **Forgot to set $\gcd(p, q) = 1$ **: Without assuming mutual prime, the contradiction cannot be derived ($p, q$ may be an even number but is not contradictory, because it can be divided by 2 at the same time to continue the reduction)
+2. **Confusing the inverse proposition and the inverse proposition**: Prove that "$p^2$ even $\implies$ $p$ even" uses the inverse proposition "$p$ odd $\implies$ $p^2$ odd", not the inverse proposition "$p$ even" $\implies$ $p^2$ even" (the latter is also correct but not required here)
+3. **Wrong Generalization**: Do not think that using similar methods can prove that all square roots are irrational. For example, $\sqrt{4} = 2$ is a rational number. Using the same reasoning, the odd-even contradiction cannot be derived after the " $p^2 = 4q^2$ " step.
+4. **Conditions for using the rational root theorem**: The rational root theorem requires that the polynomial coefficients are integers and the first coefficient is not zero. Here $x^2 - 2$ meets the conditions
diff --git a/examples/real_analysis.md b/examples/real_analysis.md
index 964ba05..9bb2a50 100644
--- a/examples/real_analysis.md
+++ b/examples/real_analysis.md
@@ -1,108 +1,108 @@
-# 一致连续性的证明与反例构造
+# Proof and counterexample construction of consistent continuity
 
-## 用户输入
-证明函数 $f(x) = x^2$ 在区间 $[0, 1]$ 上一致连续，但在 $[0, \infty)$ 上不一致连续。
+## User input
+Show that the function $f(x) = x^2$ is consistently continuous on the interval $[0, 1]$ but not on the interval $[0, \infty)$.
 
-## Skill 分类
-实分析
+## Skill Category
+real analysis
 
-## 题意解析
-- **函数**：$f(x) = x^2$，定义在实数域。
-- **需要证明的两个命题**：
-  1. $f$ 在闭区间 $[0, 1]$ 上一致连续。
-  2. $f$ 在无界区间 $[0, \infty)$ 上不一致连续。
-- **关键区别**：$[0, 1]$ 是紧集（闭且有界），闭区间上的连续函数必然一致连续（Cantor 定理，可用于佐证但不能替代构造性证明）。$[0, \infty)$ 不是紧集。
-- **一致连续的定义**：$\forall \varepsilon > 0,\ \exists \delta > 0,\ \forall x, y \in D,\ |x - y| < \delta \implies |f(x) - f(y)| < \varepsilon$。
-- **不一致连续的含义**：存在 $\varepsilon_0 > 0$，使得对任意 $\delta > 0$，总存在 $x, y$ 满足 $|x - y| < \delta$ 但 $|f(x) - f(y)| \geq \varepsilon_0$。
+## Question meaning analysis
+- **Function**: $f(x) = x^2$, defined in the real number domain.
+- **Two propositions that need to be proved**:
+1. $f$ is uniformly continuous on the closed interval $[0, 1]$.
+2. $f$ is not consistent on the unbounded interval $[0, \infty)$.
+- **Key difference**: $[0, 1]$ is a compact set (closed and bounded), and continuous functions on the closed interval must be consistent and continuous (Cantor's theorem, which can be used to support but cannot replace constructive proof). $[0, \infty)$ is not a compact set.
+- **Consistent Definition**: $\forall \varepsilon > 0,\ \exists \delta > 0,\ \forall x, y \in D,\ |x - y| < \delta \implies |f(x) - f(y)| < \varepsilon$.
+- **Inconsistent Continuous Meaning**: There exists $\varepsilon_0 > 0$ such that for any $\delta > 0$ , there is always $x, y$ that satisfies $|x - y| < \delta$ but not $|f(x) - f(y)| \geq \varepsilon_0$ .
 
-## 方法选择
-**证明一致连续**：利用恒等式 $|x^2 - y^2| = |x + y| \cdot |x - y|$，在有界区间上放缩 $|x + y|$。这是标准的 $\varepsilon$-$\delta$ 构造方法。
+## Method selection
+**Prove consistent continuity**: Use the identity $|x^2 - y^2| = |x + y| \cdot |x - y|$ and scale $|x + y|$ on the bounded interval. This is the standard $\varepsilon$ - $\delta$ constructor.
 
-**证明不一致连续**：构造序列反例，选取 $x_n$ 和 $y_n$ 使得 $|x_n - y_n| \to 0$ 但 $|f(x_n) - f(y_n)|$ 保持有下界。常用的构造是 $x_n = n + \frac{1}{n}$，$y_n = n$，利用平方差展开中的 $2n$ 因子发散。
+**Prove inconsistent continuity**: Construct a sequence counterexample, select $x_n$ and $y_n$ so that $|x_n - y_n| \to 0$ but $|f(x_n) - f(y_n)|$ remains lower bound. Commonly used constructs are $x_n = n + \frac{1}{n}$, $y_n = n$, which use the $2n$ factor divergence in the squared difference expansion.
 
-## 解题过程
+## Problem solving process
 
-### 第一部分：$f(x) = x^2$ 在 $[0, 1]$ 上一致连续
+### Part One: $f(x) = x^2$ is consistent on $[0, 1]$
 
-对任意 $x, y \in [0, 1]$：
+For any $x, y \in [0, 1]$ :
 
-$$|f(x) - f(y)| = |x^2 - y^2| = |x + y| \cdot |x - y|$$
+ $$|f(x) - f(y)| = |x^2 - y^2| = |x + y| \cdot |x - y|$$
 
-由于 $x, y \in [0, 1]$，所以 $|x + y| \leq 1 + 1 = 2$。因此：
+Since $x, y \in [0, 1]$ , so $|x + y| \leq 1 + 1 = 2$ . therefore:
 
-$$|f(x) - f(y)| \leq 2|x - y|$$
+ $$|f(x) - f(y)| \leq 2|x - y|$$
 
-给定 $\varepsilon > 0$，取 $\delta = \dfrac{\varepsilon}{2} > 0$。则当 $|x - y| < \delta$ 时：
+Given $\varepsilon > 0$ , take $\delta = \dfrac{\varepsilon}{2} > 0$ . Then when $|x - y| < \delta$:
 
-$$|f(x) - f(y)| \leq 2|x - y| < 2 \cdot \frac{\varepsilon}{2} = \varepsilon$$
+ $$|f(x) - f(y)| \leq 2|x - y| < 2 \cdot \frac{\varepsilon}{2} = \varepsilon$$
 
-这正是一致连续的定义。$\square$
+This is the very definition of consistent continuity. $\square$
 
-### 第二部分：$f(x) = x^2$ 在 $[0, \infty)$ 上不一致连续
+### Part Two: $f(x) = x^2$ is inconsistently contiguous on $[0, \infty)$
 
-证明思路：设存在一致连续的 $\delta(\varepsilon)$，则对足够大的 $x$，$|f(x+\delta/2) - f(x)|$ 应能被 $\varepsilon$ 控制。但
+Proof idea: Assuming there is a consistent and continuous $\delta(\varepsilon)$, then for sufficiently large $x$, $|f(x+\delta/2) - f(x)|$ should be controlled by $\varepsilon$. but
 
-$$|f(x + h) - f(x)| = |(x+h)^2 - x^2| = |2xh + h^2|$$
+ $$|f(x + h) - f(x)| = |(x+h)^2 - x^2| = |2xh + h^2|$$
 
-对于固定的 $h$，当 $x \to \infty$ 时，这一差值趋于无穷大，矛盾。
+For a fixed $h$ , when $x \to \infty$ , this difference tends to infinity, paradoxically.
 
-正式证明：取 $\varepsilon_0 = 1$。对任意 $\delta > 0$，选取
+Formal proof: take $\varepsilon_0 = 1$ . For any $\delta > 0$ , select
 
-$$x = \frac{1}{\delta},\quad y = \frac{1}{\delta} + \frac{\delta}{2}$$
+ $$x = \frac{1}{\delta},\quad y = \frac{1}{\delta} + \frac{\delta}{2}$$
 
-则 $|x - y| = \dfrac{\delta}{2} < \delta$，但：
+then $|x - y| = \dfrac{\delta}{2} < \delta$ , but:
 
-$$|f(x) - f(y)| = \left|\left(\frac{1}{\delta} + \frac{\delta}{2}\right)^2 - \left(\frac{1}{\delta}\right)^2\right|$$
+ $$|f(x) - f(y)| = \left|\left(\frac{1}{\delta} + \frac{\delta}{2}\right)^2 - \left(\frac{1}{\delta}\right)^2\right|$$
 
-$$= \left|\frac{2}{\delta} \cdot \frac{\delta}{2} + \left(\frac{\delta}{2}\right)^2\right| = 1 + \frac{\delta^2}{4} \geq 1 = \varepsilon_0$$
+ $$= \left|\frac{2}{\delta} \cdot \frac{\delta}{2} + \left(\frac{\delta}{2}\right)^2\right| = 1 + \frac{\delta^2}{4} \geq 1 = \varepsilon_0$$
 
-因此对任意 $\delta > 0$，均存在 $x, y \in [0, \infty)$ 满足 $|x - y| < \delta$ 但 $|f(x) - f(y)| \geq 1$。
+Therefore for any $\delta > 0$ , there exists $x, y \in [0, \infty)$ that satisfies $|x - y| < \delta$ but not $|f(x) - f(y)| \geq 1$ .
 
-这违反了 $f$ 在 $[0, \infty)$ 上一致连续的定义。$\square$
+This violates the definition of $f$ being consistently continuous over $[0, \infty)$ . $\square$
 
-**备选构造**（更优雅）：取 $x_n = n$，$y_n = n + \frac{1}{n}$。则 $|x_n - y_n| = \frac{1}{n} \to 0$，但：
+**Alternative construction** (more elegant): take $x_n = n$ , $y_n = n + \frac{1}{n}$ . then $|x_n - y_n| = \frac{1}{n} \to 0$ , but:
 
-$$|f(x_n) - f(y_n)| = \left|n^2 - \left(n + \frac{1}{n}\right)^2\right| = \left|n^2 - n^2 - 2 - \frac{1}{n^2}\right| = 2 + \frac{1}{n^2} > 2$$
+ $$|f(x_n) - f(y_n)| = \left|n^2 - \left(n + \frac{1}{n}\right)^2\right| = \left|n^2 - n^2 - 2 - \frac{1}{n^2}\right| = 2 + \frac{1}{n^2} > 2$$
 
-与一致连续矛盾（一致连续要求 $|x_n - y_n| \to 0 \implies |f(x_n) - f(y_n)| \to 0$）。
+Contradicts consistent continuity (consistent continuity requires $|x_n - y_n| \to 0 \implies |f(x_n) - f(y_n)| \to 0$ ).
 
-## 验算
+## Check calculation
 
-### 方法一：$\varepsilon$-$\delta$ 结构检查
+### Method 1: $\varepsilon$ - $\delta$ structure check
 
-**一致连续证明的结构验证**：
-1. 从不等式放缩出发：$|f(x) - f(y)| \leq L|x - y|$（Lipschitz 条件）。
-2. 找到 Lipschitz 常数 $L = 2$，不依赖于 $x, y$ 的选取。
-3. 给定 $\varepsilon$，$\delta = \varepsilon/2$ 中的 $\delta$ 仅依赖于 $\varepsilon$，不依赖于具体的 $x$ 或 $y$。这正是"一致"的含义。✓
+**Structural Verification of Consistent Continuity Proofs**:
+1. Starting from the inequality scaling: $|f(x) - f(y)| \leq L|x - y|$ (Lipschitz condition).
+2. Find the Lipschitz constant $L = 2$, independent of the selection of $x, y$.
+3. Given $\varepsilon$ , $\delta$ in $\delta = \varepsilon/2$ only depends on $\varepsilon$ , not on the specific $x$ or $y$ . This is exactly what "consistent" means. ✓
 
-**不一致连续证明的结构验证**：
-1. 选取 $\varepsilon_0 = 1$ 是具体的正数。
-2. 对"任意"$\delta$ 构造了 $x, y$ 使得 $|x - y| < \delta$ 但 $|f(x) - f(y)| \geq \varepsilon_0$。$x$ 和 $y$ 的构造依赖于 $\delta$（即 $x = 1/\delta$），这是正确的——反例当然可以依赖于给定的 $\delta$。✓
+**Structural Verification of Inconsistent Continuity Proofs**:
+1. Select $\varepsilon_0 = 1$ to be a specific positive number.
+2. Constructing $x, y$ for "any" $\delta$ makes $|x - y| < \delta$ but not $|f(x) - f(y)| \geq \varepsilon_0$ . It is correct that the construction of $x$ and $y$ depends on $\delta$ (i.e. $x = 1/\delta$ ) - counterexamples can of course rely on the given $\delta$ . ✓
 
-### 方法二：边界情况的验证
+### Method 2: Verification of boundary conditions
 
-**$[0, 1]$ 上**：当 $x = 1$，$y = 0.999$（$|x-y| = 0.001$）：
-$$|f(1) - f(0.999)| = |1 - 0.998001| = 0.001999 \approx 2 \times 0.001$$
-符合 Lipschitz 界。✓
+** $[0, 1]$ on **: When $x = 1$ , $y = 0.999$ ( $|x-y| = 0.001$ ):
+ $$|f(1) - f(0.999)| = |1 - 0.998001| = 0.001999 \approx 2 \times 0.001$$
+Meets Lipschitz circles. ✓
 
-**$[0, \infty)$ 上**：当 $n = 100$ 时，$x_n = 100$，$y_n = 100.01$：
-$$|f(100) - f(100.01)| = |10000 - 10002.0001| = 2.0001 > 2$$
-即使两点间距仅 $0.01$，函数值差已超过 $2$。✓
+** $[0, \infty)$ on**: when $n = 100$ , $x_n = 100$ , $y_n = 100.01$ :
+ $$|f(100) - f(100.01)| = |10000 - 10002.0001| = 2.0001 > 2$$
+Even though the distance between the two points is only $0.01$, the function value difference exceeds $2$. ✓
 
-### 方法三：定理对照
+### Method 3: Theorem comparison
 
-Cantor 定理：紧集上的连续函数一致连续。$[0, 1]$ 是 $\mathbb{R}$ 中的紧集，$f(x) = x^2$ 连续，因此一致连续。这与第一部分构造性证明的结论一致。✓
+Cantor's theorem: Continuous functions on compact sets are uniformly continuous. $[0, 1]$ is a compact set in $\mathbb{R}$ , $f(x) = x^2$ is continuous, and therefore uniformly continuous. This is consistent with the conclusion of the first part of the constructive proof. ✓
 
-$[0, \infty)$ 不是紧集（无界），Cantor 定理不适用，因此一致连续性不能直接从连续性推出——反例确证了这一点。✓
+$[0, \infty)$ is not a compact set (unbounded), Cantor's theorem does not apply, so consistent continuity cannot be directly derived from continuity - the counterexample confirms this. ✓
 
-## 最终答案
+## Final answer
 
-$$\boxed{f(x) = x^2 \text{ 在 } [0, 1] \text{ 上一致连续}}$$
+$$\boxed{f(x) = x^2 \text{ is consistent and continuous on } [0, 1] \text{}}$$
 
-$$\boxed{f(x) = x^2 \text{ 在 } [0, \infty) \text{ 上不一致连续}}$$
+$$\boxed{f(x) = x^2 \text{ is inconsistently continuous on } [0, \infty) \text{}}$$
 
-## 易错点
-1. **Lipschitz 常数对有界区间的作用**：放缩 $|x + y| \leq 2$ 的关键是利用了区间的有界性。如果区间无界（如 $[0, \infty)$），$|x + y|$ 没有统一的上界，因此无法导出全局的 Lipschitz 估计。
-2. **反例的构造**：$|x_n - y_n| \to 0$ 但 $|f(x_n) - f(y_n)| \not\to 0$ 是不一致连续的**等价刻画**，不是充分条件。这个条件等价于"存在 $\varepsilon_0$ 使得 $\forall \delta, \exists x, y: |x-y|<\delta \land |f(x)-f(y)|\geq\varepsilon_0$"。构造时确保差额至少为常数（不趋于零）。
-3. **$\delta$ 与 $\varepsilon$ 的依赖**：一致连续要求 $\delta$ 仅依赖于 $\varepsilon$ 而不依赖于 $x$。在 $[0,1]$ 的证明中 $\delta = \varepsilon/2$ 正是如此。不要把逐点连续与一致连续混淆。
-4. **$x = 1/\delta$ 的讨论**：当 $\delta \to 0$ 时 $x \to \infty$，说明反例确实需要跑到无穷远处。这揭示了不一致连续的根源在于函数在远处的斜率无限增长。
+## Easy to make mistakes
+1. **The effect of Lipschitz constant on bounded intervals**: The key to scaling $|x + y| \leq 2$ is to take advantage of the boundedness of the interval. If the interval is unbounded (such as $[0, \infty)$ ), $|x + y|$ does not have a uniform upper bound, so the global Lipschitz estimate cannot be derived.
+2. **Construction of counterexamples**: $|x_n - y_n| \to 0$ but $|f(x_n) - f(y_n)| \not\to 0$ is an inconsistent and continuous **equivalent characterization** and is not a sufficient condition. This condition is equivalent to "there is $\varepsilon_0$ such that $\forall \delta, \exists x, y: |x-y|<\delta \land |f(x)-f(y)|\geq\varepsilon_0$". Constructed to ensure that the difference is at least constant (does not tend to zero).
+3. ** Dependence of $\delta$ and $\varepsilon$**: Consistent continuity requires that $\delta$ only depends on $\varepsilon$ and not on $x$. In the proof of $[0,1]$ $\delta = \varepsilon/2$ does exactly this. Don't confuse point-by-point continuity with consistent continuity.
+4. ** Discussion of $x = 1/\delta$**: When $\delta \to 0$ is $x \to \infty$, it means that the counterexample does need to go to infinity. This reveals that the source of inconsistent continuity lies in the infinite growth of the slope of the function in the distance.
diff --git a/examples/sequences.md b/examples/sequences.md
index beb96f9..86727b4 100644
--- a/examples/sequences.md
+++ b/examples/sequences.md
@@ -1,109 +1,109 @@
-# 递推数列求通项与前 n 项和
+# Find the sum of the general term and the first n terms of the recursive sequence
 
-## 用户输入
-已知数列 $\{a_n\}$ 满足 $a_1 = 1$，$a_{n+1} = 2a_n + 3$，求 $\{a_n\}$ 的通项公式与前 $n$ 项和 $S_n$。
+## User input
+It is known that the sequence $\{a_n\}$ satisfies $a_1 = 1$ and $a_{n+1} = 2a_n + 3$. Find the general formula of $\{a_n\}$ and the first $n$ terms and $S_n$.
 
-## Skill 分类
-数列
+## Skill Category
+Sequence
 
-## 题意解析
-- **已知条件**：首项 $a_1 = 1$，递推关系 $a_{n+1} = 2a_n + 3$（$n \geq 1$）。
-- **求解目标**：通项 $a_n$ 与前 $n$ 项和 $S_n = \sum_{k=1}^n a_k$。
-- **递推类型识别**：一阶线性非齐次递推，形如 $a_{n+1} = pa_n + q$，其中 $p = 2$，$q = 3$，且 $p \neq 1$。
-- **定义域**：$n \in \mathbb{N}^+$。
-- **解的唯一性**：由首项和递推关系唯一确定整个数列。
+## Question meaning analysis
+- **Known conditions**: first item $a_1 = 1$, recursion relationship $a_{n+1} = 2a_n + 3$ ($n \geq 1$).
+- **Solution Objective**: General term $a_n$ and previous $n$ terms and $S_n = \sum_{k=1}^n a_k$.
+- **Recursion type identification**: First-order linear non-homogeneous recursion, in the form of $a_{n+1} = pa_n + q$, where $p = 2$, $q = 3$, and $p \neq 1$.
+- **Domain**: $n \in \mathbb{N}^+$.
+- **Uniqueness of solution**: The entire sequence is uniquely determined by the leading term and the recursion relationship.
 
-## 方法选择
-**首选方法**：特征方程法（待定系数法），通过构造等比数列将非齐次递推转化为齐次递推求解。核心思路：找一个常数 $\lambda$ 使 $a_{n+1} + \lambda = p(a_n + \lambda)$，则 $\{a_n + \lambda\}$ 为等比数列。
+## Method selection
+**Preferred method**: Characteristic equation method (undetermined coefficient method), which converts non-homogeneous recursion into homogeneous recursion solution by constructing a geometric sequence. The core idea: Find a constant $\lambda$ to make $a_{n+1} + \lambda = p(a_n + \lambda)$, then $\{a_n + \lambda\}$ is a geometric sequence.
 
-**备选方法**：迭代展开法，从 $a_n$ 逐项回推到 $a_1$；生成函数法。特征方程法最为简洁高效。
+**Alternative method**: iterative expansion method, from $a_n$ to $a_1$ item by item; generating function method. The characteristic equation method is the most concise and efficient.
 
-## 解题过程
+## Problem solving process
 
-### 第一步：求通项 $a_n$
+### Step 1: Find the general term $a_n$
 
-设存在常数 $\lambda$，使得：
+Suppose there is a constant $\lambda$ such that:
 
-$$a_{n+1} + \lambda = 2(a_n + \lambda)$$
+ $$a_{n+1} + \lambda = 2(a_n + \lambda)$$
 
-展开右边：$a_{n+1} + \lambda = 2a_n + 2\lambda$
+Expand the right side: $a_{n+1} + \lambda = 2a_n + 2\lambda$
 
-与原递推 $a_{n+1} = 2a_n + 3$ 比较得：
+Compare with the original recursion $a_{n+1} = 2a_n + 3$:
 
-$$\lambda = 2\lambda - 3 \quad \Rightarrow \quad \lambda = 3$$
+ $$\lambda = 2\lambda - 3 \quad \Rightarrow \quad \lambda = 3$$
 
-因此：
+therefore:
 
-$$a_{n+1} + 3 = 2(a_n + 3)$$
+ $$a_{n+1} + 3 = 2(a_n + 3)$$
 
-这说明数列 $\{a_n + 3\}$ 是公比为 $2$ 的等比数列。
+This shows that the sequence $\{a_n + 3\}$ is a geometric sequence whose common ratio is $2$.
 
-求 $\{a_n + 3\}$ 的首项：
+Find the first term of $\{a_n + 3\}$:
 
-$$a_1 + 3 = 1 + 3 = 4$$
+ $$a_1 + 3 = 1 + 3 = 4$$
 
-所以：
+so:
 
-$$a_n + 3 = 4 \cdot 2^{n-1} = 2^{n+1}$$
+ $$a_n + 3 = 4 \cdot 2^{n-1} = 2^{n+1}$$
 
-$$a_n = 2^{n+1} - 3 \quad (n \in \mathbb{N}^+)$$
+ $$a_n = 2^{n+1} - 3 \quad (n \in \mathbb{N}^+)$$
 
-### 第二步：求前 $n$ 项和 $S_n$
+### Step 2: Find the first $n$ items and $S_n$
 
-$$S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (2^{k+1} - 3) = \sum_{k=1}^n 2^{k+1} - \sum_{k=1}^n 3$$
+ $$S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (2^{k+1} - 3) = \sum_{k=1}^n 2^{k+1} - \sum_{k=1}^n 3$$
 
-第一项是等比数列求和（首项 $2^2 = 4$，公比 $2$，项数 $n$）：
+The first term is the sum of the geometric sequence (first term $2^2 = 4$ , common ratio $2$ , number of terms $n$ ):
 
-$$\sum_{k=1}^n 2^{k+1} = 4 \cdot \frac{2^n - 1}{2 - 1} = 4(2^n - 1) = 2^{n+2} - 4$$
+ $$\sum_{k=1}^n 2^{k+1} = 4 \cdot \frac{2^n - 1}{2 - 1} = 4(2^n - 1) = 2^{n+2} - 4$$
 
-第二项为常数级数：
+The second term is a constant series:
 
-$$\sum_{k=1}^n 3 = 3n$$
+ $$\sum_{k=1}^n 3 = 3n$$
 
-因此：
+therefore:
 
-$$S_n = (2^{n+2} - 4) - 3n = 2^{n+2} - 3n - 4$$
+ $$S_n = (2^{n+2} - 4) - 3n = 2^{n+2} - 3n - 4$$
 
-## 验算
+## Check calculation
 
-### 方法一：逐项验证
+### Method 1: Verify item by item
 
-通过通项公式计算 $a_1, a_2, a_3$，再通过递推公式交叉验证。
+Calculate $a_1, a_2, a_3$ through the general formula, and then cross-validate through the recursive formula.
 
-| $n$ | 通项 $a_n = 2^{n+1} - 3$ | 递推 $a_{n} = 2a_{n-1} + 3$ |
+| $n$ | General term $a_n = 2^{n+1} - 3$ | Recursion $a_{n} = 2a_{n-1} + 3$ |
 |-----|---------------------------|------------------------------|
-| 1   | $2^2 - 3 = 4 - 3 = 1$     | （已知 $a_1 = 1$）            |
-| 2   | $2^3 - 3 = 8 - 3 = 5$     | $2 \cdot 1 + 3 = 5$          |
-| 3   | $2^4 - 3 = 16 - 3 = 13$    | $2 \cdot 5 + 3 = 13$         |
+| 1 | $2^2 - 3 = 4 - 3 = 1$ | (known as $a_1 = 1$ ) |
+| 2   |  $2^3 - 3 = 8 - 3 = 5$      |  $2 \cdot 1 + 3 = 5$           |
+| 3   |  $2^4 - 3 = 16 - 3 = 13$     |  $2 \cdot 5 + 3 = 13$          |
 
-全部一致。
+All consistent.
 
-### 方法二：代入验证
+### Method 2: Substitution verification
 
-将 $a_n = 2^{n+1} - 3$ 代入原递推式：
+Substitute $a_n = 2^{n+1} - 3$ into the original recursion:
 
-$$a_{n+1} = 2^{(n+1)+1} - 3 = 2^{n+2} - 3$$
+ $$a_{n+1} = 2^{(n+1)+1} - 3 = 2^{n+2} - 3$$
 
-$$2a_n + 3 = 2(2^{n+1} - 3) + 3 = 2^{n+2} - 6 + 3 = 2^{n+2} - 3$$
+ $$2a_n + 3 = 2(2^{n+1} - 3) + 3 = 2^{n+2} - 6 + 3 = 2^{n+2} - 3$$
 
-左边 = 右边，递推式成立。
+Left side = right side, the recursion holds.
 
-### 方法三：前 $n$ 项和验证
+### Method 3: First $n$ items and verification
 
-$n = 1$：$S_1 = a_1 = 1$，公式给出 $2^{3} - 3 \cdot 1 - 4 = 8 - 3 - 4 = 1$。✓
+$n = 1$ : $S_1 = a_1 = 1$ , the formula gives $2^{3} - 3 \cdot 1 - 4 = 8 - 3 - 4 = 1$ . ✓
 
-$n = 2$：$S_2 = 1 + 5 = 6$，公式给出 $2^{4} - 3 \cdot 2 - 4 = 16 - 6 - 4 = 6$。✓
+$n = 2$ : $S_2 = 1 + 5 = 6$ , the formula gives $2^{4} - 3 \cdot 2 - 4 = 16 - 6 - 4 = 6$ . ✓
 
-$n = 3$：$S_3 = 1 + 5 + 13 = 19$，公式给出 $2^{5} - 3 \cdot 3 - 4 = 32 - 9 - 4 = 19$。✓
+$n = 3$ : $S_3 = 1 + 5 + 13 = 19$ , the formula gives $2^{5} - 3 \cdot 3 - 4 = 32 - 9 - 4 = 19$ . ✓
 
-## 最终答案
+## Final answer
 
-$$\boxed{a_n = 2^{n+1} - 3 \quad (n \in \mathbb{N}^+)}$$
+ $$\boxed{a_n = 2^{n+1} - 3 \quad (n \in \mathbb{N}^+)}$$
 
-$$\boxed{S_n = 2^{n+2} - 3n - 4 \quad (n \in \mathbb{N}^+)}$$
+ $$\boxed{S_n = 2^{n+2} - 3n - 4 \quad (n \in \mathbb{N}^+)}$$
 
-## 易错点
-1. **待定系数求解 $\lambda$**：方程是 $\lambda = p\lambda - q$ 而非 $\lambda = p\lambda + q$，注意符号。
-2. **$\lambda$ 和 $p\lambda - q$ 的关系**：在推导 $\lambda = 2\lambda - 3$ 时，$\lambda = a_{n+1} + \lambda$ 方程中的 $\lambda$ 既是未知数又出现在两边，不要弄混。
-3. **等比数列求和**：求 $\sum 2^{k+1}$ 时首项不是 $2$ 而是 $2^2 = 4$，公比 $2$，项数 $n$，而非 $n-1$。
-4. **$p = 1$ 的特殊情况**：当 $p = 1$ 时（如 $a_{n+1} = a_n + 3$），不能用此方法构造等比数列，应直接用等差数列处理。
+## Easy to make mistakes
+1. **Undetermined coefficient solution $\lambda$ **: The equation is $\lambda = p\lambda - q$ instead of $\lambda = p\lambda + q$, pay attention to the sign.
+2. **Relationship between $\lambda$ and $p\lambda - q$**: When deriving $\lambda = 2\lambda - 3$, $\lambda$ in the $\lambda = a_{n+1} + \lambda$ equation is both an unknown variable and appears on both sides, so don’t get confused.
+3. **Sum of geometric sequence**: When finding $\sum 2^{k+1}$, the first term is not $2$ but $2^2 = 4$, the common ratio is $2$, and the number of terms is $n$, not $n-1$.
+4. **Special case of $p = 1$**: When $p = 1$ (such as $a_{n+1} = a_n + 3$), this method cannot be used to construct a geometric sequence, and it should be processed directly with an arithmetic sequence.
diff --git a/examples/solution_checking.md b/examples/solution_checking.md
index 84ffe0b..a9d3aa1 100644
--- a/examples/solution_checking.md
+++ b/examples/solution_checking.md
@@ -1,95 +1,95 @@
-# 解题过程审查：二次方程求解
+# Problem-solving process review: solving quadratic equations
 
-## 用户输入
-一位学生解方程 $x^2 - 5x + 6 = 0$，写出 $(x-2)(x-3) = 0$，得到 $x = 2, 3$。然后学生写道："再看 $(x-6)^2 = x^2 - 12x + 36 = 0$，似乎不一样，所以 $x = 6$ 不是解。" 请检查这个解题过程。
+## User input
+A student solves the equation $x^2 - 5x + 6 = 0$, writes $(x-2)(x-3) = 0$, and gets $x = 2, 3$. Then the student writes: "Looking at $(x-6)^2 = x^2 - 12x + 36 = 0$ again, it seems different, so $x = 6$ is not the solution." Please check this problem-solving process.
 
-## Skill 分类
-代数 / 解题审核
+## Skill Category
+Algebra/Problem Solving Review
 
-## 题意解析
-- **学生答案**：$x = 2, 3$（正确）
-- **需要审核的内容**：
-  1. 主要解题步骤是否正确
-  2. 学生对 $x = 6$ 的检验是否正确（推理逻辑是否合理）
-  3. 是否有不必要的步骤
-- **目标**：指出正确之处、纠正错误推理、给出标准验证方式
+## Question meaning analysis
+- **Student Answer**: $x = 2, 3$ (Correct)
+- **Content that needs to be reviewed**:
+1. Are the main problem-solving steps correct?
+2. Whether the student’s test of $x = 6$ is correct (whether the reasoning logic is reasonable)
+3. Are there any unnecessary steps?
+- **Objective**: Point out what is correct, correct wrong reasoning, and give standard verification methods
 
-## 方法选择
-采用**逐步骤分析法**：逐一审查学生的每一步推理，区分正确步骤与错误/冗余步骤，然后给出标准答案和验证。
+## Method selection
+Use the **step-by-step analysis method**: review each step of the student's reasoning one by one, distinguish the correct steps from the wrong/redundant steps, and then give standard answers and verifications.
 
-## 解题过程
+## Problem solving process
 
-### 审查学生的解题
+### Review student solutions
 
-**步骤 1：因式分解** $(x-2)(x-3) = 0$
+**Step 1: Factorization** $(x-2)(x-3) = 0$
 
-审查：$x^2 - 5x + 6 = (x-2)(x-3)$ ✓
+Review: $x^2 - 5x + 6 = (x-2)(x-3)$ ✓
 
-验证：展开 $(x-2)(x-3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6$，完全正确。
+Verification: Expand $(x-2)(x-3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6$ , absolutely correct.
 
-**步骤 2：应用零因子性质** → $x = 2$ 或 $x = 3$
+**Step 2: Apply the zero-factor property** → $x = 2$ or $x = 3$
 
-审查：零因子性质——若两个实数乘积为零，则至少一个因子为零。即 $(x-2)(x-3) = 0 \iff x-2=0 \text{ 或 } x-3=0 \iff x=2 \text{ 或 } x=3$ ✓
+Review: Zero Factor Property - If the product of two real numbers is zero, then at least one factor is zero. That is $(x-2)(x-3) = 0 \iff x-2=0 \text{ or } x-3=0 \iff x=2 \text{ or } x=3$ ✓
 
-**步骤 3：检验 $x = 6$**
+**Step 3: Verify $x = 6$ **
 
-审查：学生写 "$(x-6)^2 = x^2 - 12x + 36 = 0$ 似乎不一样，所以 $x = 6$ 不是解"
+Review: Student writes "$(x-6)^2 = x^2 - 12x + 36 = 0$ seems different, so $x = 6$ is not the solution"
 
-这段推理有问题：
-1. **逻辑错误**：$(x-6)$ 不是原始多项式 $x^2-5x+6$ 的因式，展开 $(x-6)^2$ 与原始方程没有任何关系（除非试图比较两个不同的多项式，而这本身不是有效的验证方法）
-2. **不必要的步骤**：正确答案 $x=2,3$ 已由因式分解得出，额外检验 $x=6$ 纯属多余
-3. **验证方法不对**：正确的验证是将 $x=6$ 代入原方程计算 $6^2 - 5 \cdot 6 + 6$ 是否等于 $0$
+There is a problem with this reasoning:
+1. **Logic Error**: $(x-6)$ is not a factor of the original polynomial $x^2-5x+6$, and the expansion $(x-6)^2$ has nothing to do with the original equation (unless you are trying to compare two different polynomials, which in itself is not a valid verification method)
+2. **Unnecessary step**: The correct answer $x=2,3$ is already factored, and the additional check $x=6$ is purely redundant
+3. **Incorrect verification method**: The correct verification is to substitute $x=6$ into the original equation to calculate whether $6^2 - 5 \cdot 6 + 6$ is equal to $0$
 
-### 正确的验证方法
+### Correct verification method
 
-**验证 $x = 2$：**
-$$2^2 - 5 \cdot 2 + 6 = 4 - 10 + 6 = 0 \quad \checkmark$$
+**Verify $x = 2$:**
+ $$2^2 - 5 \cdot 2 + 6 = 4 - 10 + 6 = 0 \quad \checkmark$$
 
-**验证 $x = 3$：**
-$$3^2 - 5 \cdot 3 + 6 = 9 - 15 + 6 = 0 \quad \checkmark$$
+**Verify $x = 3$:**
+ $$3^2 - 5 \cdot 3 + 6 = 9 - 15 + 6 = 0 \quad \checkmark$$
 
-**验证 $x = 6$（说明为何不是解）：**
-$$6^2 - 5 \cdot 6 + 6 = 36 - 30 + 6 = 12 \neq 0 \quad \text{确实不是解}$$
+**Verify $x = 6$ (explain why it is not a solution):**
+$$6^2 - 5 \cdot 6 + 6 = 36 - 30 + 6 = 12 \neq 0 \quad \text{It is indeed not a solution}$$
 
-### 为什么因式分解就够了？
+### Why is factoring enough?
 
-二次多项式 $x^2 - 5x + 6$ 最多有两个根（代数学基本定理）。因式分解得到两个根 $2$ 和 $3$，它们**已经穷举了所有可能的根**。不需要（也没有理由）检验 $x=6$——除非你怀疑因式分解有误（此时应验证 $x=2$ 和 $x=3$，而非随便选一个其他数）。
+A quadratic polynomial $x^2 - 5x + 6$ has at most two roots (Fundamental Theorem of Algebra). Factoring yields two roots $2$ and $3$ , which exhaustively exhaust all possible roots. There is no need (and no reason) to check $x=6$ - unless you suspect that the factorization is wrong (in which case you should check $x=2$ and $x=3$ rather than picking any other number).
 
-## 验算
+## Check calculation
 
-**验算方法 1：直接代入验证**
+**Verification method 1: Directly enter the verification**
 
-| $x$ | $x^2$ | $-5x$ | $+6$ | 合计 | 判定 |
+| $x$ | $x^2$ | $-5x$ | $+6$ | Total | Judgment |
 |-----|-------|-------|------|------|------|
-| $2$ | $4$ | $-10$ | $6$ | $0$ | ✓ 是根 |
-| $3$ | $9$ | $-15$ | $6$ | $0$ | ✓ 是根 |
-| $6$ | $36$ | $-30$ | $6$ | $12$ | ✗ 不是根 |
+| $2$ | $4$ | $-10$ | $6$ | $0$ | ✓ is the root |
+| $3$ | $9$ | $-15$ | $6$ | $0$ | ✓ is the root |
+| $6$ | $36$ | $-30$ | $6$ | $12$ | ✗ Not a root |
 
-**验算方法 2：求根公式验证**
+**Verification method 2: Verification of root formula**
 
-对于 $ax^2 + bx + c = 0$（$a=1, b=-5, c=6$）：
+For $ax^2 + bx + c = 0$ ( $a=1, b=-5, c=6$ ):
 
-$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2}$$
+ $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2}$$
 
-得 $x = 3$ 或 $x = 2$，与因式分解一致 ✓
+Get $x = 3$ or $x = 2$, consistent with factoring ✓
 
-**验算方法 3：韦达定理验证**
+**Check Calculation Method 3: Verification of Vedic Theorem**
 
-设两根为 $x_1, x_2$，则：
-- $x_1 + x_2 = 5$ ✓（$2+3=5$，等于 $-b/a$）
-- $x_1 \cdot x_2 = 6$ ✓（$2 \times 3 = 6$，等于 $c/a$）
+Assume the two roots are $x_1, x_2$ , then:
+- $x_1 + x_2 = 5$ ✓ ( $2+3=5$ , equal to $-b/a$ )
+- $x_1 \cdot x_2 = 6$ ✓ ( $2 \times 3 = 6$ , equal to $c/a$ )
 
-## 最终答案
+## Final answer
 
-学生得出的答案 $x = 2$ 或 $x = 3$ **正确**。但解题过程中：
-- 因式分解和零因子性质的应用 **正确** ✓
-- 对 $x=6$ 的检验 **不必要且推理不当** ✗（$(x-6)^2$ 与原始方程无关，不是有效验证）
+The student's answer $x = 2$ or $x = 3$ is **correct**. But in the process of solving the problem:
+- Application of factorization and zero factor property **Correct** ✓
+- Test of $x=6$ **unnecessary and poorly reasoned** ✗ ($(x-6)^2$ has nothing to do with the original equation and is not a valid verification)
 
-正确的验证方法是直接将候选解代入原方程计算。
+The correct verification method is to directly substitute the candidate solution into the original equation for calculation.
 
-## 易错点
-1. **混淆因式与验证**：用 $(x-6)^2$ 去"比较"是无效方法——验证应始终回到**原方程**
-2. **画蛇添足**：根已经正确求出后，无需额外检验与问题无关的数字
-3. **零因子性质的适用范围**：$(x-2)(x-3)=0 \iff x=2 \text{ 或 } x=3$ 正确的前提是在实数（或复数）域中，且因式分解是精确的（不是近似的）
-4. **验证的通用原则**：任何声称的解 $x_0$ 必须满足 $f(x_0) = 0$，只需代入验证即可，无需构造无关多项式
-5. **二次方程根的个数**：n 次多项式最多有 n 个不同的根，找到 n 个后即穷举完毕
+## Easy to make mistakes
+1. **Confusing Factors and Verification**: Using $(x-6)^2$ to "compare" is an invalid method - verification should always go back to the **original equation**
+2. **Superfluous**: After the roots have been found correctly, there is no need to additionally check numbers that are irrelevant to the problem.
+3. **Applicable scope of zero factor property**: $(x-2)(x-3)=0 \iff x=2 \text{ or } x=3$ The correct premise is that it is in the real number (or complex number) domain, and the factorization is exact (not approximate)
+4. **General principles of verification**: Any claimed solution $x_0$ must satisfy $f(x_0) = 0$, which can be verified by substituting it without constructing an irrelevant polynomial.
+5. **The number of roots of a quadratic equation**: A polynomial of degree n can have at most n different roots. Once n are found, the exhaustive list is complete.
diff --git a/examples/topology.md b/examples/topology.md
index 5e1d40d..00b1689 100644
--- a/examples/topology.md
+++ b/examples/topology.md
@@ -1,87 +1,87 @@
-# 度量空间中有限集的闭性证明
+# Proof of closure of finite sets in metric space
 
-## 用户输入
-在度量空间 $(X, d)$ 中，证明任意有限集是闭集。
+## User input
+In the metric space $(X, d)$, show that any finite set is closed.
 
-## Skill 分类
-拓扑学 / 点集拓扑基础
+## Skill Category
+Topology/Basics of Point Set Topology
 
-## 题意解析
-- **条件**：$(X, d)$ 是任意度量空间，$F = \{x_1, x_2, \ldots, x_n\}$ 是 $X$ 中的有限子集（$n \in \mathbb{N}^+$）
-- **目标**：证明 $F$ 是闭集，即 $F$ 的补集 $X \setminus F$ 是开集
-- **开集定义**：$U \subseteq X$ 是开集，当且仅当对任意 $y \in U$，存在 $\varepsilon > 0$ 使得开球 $B(y, \varepsilon) \subseteq U$
-- **度量空间任意性**：该结论不依赖于 $\mathbb{R}^n$ 的特殊性质，对任意度量空间都成立
-- **解的唯一性**：证明路径唯一（补集为开），验证需要检查多个细节
+## Question meaning analysis
+- **Condition**: $(X, d)$ is an arbitrary metric space, $F = \{x_1, x_2, \ldots, x_n\}$ is a finite subset of $X$ ( $n \in \mathbb{N}^+$ )
+- **Goal**: Prove that $F$ is a closed set, that is, the complement of $F$ and $X \setminus F$ is an open set
+- **Open set definition**: $U \subseteq X$ is an open set if and only if for any $y \in U$, there exists $\varepsilon > 0$ such that tee $B(y, \varepsilon) \subseteq U$
+- **Metric space arbitrary**: This conclusion does not depend on the special properties of $\mathbb{R}^n$ and is true for any metric space
+- **Uniqueness of solution**: Prove that the path is unique (the complement is open), and verification requires checking multiple details
 
-## 方法选择
-采用**直接证明法**：通过证明补集是开集来证明 $F$ 是闭集。这是度量空间中处理闭集的标准方法，直接利用了开/闭集的对偶定义。
+## Method selection
+Use **direct proof**: prove that $F$ is a closed set by proving that the complement is an open set. This is the standard way to deal with closed sets in metric spaces, directly exploiting the dual definition of open/closed sets.
 
-备选方法：也可通过证明 $F$ 包含其所有极限点来证明闭性（等价的刻画），但补集为开的方法更直观且计算更明确。
+Alternative: It is also possible to prove closure (equivalent characterization) by showing that $F$ contains all its limit points, but the approach of having the complement open is more intuitive and computationally explicit.
 
-## 解题过程
+## Problem solving process
 
-**Step 1：转化问题**
+**Step 1: Conversion problem**
 
-要证明 $F$ 是闭集，等价于证明 $X \setminus F$ 是开集。由开集定义，需证明：对任意 $y \in X \setminus F$，存在 $r > 0$ 使得 $B(y, r) \subseteq X \setminus F$，即 $B(y, r) \cap F = \varnothing$。
+Proving that $F$ is a closed set is equivalent to proving that $X \setminus F$ is an open set. Defined by an open set, it needs to be proved: for any $y \in X \setminus F$, there exists $r > 0$ such that $B(y, r) \subseteq X \setminus F$, that is, $B(y, r) \cap F = \varnothing$.
 
-**Step 2：构造半径 $r$**
+**Step 2: Construct radius $r$ **
 
-任取 $y \notin F$。由于 $y \neq x_i$（$i = 1, 2, \ldots, n$），由度量定义有 $d(y, x_i) > 0$。
+Take any $y \notin F$ . Since $y \neq x_i$ ( $i = 1, 2, \ldots, n$ ), defined by the metric there is $d(y, x_i) > 0$ .
 
-令
-$$r = \min\{d(y, x_1), d(y, x_2), \ldots, d(y, x_n)\}$$
+make
+ $$r = \min\{d(y, x_1), d(y, x_2), \ldots, d(y, x_n)\}$$
 
-**关键点**：由于 $F$ 是有限集，这个最小值一定存在（有限个正实数的最小值仍为正实数）。如果是无限集，下确界可能为零，最小值可能不存在——这是"有限性"条件发挥作用的地方。
+**Key point**: Since $F$ is a finite set, this minimum value must exist (the minimum value of a finite number of positive real numbers is still a positive real number). In the case of an infinite set, the lower bound may be zero and the minimum may not exist - this is where the "finiteness" condition comes into play.
 
-因为每个 $d(y, x_i) > 0$，且取的是有限个正数的最小值，所以 $r > 0$。
+Because each $d(y, x_i) > 0$ takes the minimum value of a finite number of positive numbers, so $r > 0$ .
 
-**Step 3：验证 $B(y, r) \subseteq X \setminus F$**
+**Step 3: Verify $B(y, r) \subseteq X \setminus F$ **
 
-任取 $z \in B(y, r)$，即 $d(y, z) < r$。需证明 $z \notin F$。
+Take any $z \in B(y, r)$, which is $d(y, z) < r$. Need to prove $z \notin F$ .
 
-对任意 $x_i \in F$，由 $r$ 的定义有 $r \leq d(y, x_i)$。因此：
-$$d(y, z) < r \leq d(y, x_i)$$
+For any $x_i \in F$, the definition of $r$ has $r \leq d(y, x_i)$. therefore:
+ $$d(y, z) < r \leq d(y, x_i)$$
 
-所以 $d(y, z) < d(y, x_i)$，故 $z \neq x_i$（否则 $d(y, z) = d(y, x_i)$）。
+So $d(y, z) < d(y, x_i)$ , therefore $z \neq x_i$ (otherwise $d(y, z) = d(y, x_i)$ ).
 
-由于这对所有 $i = 1, \ldots, n$ 都成立，所以 $z \notin F$，即 $z \in X \setminus F$。
+Since this is true for all $i = 1, \ldots, n$ , $z \notin F$ is also $z \in X \setminus F$ .
 
-因此 $B(y, r) \subseteq X \setminus F$，$X \setminus F$ 是开集。
+Therefore $B(y, r) \subseteq X \setminus F$ , $X \setminus F$ are open sets.
 
-**Step 4：结论**
+**Step 4: Conclusion**
 
-$X \setminus F$ 是开集 $\implies$ $F$ 是闭集。证毕。
+$X \setminus F$ is an open set $\implies$ $F$ is a closed set. Certification completed.
 
-**Step 5：反例——无限集不一定闭**
+**Step 5: Counterexample - infinite sets are not necessarily closed**
 
-考虑度量空间 $\mathbb{R}$（通常度量）中的无限集 $S = \left\{\frac{1}{n} \mid n \in \mathbb{N}^+\right\}$。$0 \notin S$，但对任意 $r > 0$，取 $n > 1/r$，有 $\frac{1}{n} \in B(0, r)$，所以不存在以 $0$ 为中心的球完全在补集中。因此 $\mathbb{R} \setminus S$ 不是开集，$S$ 不是闭集。
+Consider the infinite set $S = \left\{\frac{1}{n} \mid n \in \mathbb{N}^+\right\}$ in the metric space $\mathbb{R}$ (the usual metric). $0 \notin S$ , but for any $r > 0$ , taking $n > 1/r$ , there is $\frac{1}{n} \in B(0, r)$ , so there is no ball centered on $0$ that is completely in the complement set. Therefore $\mathbb{R} \setminus S$ is not an open set and $S$ is not a closed set.
 
-关键对比：对于无限集 $S$，$\inf\{d(0, s) : s \in S\} = 0$，但最小值不存在（因为 $0 \notin S$）。这正是证明中"有限集确保最小值存在"对应的反例。
+Key comparison: for the infinite set $S$ , $\inf\{d(0, s) : s \in S\} = 0$ , but the minimum does not exist (because $0 \notin S$ ). This is exactly the counterexample corresponding to "finite sets ensure the existence of minimum values" in the proof.
 
-## 验算
+## Check calculation
 
-**验算方法 1：验证开集定义的每个条件**
+**Verification method 1: Verify each condition defined by the open set**
 
-1. $y \notin F$ $\checkmark$ — 前提
-2. $r > 0$ $\checkmark$ — 每个 $d(y, x_i) > 0$，有限个正数的最小值为正
-3. $B(y, r) \subseteq X \setminus F$ $\checkmark$ — 如上所证，球内任意点到 $y$ 的距离小于到每个 $x_i$ 的距离，故不在 $F$ 中
+1. $y \notin F$ $\checkmark$ — Prerequisite
+2. $r > 0$ $\checkmark$ — For each $d(y, x_i) > 0$ , the smallest of a finite number of positive numbers is positive
+3. $B(y, r) \subseteq X \setminus F$ $\checkmark$ — As proved above, the distance from any point in the ball to $y$ is less than the distance to each $x_i$, so it is not in $F$
 
-**验算方法 2：通过极限点等价条件验证**
+**Verification method 2: Verification through limit point equivalent conditions**
 
-$F$ 为闭集的等价条件是 $F$ 包含其所有极限点。对于有限集 $F$：假设 $\{z_k\}$ 是 $F$ 中的收敛序列，$\lim z_k = z$。由于 $F$ 有限，序列必然有无限多项取某个固定的 $x_i$，故 $z = x_i \in F$。因此 $F$ 包含所有极限点，是闭集。
+The equivalent condition for $F$ to be a closed set is that $F$ contains all its limit points. For the finite set $F$ : Assume $\{z_k\}$ is a convergent sequence in $F$ , $\lim z_k = z$ . Since $F$ is finite, the sequence must have an infinite number of items taking a fixed $x_i$, so $z = x_i \in F$. Therefore $F$ contains all limit points and is a closed set.
 
-**验算方法 3：特例验证**
+**Verification method 3: Special case verification**
 
-取 $X = \mathbb{R}$（通常度量），$F = \{1, 3, 5\}$。$y = 2 \notin F$，$r = \min\{|2-1|, |2-3|, |2-5|\} = \min\{1, 1, 3\} = 1$，$B(2, 1) = (1, 3)$，而 $(1, 3) \cap F = \varnothing$。✓
+Takes $X = \mathbb{R}$ (the usual measure), $F = \{1, 3, 5\}$ . $y = 2 \notin F$ , $r = \min\{|2-1|, |2-3|, |2-5|\} = \min\{1, 1, 3\} = 1$ , $B(2, 1) = (1, 3)$ , and $(1, 3) \cap F = \varnothing$ . ✓
 
-## 最终答案
+## Final answer
 
-在任意度量空间 $(X, d)$ 中，任意有限子集 $F = \{x_1, \ldots, x_n\}$ 是闭集。
+In any metric space $(X, d)$, any finite subset $F = \{x_1, \ldots, x_n\}$ is a closed set.
 
-证明核心：对 $y \notin F$，取 $r = \min_{1 \leq i \leq n} d(y, x_i) > 0$，则 $B(y, r) \subseteq X \setminus F$。
+Proof core: For $y \notin F$, take $r = \min_{1 \leq i \leq n} d(y, x_i) > 0$, then $B(y, r) \subseteq X \setminus F$.
 
-## 易错点
-1. **忽略有限性的关键作用**：最小值的存在依赖于集合的有限性。对无限集只能取下确界，而下确界可能为零
-2. **混淆 $r > 0$ 的推理**：$r > 0$ 不仅因为每个 $d(y, x_i) > 0$，更因为取的是**有限**个正数的**最小值**（而非下确界）
-3. **误以为"任意无限集都不是闭集"**：无限集可以是闭集（如 $\mathbb{R}$ 中的闭区间 $[0,1]$），只是在证明中有限性的论证不适用于无限集
-4. **将结论局限于 $\mathbb{R}^n$**：证明仅使用度量空间的三条公理（正定性、对称性、三角不等式），对任意度量空间均成立
+## Easy to make mistakes
+1. **Ignore the key role of finiteness**: The existence of the minimum depends on the finiteness of the set. For infinite sets, only the indeterminate can be taken, and the indeterminate may be zero.
+2. **Confusing reasoning** for $r > 0$: $r > 0$ is not only because of each $d(y, x_i) > 0$, but also because it takes the **minimum** of **finite** positive numbers (rather than the lower bound)
+3. **Misunderstood that "any infinite set is not a closed set"**: Infinite sets can be closed sets (such as the closed interval $[0,1]$ in $\mathbb{R}$), but the argument of finiteness in the proof does not apply to infinite sets
+4. **Limit the conclusion to $\mathbb{R}^n$ **: Prove that only using three axioms of metric space (positive certainty, symmetry, triangle inequality), they are true for any metric space
diff --git a/examples/trigonometry.md b/examples/trigonometry.md
index b5f7d3a..771d1c1 100644
--- a/examples/trigonometry.md
+++ b/examples/trigonometry.md
@@ -1,237 +1,237 @@
-# 三角恒等式证明
+# Trigonometric identity proof
 
-## 用户输入
-证明恒等式：$\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)$
+## User input
+Prove the identity: $\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)$
 
-## Skill 分类
-三角学（Trigonometry）— 三角恒等式证明
+## Skill Category
+Trigonometry - Proof of trigonometric identities
 
-## 题意解析
+## Question meaning analysis
 
-**已知条件：**
-- 需证明的恒等式：$\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)$
+**Known conditions:**
+- Identity to be proved: $\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)$
 
-**求解目标：**
-- 通过代数与三角恒等变换证明等号左右两边恒等。
+**Solution goal:**
+- Prove that the left and right sides of the equal sign are identical through algebraic and trigonometric identity transformations.
 
-**变量与定义域：**
-- $\theta \in \mathbb{R}$，恒等式对一切实数 $\theta$ 成立（左式和右式的定义域均为 $\mathbb{R}$）。
+**Variables and Domains:**
+- $\theta \in \mathbb{R}$, the identity holds for all real numbers $\theta$ (the domain of both the left and right expressions is $\mathbb{R}$).
 
-**隐式条件：**
-- $\sin^3\theta$ 可视为 $(\sin\theta)^3$，同理 $\cos^3\theta = (\cos\theta)^3$。
-- 这是立方和公式 $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ 在三角学中的应用。
-- 三角基本恒等式 $\sin^2\theta + \cos^2\theta = 1$ 将发挥关键作用。
+**Implicit condition:**
+- $\sin^3\theta$ can be treated as $(\sin\theta)^3$, and similarly $\cos^3\theta = (\cos\theta)^3$.
+- This is an application of the sum of cubes formula $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ in trigonometry.
+- The fundamental trigonometric identity $\sin^2\theta + \cos^2\theta = 1$ will play a key role.
 
-**解的个数：**
-- 恒等式证明不是解方程，目标不是求 $\theta$ 的值，而是证明对一切 $\theta$ 等式成立。
+**Number of solutions:**
+- The identity proof is not about solving equations. The goal is not to find the value of $\theta$, but to prove that the equation holds for all $\theta$.
 
-## 方法选择
+## Method selection
 
-**选择方法：** 从左边出发，应用立方和公式 + 基本恒等式化简
+**Selection method:** Starting from the left, apply the sum of cubes formula + simplify the basic identity
 
-**理由：**
-- 左边的 $\sin^3\theta + \cos^3\theta$ 结构上正是立方和 $a^3+b^3$，联想到因式分解。
-- 分解后出现 $\sin^2\theta - \sin\theta\cos\theta + \cos^2\theta$，其中 $\sin^2\theta + \cos^2\theta = 1$ 恰好将表达式化简为 $1 - \sin\theta\cos\theta$，完美匹配右边的因子。
+**reason:**
+- The $\sin^3\theta + \cos^3\theta$ on the left is structurally exactly the sum of cubes $a^3+b^3$, which reminds us of factorization.
+- After decomposition, $\sin^2\theta - \sin\theta\cos\theta + \cos^2\theta$ appears, where $\sin^2\theta + \cos^2\theta = 1$ happens to reduce the expression to $1 - \sin\theta\cos\theta$ , perfectly matching the factors on the right.
 
-**备选方法：**
-- 从右边展开验证：$(\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta) = \sin\theta + \cos\theta - \sin^2\theta\cos\theta - \sin\theta\cos^2\theta$，需要进一步变形才能配对成 $\sin^3\theta + \cos^3\theta$，步骤稍多但也可行。
-- 左右同时减去某项化简：不如直接单向化简清晰。
+**Alternative method:**
+- Expand the verification from the right: $(\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta) = \sin\theta + \cos\theta - \sin^2\theta\cos\theta - \sin\theta\cos^2\theta$, which needs further deformation to be paired into $\sin^3\theta + \cos^3\theta$. The steps are slightly more but feasible.
+- Simplify by subtracting a certain item from the left and right at the same time: not as clear as a direct one-way simplification.
 
-## 解题过程
+## Problem solving process
 
-### 从左边出发推导
+### Derive from the left
 
-**起点：** $\sin^3\theta + \cos^3\theta$
+**Starting point:** $\sin^3\theta + \cos^3\theta$
 
-**第一步：应用立方和公式。**
+**Step 1: Apply the Sum of Cubes formula. **
 
-令 $a = \sin\theta$，$b = \cos\theta$，则
+Let $a = \sin\theta$ , $b = \cos\theta$ , then
 
-$$
+ $$
 a^3 + b^3 = (a+b)(a^2 - ab + b^2)
 $$
 
-代入：
+Substitute:
 
-$$
+ $$
 \sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(\sin^2\theta - \sin\theta\cos\theta + \cos^2\theta) \tag{1}
 $$
 
-**第二步：应用基本三角恒等式。**
+**Step 2: Apply basic trigonometric identities. **
 
-由 $\sin^2\theta + \cos^2\theta = 1$，可将括号内的 $\sin^2\theta + \cos^2\theta$ 替换为 $1$：
+From $\sin^2\theta + \cos^2\theta = 1$, $\sin^2\theta + \cos^2\theta$ in parentheses can be replaced with $1$:
 
-$$
+ $$
 \sin^2\theta - \sin\theta\cos\theta + \cos^2\theta = (\sin^2\theta + \cos^2\theta) - \sin\theta\cos\theta = 1 - \sin\theta\cos\theta \tag{2}
 $$
 
-**第三步：代入 (2) 到 (1)。**
+**Step 3: Substitute (2) to (1). **
 
-$$
+ $$
 \sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)
 $$
 
-这就是要证明的右边。等式得证。
+This is the right side to prove. The equation is proved.
 
-### 反向推导（验算用）
+### Reverse derivation (for verification)
 
-从右边出发：
+Starting from the right:
 
-$$
+ $$
 (\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)
 $$
 
-展开：
+Expand:
 
-$$
+ $$
 = (\sin\theta + \cos\theta) \cdot 1 - (\sin\theta + \cos\theta) \cdot \sin\theta\cos\theta
 $$
 
-$$
+ $$
 = \sin\theta + \cos\theta - \sin^2\theta\cos\theta - \sin\theta\cos^2\theta
 $$
 
-提取 $\sin\theta\cos\theta$（从后两项）：
+Extract $\sin\theta\cos\theta$ (from the last two items):
 
-$$
+ $$
 = \sin\theta + \cos\theta - \sin\theta\cos\theta(\sin\theta + \cos\theta)
 $$
 
-将 $(\sin\theta + \cos\theta)$ 视为公因子：
+Treat $(\sin\theta + \cos\theta)$ as a common factor:
 
-$$
+ $$
 = (\sin\theta + \cos\theta)\left[1 - \sin\theta\cos\theta\left(\frac{\sin\theta + \cos\theta}{\sin\theta + \cos\theta}\right)\right]
 $$
 
-这里出错了。让我重新做。
+Something went wrong here. Let me do it again.
 
-实际上从右边展开回去也很直接。让我用另一种方式：
+In fact, it is very straightforward to unfold it from the right side. Let me do it another way:
 
-$(\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)$
+ $(\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)$
 
-$= \sin\theta \cdot 1 - \sin\theta \cdot \sin\theta\cos\theta + \cos\theta \cdot 1 - \cos\theta \cdot \sin\theta\cos\theta$
+ $= \sin\theta \cdot 1 - \sin\theta \cdot \sin\theta\cos\theta + \cos\theta \cdot 1 - \cos\theta \cdot \sin\theta\cos\theta$
 
-$= \sin\theta + \cos\theta - \sin^2\theta\cos\theta - \sin\theta\cos^2\theta$
+ $= \sin\theta + \cos\theta - \sin^2\theta\cos\theta - \sin\theta\cos^2\theta$
 
-$= \sin\theta(1 - \sin\theta\cos\theta) + \cos\theta(1 - \sin\theta\cos\theta)$  -- no that's just re-writing the original.
+ $= \sin\theta(1 - \sin\theta\cos\theta) + \cos\theta(1 - \sin\theta\cos\theta)$   -- no that's just re-writing the original.
 
-Let me try: $= \sin\theta + \cos\theta - \cos\theta\sin^2\theta - \sin\theta\cos^2\theta$
+Let me try:  $= \sin\theta + \cos\theta - \cos\theta\sin^2\theta - \sin\theta\cos^2\theta$
 
-Replace $\sin^2\theta = 1 - \cos^2\theta$ and $\cos^2\theta = 1 - \sin^2\theta$:
+Replace  $\sin^2\theta = 1 - \cos^2\theta$  and  $\cos^2\theta = 1 - \sin^2\theta$ :
 
 Actually, this approach is getting messy. Let me reformulate.
 
 Better: Group terms differently:
 
-$= \sin\theta - \sin\theta\cos^2\theta + \cos\theta - \sin^2\theta\cos\theta$
+ $= \sin\theta - \sin\theta\cos^2\theta + \cos\theta - \sin^2\theta\cos\theta$
 
-$= \sin\theta(1 - \cos^2\theta) + \cos\theta(1 - \sin^2\theta)$
+ $= \sin\theta(1 - \cos^2\theta) + \cos\theta(1 - \sin^2\theta)$
 
-$= \sin\theta \cdot \sin^2\theta + \cos\theta \cdot \cos^2\theta$
+ $= \sin\theta \cdot \sin^2\theta + \cos\theta \cdot \cos^2\theta$
 
-$= \sin^3\theta + \cos^3\theta$, ✓
+ $= \sin^3\theta + \cos^3\theta$ , ✓
 
 This is clean.
 
-## 验算
+## Check calculation
 
-### 验算方法一：数值抽样检验
+### Verification method one: numerical sampling inspection
 
-选取五个代表性的 $\theta$ 值，分别计算左右两边。
+Select five representative $\theta$ values ​​and calculate the left and right sides respectively.
 
-**① $\theta = 0$：**
+**①  $\theta = 0$ ：**
 
-$$
+ $$
 \sin 0 = 0,\ \cos 0 = 1
 $$
 
-左式：$0^3 + 1^3 = 1$
+Left style: $0^3 + 1^3 = 1$
 
-右式：$(0 + 1)(1 - 0 \times 1) = 1 \times 1 = 1$ ✓
+Right form: $(0 + 1)(1 - 0 \times 1) = 1 \times 1 = 1$ ✓
 
-**② $\theta = \frac{\pi}{6}$（$30^\circ$）：**
+**②  $\theta = \frac{\pi}{6}$ （ $30^\circ$ ）：**
 
-$$
+ $$
 \sin\frac{\pi}{6} = \frac{1}{2},\ \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}
 $$
 
-左式：
+Left style:
 
-$$
+ $$
 \left(\frac{1}{2}\right)^3 + \left(\frac{\sqrt{3}}{2}\right)^3 = \frac{1}{8} + \frac{3\sqrt{3}}{8} = \frac{1 + 3\sqrt{3}}{8}
 $$
 
-右式：
+Right form:
 
-$$
+ $$
 \left(\frac{1}{2} + \frac{\sqrt{3}}{2}\right)\left(1 - \frac{1}{2} \times \frac{\sqrt{3}}{2}\right) = \frac{1+\sqrt{3}}{2} \times \left(1 - \frac{\sqrt{3}}{4}\right)
 $$
 
-$$
+ $$
 = \frac{1+\sqrt{3}}{2} \times \frac{4-\sqrt{3}}{4} = \frac{(1+\sqrt{3})(4-\sqrt{3})}{8}
 $$
 
-展开分子：$(1+\sqrt{3})(4-\sqrt{3}) = 4 - \sqrt{3} + 4\sqrt{3} - 3 = 1 + 3\sqrt{3}$
+Expand numerator: $(1+\sqrt{3})(4-\sqrt{3}) = 4 - \sqrt{3} + 4\sqrt{3} - 3 = 1 + 3\sqrt{3}$
 
-因此右式 $= \dfrac{1 + 3\sqrt{3}}{8}$，与左式相等。✓
+Therefore, the right expression $= \dfrac{1 + 3\sqrt{3}}{8}$ is equal to the left expression. ✓
 
-**③ $\theta = \frac{\pi}{4}$（$45^\circ$）：**
+**③  $\theta = \frac{\pi}{4}$ （ $45^\circ$ ）：**
 
-$$
+ $$
 \sin\frac{\pi}{4} = \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}
 $$
 
-左式：
+Left style:
 
-$$
+ $$
 \left(\frac{\sqrt{2}}{2}\right)^3 + \left(\frac{\sqrt{2}}{2}\right)^3 = 2 \times \frac{2\sqrt{2}}{8} = \frac{4\sqrt{2}}{8} = \frac{\sqrt{2}}{2}
 $$
 
-右式：
+Right form:
 
-$$
+ $$
 \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right)\left(1 - \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2}\right) = \sqrt{2} \times \left(1 - \frac{2}{4}\right) = \sqrt{2} \times \frac{1}{2} = \frac{\sqrt{2}}{2}
 $$
 
 ✓
 
-**④ $\theta = \frac{\pi}{3}$（$60^\circ$）：**
+**④  $\theta = \frac{\pi}{3}$ （ $60^\circ$ ）：**
 
-$$
+ $$
 \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2},\ \cos\frac{\pi}{3} = \frac{1}{2}
 $$
 
-左式：
+Left style:
 
-$$
+ $$
 \left(\frac{\sqrt{3}}{2}\right)^3 + \left(\frac{1}{2}\right)^3 = \frac{3\sqrt{3}}{8} + \frac{1}{8} = \frac{3\sqrt{3}+1}{8}
 $$
 
-右式：
+Right form:
 
-$$
+ $$
 \left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right)\left(1 - \frac{\sqrt{3}}{2} \times \frac{1}{2}\right) = \frac{\sqrt{3}+1}{2} \times \frac{4-\sqrt{3}}{4} = \frac{(\sqrt{3}+1)(4-\sqrt{3})}{8}
 $$
 
-展开：$(4\sqrt{3} - 3 + 4 - \sqrt{3}) = 3\sqrt{3} + 1$
+Expand: $(4\sqrt{3} - 3 + 4 - \sqrt{3}) = 3\sqrt{3} + 1$
 
-右式 $= \dfrac{3\sqrt{3}+1}{8}$ = 左式 ✓
+Right $= \dfrac{3\sqrt{3}+1}{8}$ = Left ✓
 
-**⑤ $\theta = \frac{\pi}{2}$（$90^\circ$）：**
+**⑤  $\theta = \frac{\pi}{2}$ （ $90^\circ$ ）：**
 
-$$
+ $$
 \sin\frac{\pi}{2} = 1,\ \cos\frac{\pi}{2} = 0
 $$
 
-左式：$1^3 + 0^3 = 1$
+Left style: $1^3 + 0^3 = 1$
 
-右式：$(1 + 0)(1 - 1 \times 0) = 1$ ✓
+Right form: $(1 + 0)(1 - 1 \times 0) = 1$ ✓
 
-五个特殊角全部验证通过。
+All five special corners have been verified.
 
-### 验算方法二：反向推导（从右边到左边）
+### Calculation method two: reverse derivation (from right to left)
 
-$$
+ $$
 \begin{aligned}
 (\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)
 &= \sin\theta - \sin^2\theta\cos\theta + \cos\theta - \sin\theta\cos^2\theta \\[6pt]
@@ -241,19 +241,19 @@ $$
 \end{aligned}
 $$
 
-其中第二步使用了 $\sin^2\theta = 1 - \cos^2\theta$ 和 $\cos^2\theta = 1 - \sin^2\theta$。✓
+The second step uses $\sin^2\theta = 1 - \cos^2\theta$ and $\cos^2\theta = 1 - \sin^2\theta$. ✓
 
-## 最终答案
+## Final answer
 
-$$
+ $$
 \boxed{\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)}
 $$
 
-对一切 $\theta \in \mathbb{R}$ 成立。
+True for everything $\theta \in \mathbb{R}$.
 
-## 易错点
-1. **立方和公式记错：** $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$，注意中间是 $-ab$ 而非 $+ab$。如果与立方差公式 $a^3-b^3 = (a-b)(a^2+ab+b^2)$ 混淆会导致错误。
-2. **忘记基本恒等式：** $\sin^2\theta + \cos^2\theta = 1$ 是三角学最基本的恒等式，本题中在因式分解的后一步即需要这个替换。
-3. **展开右边时符号错误：** 从右边展开 $(\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)$ 时，负号要在每项都体现。
-4. **混淆 $\sin^3\theta$ 与 $\sin\theta^3$：** $\sin^3\theta$ 表示 $(\sin\theta)^3$，而 $\sin\theta^3$ 通常表示 $\sin(\theta^3)$，两者完全不同。在等式中始终使用 $\sin^3\theta$ 的写法。
-5. **误以为需要解方程：** 恒等式证明不是解方程——不需要求出具体的 $\theta$ 值。证明的目标是展示等式对一切 $\theta$ 都成立。
+## Easy to make mistakes
+1. **The formula for the sum of cubes is incorrectly remembered:** $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. Note that the middle is $-ab$ instead of $+ab$. An error will result if confused with the cubic difference formula $a^3-b^3 = (a-b)(a^2+ab+b^2)$.
+2. **Forgot the basic identity:** $\sin^2\theta + \cos^2\theta = 1$ is the most basic identity in trigonometry. This replacement is required in the last step of factorization in this question.
+3. **Sign error when expanding the right side:** When expanding $(\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)$ from the right side, the negative sign must be reflected in each item.
+4. **Confusing $\sin^3\theta$ with $\sin\theta^3$:** $\sin^3\theta$ means $(\sin\theta)^3$, while $\sin\theta^3$ usually means $\sin(\theta^3)$, which are completely different. Always use $\sin^3\theta$ in equations.
+5. **Mistaken belief that you need to solve an equation:** The identity proof is not the solution of the equation - there is no need to find the specific value of $\theta$. The goal of the proof is to show that the equation holds for everything $\theta$.
diff --git a/modules/classification.md b/modules/classification.md
index 982e346..002ac8c 100644
--- a/modules/classification.md
+++ b/modules/classification.md
@@ -1,1567 +1,1567 @@
-# 输入分类模块
+#Input classification module
 
-## 概述
+## Overview
 
-本模块定义了 Math.skill 中输入问题的分类体系。对于每一个输入，系统首先将其归入以下类别之一，然后根据类别的特性选择合适的求解策略和验证方法。
+This module defines the classification system for input problems in Math.skill. For each input, the system first classifies it into one of the following categories and then selects an appropriate solution strategy and verification method based on the characteristics of the category.
 
 ---
 
 ## calculation
 
-**识别特征：**
-- 输入仅包含数值和运算符（$+,-,\times,\div,\hat{\ },\sqrt{}$ 等）
-- 无变量、无等号或不等号
-- 常见表述："计算"、"求值"、"结果是多少"
-- 示例：$\sqrt{169} + 3^4 - 12 \div 4$
-
-**推荐求解策略：**
-- 按运算优先级逐步计算
-- 使用括号明确运算顺序
-- 对复杂表达式进行分步拆解，每步独立验证
-
-**必须检查的条件：**
-- 表达式中的除法分母不为零
-- 偶数次根号下表达式 $\geq 0$
-- 对数底数 $>0$ 且 $\neq 1$，真数 $>0$
-- 指数运算结果是否在合理范围内
-
-**常见错误：**
-- 运算优先级错误（如先加后乘）
-- 负号处理错误（$(-3)^2 = 9$，不是 $-9$）
-- 分数运算中通分错误
-- 小数与分数混合时精度丢失
-- 大数运算溢出或舍入错误
-
-**推荐验证方法：**
-- 逆运算验证（用结果反推）
-- 估算量级是否合理
-- 使用不同计算路径重算
-- 对关键中间结果交叉验证
-
-**输出格式：**
-```
-结果: [精确值]
-逐步计算过程: [每步运算及中间结果]
+**Identification Features:**
+- Input contains only numeric values ​​and operators ($+,-,\times,\div,\hat{\ },\sqrt{}$ etc.)
+- No variables, no equal sign or inequality sign
+- Common expressions: "calculation", "evaluation", "what is the result"
+- Example: $\sqrt{169} + 3^4 - 12 \div 4$
+
+**Recommended solution strategy:**
+- Calculate step by step according to operation priority
+- Use parentheses to clarify the order of operations
+- Decompose complex expressions step by step, and verify each step independently
+
+**Conditions that must be checked:**
+- The denominator of the division in the expression is not zero
+- Expression under even-numbered radicals $\geq 0$
+- base of logarithm $>0$ and $\neq 1$ , real number $>0$
+- Whether the exponential operation result is within a reasonable range
+
+**Common mistakes:**
+- Incorrect operation priority (such as addition first and then multiplication)
+- Negative sign handling error ($(-3)^2 = 9$, not $-9$)
+- Common division errors in fraction operations
+- Precision is lost when decimals and fractions are mixed
+- Large number operation overflow or rounding error
+
+**Recommended verification method:**
+- Verification of inverse operation (use the result to infer)
+- Is the estimated magnitude reasonable?
+- Recalculate using different calculation paths
+- Cross-validation of key intermediate results
+
+**Output format:**
+ ```
+Result: [exact value]
+Step-by-step calculation process: [Each step operation and intermediate results]
 ```
 
-**是否追问用户：** 否
+**Whether to ask the user:** No
 
 ---
 
 ## algebra_simplification
 
-**识别特征：**
-- 包含代数变量的表达式，要求化简、因式分解或展开
-- 关键词："化简"、"因式分解"、"展开"、"合并同类项"
-- 示例：化简 $(x+2)(x-3) - (x-1)^2$
-
-**推荐求解策略：**
-- 识别表达式结构（多项式、分式、根式等）
-- 应用分配律展开括号
-- 合并同类项
-- 因式分解时先提取公因式，再应用公式
-
-**必须检查的条件：**
-- 分母不为零的条件
-- 偶数次根号下表达式 $\geq 0$
-- 因式分解结果是否完全分解
-- 展开结果是否已合并所有同类项
-
-**常见错误：**
-- 符号处理错误（展开时的正负号）
-- 漏项（忘记展开所有项）
-- 因式分解不完全
-- 平方差/完全平方公式误用
-- 分式化简时忘记定义域约束
-
-**推荐验证方法：**
-- 代入特殊值检验两端是否相等
-- 对化简结果重新展开，验证是否等于原式
-- 检查因式分解结果乘回去是否等于原式
-- 验证定义域约束是否一致
-
-**输出格式：**
-```
-化简结果: [最简形式]
-关键步骤: [每一步的变换]
-注意: [定义域约束条件]
-```
-
-**是否追问用户：** 仅当表达式有歧义时
+**Identification Features:**
+- Expressions containing algebraic variables requiring reduction, factorization or expansion
+- Keywords: "simplify", "factorization", "expand", "merge similar terms"
+- Example: Simplify $(x+2)(x-3) - (x-1)^2$
+
+**Recommended solution strategy:**
+- Recognize expression structures (polynomials, fractions, radicals, etc.)
+- Apply the distributive property to expand the brackets
+- Merge similar items
+- When factoring, first extract the common factors and then apply the formula
+
+**Conditions that must be checked:**
+- Conditions when the denominator is not zero
+- Expression under even-numbered roots $\geq 0$
+- Whether the factorization result is completely decomposed
+- Whether the expansion result has merged all similar items
+
+**Common mistakes:**
+- Sign handling error (positive and negative signs on expansion)
+- Missing items (forgot to expand all items)
+- Incomplete factorization
+- Misuse of square difference/perfect square formula
+- Domain constraints are forgotten when simplifying fractions
+
+**Recommended verification method:**
+- Substitute special values ​​to check whether both ends are equal
+- Re-expand the simplified results to verify whether they are equal to the original formula
+- Check whether the factored result multiplied back is equal to the original formula
+- Verify that domain constraints are consistent
+
+**Output format:**
+ ```
+Simplification result: [simplest form]
+Key steps: [Transformation of each step]
+Note: [Domain constraints]
+```
+
+**Whether to ask the user:** Only if the expression is ambiguous
 
 ---
 
 ## equation_solving
 
-**识别特征：**
-- 单个等式，包含一个未知数
-- 关键词："解方程"、"求 $x$"、等式
-- 示例：解方程 $2x^2 - 5x + 2 = 0$
-
-**推荐求解策略：**
-- 判断方程类型（线性、二次、高次、分式、根式、指数、对数、三角等）
-- 线性：移项法求解
-- 二次：判别式法 $\Delta = b^2 - 4ac$，求根公式 $x = \frac{-b \pm \sqrt{\Delta}}{2a}$
-- 高次：因式分解或试根法
-- 分式：去分母后求解，注意增根
-- 根式：平方消根，注意增根
-
-**必须检查的条件：**
-- 是否有增根（代入原方程验证）
-- 二次方程的判别式 $\Delta$ 是否 $\geq 0$
-- 分式方程分母不为零
-- 根式方程被开方数非负
-- 对数方程真数 $>0$，底数 $>0$ 且 $\neq 1$
-
-**常见错误：**
-- 忘记验证增根
-- 二次方程漏解（$\Delta > 0$ 时有两个根）
-- 除以可能为零的表达式
-- 根式方程平方后未验根
-- 移项时符号错误
-
-**推荐验证方法：**
-- 将所有解代入原方程验证成立
-- 检查解是否在定义域内
-- 对二次方程验证 $x_1 + x_2 = -b/a$，$x_1 x_2 = c/a$
-- 图形法验证（函数图像与 $x$ 轴交点）
-
-**输出格式：**
-```
-方程类型: [线性/二次/高次/分式/根式/...]
-解: $x = [值]$ (或 $x_1 = [值], x_2 = [值]$)
-求解步骤: [关键变换步骤]
-定义域约束: [排除的非法值]
-验根: [代入验证结果]
-```
-
-**是否追问用户：** 仅当方程有歧义或缺少约束时
+**Identification Features:**
+- A single equation with one unknown
+- Keywords: "solve equations", "find $x$", equation
+- Example: Solve the equation $2x^2 - 5x + 2 = 0$
+
+**Recommended solution strategy:**
+- Determine the type of equation (linear, quadratic, higher degree, fraction, radical, exponential, logarithmic, trigonometric, etc.)
+- Linear: solution using the transfer method
+- Quadratic: discriminant method $\Delta = b^2 - 4ac$, root formula $x = \frac{-b \pm \sqrt{\Delta}}{2a}$
+- Higher order: factorization or root trial method
+- Fraction: solve after removing the denominator, pay attention to adding roots
+- Radical formula: square root elimination, pay attention to root addition
+
+**Conditions that must be checked:**
+- Whether there are increasing roots (substitute into the original equation to verify)
+- The discriminant of the quadratic equation $\Delta$ is $\geq 0$
+- The denominator of a fractional equation is not zero
+- The radicand of a radical equation is non-negative
+- Logarithmic equations have real numbers $>0$, bases $>0$ and $\neq 1$
+
+**Common mistakes:**
+- Forgot to verify root addition
+- Missing solution to quadratic equation (two roots when $\Delta > 0$)
+- divide by an expression that may be zero
+- The root of the radical equation is not tested after squaring it
+- Wrong sign when moving items
+
+**Recommended verification method:**
+- Substitute all solutions into the original equation to verify that they are established
+- Check if the solution is within the domain
+- Verify $x_1 + x_2 = -b/a$, $x_1 x_2 = c/a$ for quadratic equations
+- Graphical method verification (intersection of function graph and $x$ axis)
+
+**Output format:**
+ ```
+Equation type: [linear/quadratic/higher degree/fractional/radical/...]
+Solution: $x = [value]$ (or $x_1 = [value], x_2 = [value]$)
+Solution steps: [Key transformation steps]
+Domain constraint: [illegal values ​​to exclude]
+Root verification: [Insert verification result]
+```
+
+**Whether to ask the user:** Only if the equation is ambiguous or missing constraints
 
 ---
 
 ## system_of_equations
 
-**识别特征：**
-- 多个等式，包含多个未知数
-- 关键词："方程组"、多个方程用大括号连接
-- 示例：$\begin{cases} 2x + y = 5 \\ x - 3y = -1 \end{cases}$
-
-**推荐求解策略：**
-- 线性方程组：代入法、消元法、矩阵法
-- 非线性方程组：代入法、消元法、因式分解
-- 对称方程组：利用对称性简化
-- 齐次方程组：设比值法
-
-**必须检查的条件：**
-- 方程的个数与未知数个数的关系
-- 线性方程组系数矩阵是否奇异
-- 是否有无穷多解或无解
-- 非线性方程组代入后的增根
-
-**常见错误：**
-- 消元时系数相乘/加的错误
-- 代入不完整
-- 忽略无解或无穷多解的情况
-- 非线性方程组转换后引入增根
-- 对称性利用时的遗漏
-
-**推荐验证方法：**
-- 将所有解代入每个方程验证
-- 矩阵法求线性方程组后验证 $A\vec{x} = \vec{b}$
-- 对二元一次方程组用克莱姆法则交叉验证
-
-**输出格式：**
-```
-方程组类型: [线性/非线性]
-解: $(x, y) = ([值], [值])$（或 $x = [值], y = [值], z = [值]$）
-求解方法: [代入法/消元法/矩阵法]
-解的个数: [唯一解/无穷多解/无解]
-验证: [代入各方程的结果]
-```
-
-**是否追问用户：** 仅当方程不满秩时确认处理方式
+**Identification Features:**
+- Multiple equations with multiple unknowns
+- Keywords: "system of equations", multiple equations connected with braces
+- Example: $\begin{cases} 2x + y = 5 \\ x - 3y = -1 \end{cases}$
+
+**Recommended solution strategy:**
+- System of linear equations: substitution method, elimination method, matrix method
+- Nonlinear equations: substitution method, elimination method, factorization
+- Symmetric systems of equations: Simplification using symmetry
+- System of homogeneous equations: Ratio method
+
+**Conditions that must be checked:**
+- The relationship between the number of equations and the number of unknowns
+- Is the coefficient matrix of the linear equation system singular?
+- Whether there are infinitely many solutions or no solutions
+- Root augmentation after substitution of nonlinear equations
+
+**Common mistakes:**
+- Error in coefficient multiplication/addition during elimination
+- Incomplete substitution
+- Ignore cases where there are no solutions or infinite solutions
+- Introduction of increasing roots after transformation of nonlinear equations
+- Omissions when exploiting symmetry
+
+**Recommended verification method:**
+- Verify by substituting all solutions into each equation
+- Verification after solving linear equations using matrix method $A\vec{x} = \vec{b}$
+- Cross-validation using Clem's rule for systems of linear equations in two variables
+
+**Output format:**
+ ```
+System type: [Linear/Nonlinear]
+Solution: $(x, y) = ([value], [value])$ (or $x = [value], y = [value], z = [value]$)
+Solution method: [Substitution method/Elimination method/Matrix method]
+Number of solutions: [unique solution/infinitely many solutions/no solutions]
+Verification: [Results inserted into each equation]
+```
+
+**Whether to ask the user:** Confirm the processing method only when the equation is not satisfied with the rank
 
 ---
 
 ## inequality_solving
 
-**识别特征：**
-- 包含不等号（$>, <, \geq, \leq$）的式子
-- 关键词："解不等式"、"求 $x$ 的范围"、"不等式组"
-- 示例：解不等式 $\frac{x-1}{x+2} \geq 0$
-
-**推荐求解策略：**
-- 一元一次不等式：移项，注意乘除负号方向反转
-- 一元二次不等式：判别式 + 根分布，数轴穿根法
-- 分式不等式：移项通分后分子分母同号分析
-- 绝对值不等式：分类讨论或几何意义
-- 不等式组：分别求解后取交集
-- 高次不等式：数轴穿根法（奇穿偶不穿）
-
-**必须检查的条件：**
-- 乘除负数时不等式方向是否反转
-- 分式不等式分母不为零
-- 绝对值不等式所有分支是否完整
-- 区间端点是否包含（等号是否成立）
-
-**常见错误：**
-- 乘以负数忘记反转不等号方向
-- 分式不等式直接交叉相乘
-- 平方两边时忘记讨论符号
-- 绝对值分类讨论不完整
-- 区间表示错误（开闭区间混淆）
-
-**推荐验证方法：**
-- 在解区间内取测试点验证原不等式
-- 在解区间外取测试点确认不成立
-- 检查区间端点（代入等于的情况）
-- 数轴验证法
-
-**输出格式：**
-```
-不等式类型: [一次/二次/分式/绝对值/不等式组]
-解集: $x \in [区间表达式]$
-关键步骤: [分类讨论过程]
-```
-
-**是否追问用户：** 否
+**Identification Features:**
+- Expressions containing the inequality sign ($>, <, \geq, \leq$)
+- Keywords: "solve inequalities", "find the range of $x$", "inequality group"
+- Example: Solving Inequalities $\frac{x-1}{x+2} \geq 0$
+
+**Recommended solution strategy:**
+- First degree inequalities of one variable: shift terms, pay attention to the reverse direction of the negative signs in multiplication and division
+- Quadratic inequalities of one variable: discriminant + root distribution, number axis through root method
+- Fractional inequality: analysis of the same sign in the numerator and denominator after the common denominator is transferred.
+- Absolute value inequalities: classification discussion or geometric significance
+- Group of inequalities: solve them separately and find the intersection
+- Higher-order inequalities: Number axis threading method (odd threads and even threads are not threaded)
+
+**Conditions that must be checked:**
+- Whether the direction of the inequality is reversed when multiplying or dividing negative numbers
+- The denominator of the fraction inequality is not zero
+- Are all branches of the absolute value inequality complete?
+- Whether the endpoint of the interval is included (whether the equal sign is true)
+
+**Common mistakes:**
+- Forgetting to reverse the direction of the inequality sign when multiplying negative numbers
+- Direct cross multiplication of fractional inequalities
+- Forgetting to discuss symbols when squaring both sides
+- Incomplete discussion of absolute value classification
+- Wrong interval representation (confusion of open and closed intervals)
+
+**Recommended verification method:**
+- Take test points within the solution interval to verify the original inequality
+- Take test points outside the solution interval to confirm that it is not established.
+- Check the endpoints of the interval (substitute the equal case)
+- Number line verification method
+
+**Output format:**
+ ```
+Inequality type: [primary/quadratic/fraction/absolute value/inequality group]
+Solution set: $x \in [interval expression]$
+Key steps: [Classification discussion process]
+```
+
+**Whether to ask the user:** No
 
 ---
 
 ## function_analysis
 
-**识别特征：**
-- 给定函数表达式，要求分析其性质
-- 关键词："定义域"、"值域"、"单调性"、"奇偶性"、"极值"、"最值"、"图像"
-- 示例：分析函数 $f(x) = \frac{x^2-1}{x-2}$ 的性质
-
-**推荐求解策略：**
-- 定义域：排除分母为零、根号下为负、对数真数 $\leq 0$ 等
-- 值域：求反函数、配方法、求导法、图像法
-- 单调性：求导判断符号、复合函数单调性法则
-- 奇偶性：验证 $f(-x) = f(x)$（偶）或 $f(-x) = -f(x)$（奇）
-- 极值：求导找临界点，二阶导判断极值类型
-- 图像：找关键点（零点、极值点、拐点、渐近线），描绘趋势
-
-**必须检查的条件：**
-- 定义域内所有约束条件
-- 导数存在性（不可导点可能是极值点）
-- 渐近线：水平、垂直、斜渐近线
-- 分段函数的连续性
-
-**常见错误：**
-- 定义域遗漏（只考虑分母，忽略根号或对数）
-- 复合函数定义域错误
-- 极值点混淆（驻点不一定是极值点）
-- 单调区间端点处理错误
-- 忽略奇函数在 $x=0$ 处 $f(0)=0$
-
-**推荐验证方法：**
-- 取特殊点验证函数值
-- 求导结果对积分验证
-- 利用对称性验证奇偶性
-- 用单调性验证极值点
-
-**输出格式：**
-```
-函数: $f(x) = [表达式]$
-定义域: $x \in [区间]$
-值域: $f(x) \in [区间]$
-单调性: [增区间和减区间]
-奇偶性: [奇函数/偶函数/非奇非偶]
-极值: [极值点坐标]
-渐近线: [渐近线方程]
-```
-
-**是否追问用户：** 仅当函数定义不清晰时
+**Identification Features:**
+- Given a function expression, ask to analyze its properties
+- Keywords: "definition domain", "range", "monotonicity", "parity", "extreme value", "maximum value", "image"
+- Example: Analyze the properties of function $f(x) = \frac{x^2-1}{x-2}$
+
+**Recommended solution strategy:**
+- Domain: exclude denominator zero, negative root sign, true logarithm number $\leq 0$, etc.
+- Value range: inverse function, combination method, derivation method, image method
+- Monotonicity: derivation judgment symbol, monotonicity rule of composite functions
+- Parity: Verify $f(-x) = f(x)$ (even) or $f(-x) = -f(x)$ (odd)
+- Extreme value: Find the critical point by taking the derivative, and determine the type of extreme value by using the second-order derivative.
+- Image: find key points (zero points, extreme points, inflection points, asymptotes) and depict trends
+
+**Conditions that must be checked:**
+- Define all constraints within the domain
+- Existence of derivatives (non-differentiable points may be extreme points)
+- Asymptotes: horizontal, vertical, oblique asymptotes
+- Continuity of piecewise functions
+
+**Common mistakes:**
+- Missing domain (only denominator is considered, root sign or logarithm is ignored)
+- Composite function domain error
+- Confusion of extreme points (stationary points are not necessarily extreme points)
+- Error in endpoint handling of monotonic intervals
+- Ignore odd functions at $x=0$ $f(0)=0$
+
+**Recommended verification method:**
+- Take special points to verify function values
+- Validation of derivation results against integrals
+- Use symmetry to verify parity
+- Verify extreme points using monotonicity
+
+**Output format:**
+ ```
+Function: $f(x) = [expression]$
+Domain: $x \in [interval]$
+Range: $f(x) \in [interval]$
+Monotonicity: [increasing and decreasing intervals]
+Parity: [odd function/even function/neither odd nor even]
+Extreme value: [Extreme point coordinates]
+Asymptote: [Asymptote equation]
+```
+
+**Whether to ask the user:** Only when the function definition is not clear
 
 ---
 
 ## geometry
 
-**识别特征：**
-- 涉及平面几何图形：三角形、四边形、圆等
-- 关键词："证明"、"求证"、"求面积"、"全等"、"相似"、"角度"
-- 示例：在 $\triangle ABC$ 中，$AB=AC$，求证 $\angle B = \angle C$
-
-**推荐求解策略：**
-- 画图辅助理解
-- 标识已知条件和求证目标
-- 全等判定：SSS、SAS、ASA、AAS、HL
-- 相似判定：AA、SAS、SSS
-- 面积法、向量法、坐标法
-- 辅助线构造（中点连线、垂线、平行线）
-
-**必须检查的条件：**
-- 三角形不等式（任意两边之和大于第三边）
-- 角度和约束（三角形内角和 $180^\circ$）
-- 全等/相似的对应关系是否正确
-- 辅助线构造的合理性
-
-**常见错误：**
-- 全等/相似对应顶点顺序错误
-- 辅助线构造不合理或多余
-- 忽略共线、共点等特殊情况
-- 角度计算时单位混淆（度与弧度）
-- 面积公式误用
-
-**推荐验证方法：**
-- 构造具体数值验证（特殊三角形）
-- 使用不同方法交叉验证
-- 量角器或坐标法验证角度
-- 逆推法验证
-
-**输出格式：**
-```
-已知: [条件]
-求证/求解: [目标]
-证明/求解步骤: [推理过程]
-辅助线: [添加的辅助线说明]
-结论: [最终结果]
-```
-
-**是否追问用户：** 是，需要确认是否有附图或补充条件
+**Identification Features:**
+- Involving plane geometric figures: triangles, quadrilaterals, circles, etc.
+- Keywords: "proof", "verification", "area", "congruence", "similarity", "angle"
+- Example: In $\triangle ABC$, $AB=AC$, verify $\angle B = \angle C$
+
+**Recommended solution strategy:**
+- Draw pictures to aid understanding
+- Identify known conditions and verification goals
+- Congruent judgment: SSS, SAS, ASA, AAS, HL
+- Similarity determination: AA, SAS, SSS
+- Area method, vector method, coordinate method
+- Auxiliary line construction (midpoint line, perpendicular line, parallel line)
+
+**Conditions that must be checked:**
+- Triangle inequality (the sum of any two sides is greater than the third side)
+- Angle sum constraints (triangle angle sum $180^\circ$ )
+- Is the congruent/similar correspondence correct?
+- The rationality of the auxiliary line construction
+
+**Common mistakes:**
+- Congruent/similar corresponding vertices are in the wrong order
+- The construction of auxiliary lines is unreasonable or redundant
+- Ignore special cases such as collinearity and common points
+- Unit confusion (degrees and radians) when calculating angles
+- Misuse of area formula
+
+**Recommended verification method:**
+- Construct specific numerical verification (special triangle)
+- Cross-validation using different methods
+- Protractor or coordinate method to verify angles
+- Backward verification
+
+**Output format:**
+ ```
+Known: [condition]
+Proof/Solution: [Objective]
+Proof/Solution Steps: [Reasoning Process]
+Auxiliary lines: [Description of added auxiliary lines]
+Conclusion: [Final Result]
+```
+
+**Whether to ask the user:** Yes, you need to confirm whether there are attached pictures or supplementary conditions
 
 ---
 
 ## analytic_geometry
 
-**识别特征：**
-- 坐标系中的几何问题，涉及点、直线、圆、圆锥曲线
-- 关键词："坐标"、"直线方程"、"圆的方程"、"椭圆"、"双曲线"、"抛物线"
-- 示例：求过点 $(1,2)$ 且与直线 $2x-y+3=0$ 平行的直线方程
-
-**推荐求解策略：**
-- 建立合适的坐标系
-- 利用距离公式、斜率公式
-- 圆的方程：$(x-a)^2 + (y-b)^2 = r^2$
-- 直线位置关系：平行（斜率相等）、垂直（斜率乘积为 $-1$）
-- 距离：点到直线 $d = \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$
-- 圆锥曲线定义和标准方程
-
-**必须检查的条件：**
-- 斜率不存在的情况（竖直线）
-- 两点重合时无法确定直线
-- 圆与直线相切/相交/相离的判断
-- 圆锥曲线的标准形式是否正确
-
-**常见错误：**
-- 斜率公式分母为零时未处理
-- 距离公式忘加绝对值
-- 两直线夹角公式用错
-- 圆的方程配方错误
-- 椭圆/双曲线中 $a,b,c$ 关系混淆
-
-**推荐验证方法：**
-- 代入点验证是否在曲线上
-- 对称性验证
-- 不同方法求解同一问题交叉验证
-- 画图直观检验
-
-**输出格式：**
-```
-坐标系: [直角坐标/极坐标]
-所求几何对象方程: [方程]
-求解步骤: [推导过程]
-关键性质: [焦点/准线/圆心/半径等]
-```
-
-**是否追问用户：** 仅当坐标系或条件不明确时
+**Identification Features:**
+- Geometric problems in coordinate systems, involving points, straight lines, circles, and conics
+- Keywords: "coordinates", "equation of a straight line", "equation of a circle", "ellipse", "hyperbola", "parabola"
+- Example: Find the equation of the straight line passing through the point $(1,2)$ and parallel to the straight line $2x-y+3=0$
+
+**Recommended solution strategy:**
+- Establish a suitable coordinate system
+- Use distance formula and slope formula
+- Equation of a circle: $(x-a)^2 + (y-b)^2 = r^2$
+- Line position relationship: parallel (slopes are equal), perpendicular (the product of slopes is $-1$ )
+- Distance: point to line $d = \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$
+- Definition and standard equations of conic sections
+
+**Conditions that must be checked:**
+- Case where slope does not exist (vertical line)
+- A straight line cannot be determined when two points coincide
+- Judgment of tangency/intersection/separation between circles and straight lines
+- Is the standard form of a conic section correct?
+
+**Common mistakes:**
+- The denominator of the slope formula is not processed when it is zero.
+- Forgot to add the absolute value in the distance formula
+- The formula for the angle between two straight lines is used incorrectly
+- Wrong formula for circle equation
+- $a,b,c$ relation confusion in ellipse/hyperbola
+
+**Recommended verification method:**
+- Verify whether the substitution point is on the curve
+- Symmetry verification
+- Cross-validation of different methods to solve the same problem
+- Visual inspection of drawings
+
+**Output format:**
+ ```
+Coordinate system: [Rectangular coordinates/polar coordinates]
+The equation of the geometric object sought: [equation]
+Solution steps: [Derivation process]
+Key properties: [Focus/Directrix/Circle Center/Radius, etc.]
+```
+
+**Whether to ask the user:** Only when the coordinate system or conditions are unclear
 
 ---
 
 ## trigonometry
 
-**识别特征：**
-- 涉及三角函数 $\sin, \cos, \tan$ 等
-- 关键词："三角函数"、"三角恒等式"、"解三角形"、"正弦定理"、"余弦定理"
-- 示例：证明 $\sin^2\theta + \cos^2\theta = 1$
-
-**推荐求解策略：**
-- 记忆和应用基本恒等式
-- 和差公式：$\sin(A \pm B), \cos(A \pm B)$
-- 倍角公式：$\sin 2\theta = 2\sin\theta\cos\theta$
-- 解三角形：正弦定理 $\frac{a}{\sin A} = 2R$，余弦定理 $a^2 = b^2 + c^2 - 2bc\cos A$
-- 三角方程：利用周期性求通解
-
-**必须检查的条件：**
-- 角度范围（锐角、$[0, 2\pi)$ 等）
-- 三角函数定义域（$\tan\theta$ 中 $\theta \neq \frac{\pi}{2}+k\pi$）
-- 三角形的解的情况（一解/两解/无解）
-- 弧度与角度的单位统一
-
-**常见错误：**
-- 正弦定理/余弦定理套用错误
-- 三角形解的个数判断错误（SSA 情况）
-- 诱导公式符号错误
-- 通解遗漏
-- 弧度与度混用
-
-**推荐验证方法：**
-- 代入特殊角验证恒等式
-- 三角恒等式两边对同一角度计算数值
-- 验证三角形内角和 $180^\circ$
-- 用不同方法求同一量交叉验证
-
-**输出格式：**
-```
-求解/证明: [目标]
-使用公式: [正弦定理/余弦定理/和差公式/...]
-步骤: [推导过程]
-结果: [角度值/边长/恒等式证明]
-解的个数: [一解/两解/无解]
-```
-
-**是否追问用户：** 仅当角度范围不明确时
+**Identification Features:**
+- Involves trigonometric functions $\sin, \cos, \tan$ etc.
+- Keywords: "Trigonometric functions", "Trigonometric identities", "Solution of triangles", "Sine theorem", "Cosine theorem"
+- Example: Proof $\sin^2\theta + \cos^2\theta = 1$
+
+**Recommended solution strategy:**
+- Memorize and apply basic identities
+- Sum and difference formula: $\sin(A \pm B), \cos(A \pm B)$
+- Double angle formula: $\sin 2\theta = 2\sin\theta\cos\theta$
+- Solve triangles: sine theorem $\frac{a}{\sin A} = 2R$, cosine theorem $a^2 = b^2 + c^2 - 2bc\cos A$
+- Trigonometric equations: using periodicity to find general solutions
+
+**Conditions that must be checked:**
+- Angle range (acute angle, $[0, 2\pi)$, etc.)
+- Trigonometric function domain ($\tan\theta$ in $\theta \neq \frac{\pi}{2}+k\pi$)
+- Solutions of triangles (one solution/two solutions/no solution)
+- Units for radians and angles are unified
+
+**Common mistakes:**
+- Sine theorem/cosine theorem application errors
+- Incorrect determination of the number of triangle solutions (SSA case)
+-Induced formula symbol errors
+- Missing general explanation
+- Mixing radians and degrees
+
+**Recommended verification method:**
+- Verify the identity by substituting special angles
+- Calculate the same angle on both sides of the trigonometric identity
+- Verify triangle angle sum $180^\circ$
+- Use different methods to find the same amount of cross-validation
+
+**Output format:**
+ ```
+Solve/Prove: [Objective]
+Use the formula: [sine theorem/cosine theorem/sum and difference formula/...]
+Steps: [Derivation process]
+Result: [angle value/side length/identity proof]
+Number of solutions: [One solution/Two solutions/No solution]
+```
+
+**Whether to ask the user:** Only when the angle range is unclear
 
 ---
 
 ## sequence
 
-**识别特征：**
-- 涉及数列的定义、通项、求和
-- 关键词："等差数列"、"等比数列"、"通项公式"、"前 $n$ 项和"、"递推"
-- 示例：已知 $\{a_n\}$ 为等差数列，$a_3=5$，$a_7=13$，求 $a_n$ 和 $S_n$
-
-**推荐求解策略：**
-- 等差：$a_n = a_1 + (n-1)d$，$S_n = \frac{n(a_1+a_n)}{2} = na_1 + \frac{n(n-1)}{2}d$
-- 等比：$a_n = a_1 q^{n-1}$，$S_n = \frac{a_1(1-q^n)}{1-q}$（$q \neq 1$）
-- 递推数列：特征根法、累加/累乘法、不动点法
-- 通项与求和关系：$a_n = S_n - S_{n-1}$（$n \geq 2$）
-
-**必须检查的条件：**
-- $n$ 的范围（$n \in \mathbb{N}^*$）
-- 等比数列公比 $q=1$ 的特殊情况
-- 递推公式的初始条件
-- 特征根为复根的情况
-- 无穷递缩等比数列（$|q| < 1$）的和
-
-**常见错误：**
-- 等差/等比公式混淆
-- 递推数列初始项代入范围错误
-- 等比求和漏掉 $q=1$ 的特例
-- 分组求和时项数计算错误
-- 错位相减法符号错误
-
-**推荐验证方法：**
-- 代入前几项验证通项公式
-- 直接列举前几项求和对比
-- 利用 $a_n = S_n - S_{n-1}$ 验证
-- 数学归纳法验证
-
-**输出格式：**
-```
-数列类型: [等差/等比/递推]
-通项公式: $a_n = [表达式]$
-前 $n$ 项和: $S_n = [表达式]$
-推导步骤: [过程]
-```
-
-**是否追问用户：** 仅当初始条件不完整时
+**Identification Features:**
+- Involving the definition, general terms and summation of sequence
+- Keywords: "arithmetic sequence", "geometric sequence", "general formula", "sum of first $n$ terms", "recursion"
+- Example: Given that $\{a_n\}$ is an arithmetic sequence, $a_3=5$ , $a_7=13$ , find $a_n$ and $S_n$
+
+**Recommended solution strategy:**
+- Equal differences: $a_n = a_1 + (n-1)d$, $S_n = \frac{n(a_1+a_n)}{2} = na_1 + \frac{n(n-1)}{2}d$
+- Equivalent: $a_n = a_1 q^{n-1}$, $S_n = \frac{a_1(1-q^n)}{1-q}$ ($q \neq 1$)
+- Recursive sequence: characteristic root method, accumulation/accumulation multiplication method, fixed point method
+- General term and summation relationship: $a_n = S_n - S_{n-1}$ ($n \geq 2$)
+
+**Conditions that must be checked:**
+- Range of $n$ ( $n \in \mathbb{N}^*$ )
+- Special case of geometric sequence common ratio $q=1$
+- Initial conditions for recursive formulas
+- The case where the characteristic root is a complex root
+- The sum of infinite decreasing geometric sequences ($|q| < 1$)
+
+**Common mistakes:**
+- Confusion with equal difference/equal ratio formulas
+- The initial item of the recursive sequence is incorrectly substituted into the range
+- The special case of missing $q=1$ in proportional summation
+- Wrong calculation of number of terms when summing by groups
+- Misplaced subtraction sign error
+
+**Recommended verification method:**
+- Substitute the first few items to verify the general formula
+- Directly list the first few items and compare them
+- Verify using $a_n = S_n - S_{n-1}$
+- Mathematical induction verification
+
+**Output format:**
+ ```
+Sequence type: [arithmetic/equal ratio/recursion]
+General formula: $a_n = [expression]$
+Sum of first $n$ terms: $S_n = [expression]$
+Derivation steps: [Process]
+```
+
+**Whether to ask the user:** Only if the initial conditions are incomplete
 
 ---
 
 ## combinatorics
 
-**识别特征：**
-- 涉及计数问题、排列组合
-- 关键词："排列"、"组合"、"有多少种"、"选法"、"排法"
-- 示例：从 5 个男生和 4 个女生中选 3 人组成委员会，要求至少 1 名女生，有多少种选法？
-
-**推荐求解策略：**
-- 排列：有顺序，$A_n^m = \frac{n!}{(n-m)!}$ 或 $P_n^m$
-- 组合：无顺序，$C_n^m = \binom{n}{m} = \frac{n!}{m!(n-m)!}$
-- 分类计数原理（加法原理）和分步计数原理（乘法原理）
-- 补集法：总数减去不满足条件的
-- 捆绑法、插空法处理相邻/不相邻问题
-
-**必须检查的条件：**
-- 是否有重复元素
-- 是否考虑顺序（排列 vs 组合）
-- 是否允许重复选取
-- 分类是否完备且不重叠
-
-**常见错误：**
-- 排列和组合混淆
-- 分类有重叠导致重复计数
-- 分类遗漏
-- 相邻/不相邻问题的插入位置计算错误
-- 隔板法适用条件判断错误
-
-**推荐验证方法：**
-- 对较小规模枚举验证
-- 补集法交叉验证
-- 递推公式验证
-- 使用不同分类方式得到相同结果
-
-**输出格式：**
-```
-问题类型: [排列/组合/混合]
-方法: [分类计数/分步计数/补集法/捆绑法/...]
-计算: [公式代入]
-结果: [数值]
-枚举验证 (小规模): [枚举结果]
-```
-
-**是否追问用户：** 仅当"相同"和"不同"元素有歧义时
+**Identification Features:**
+- Involving counting problems, permutations and combinations
+- Keywords: "arrangement", "combination", "how many types", "selection method", "arrangement method"
+- Example: Select 3 people from 5 boys and 4 girls to form a committee, requiring at least 1 girl. How many choices are there?
+
+**Recommended solution strategy:**
+- Arrangement: in order, $A_n^m = \frac{n!}{(n-m)!}$ or $P_n^m$
+- Combination: no order, $C_n^m = \binom{n}{m} = \frac{n!}{m!(n-m)!}$
+- Principle of classification counting (addition principle) and step counting principle (multiplication principle)
+- Complement method: subtract the total number that does not meet the conditions
+- Bundling method and interpolation method to deal with adjacent/non-adjacent problems
+
+**Conditions that must be checked:**
+- Are there duplicate elements?
+- Whether to consider order (permutation vs combination)
+- Whether to allow repeated selection
+- Whether the classification is complete and non-overlapping
+
+**Common mistakes:**
+- Confusion of permutations and combinations
+- Overlapping categories lead to double counting
+- Classification missing
+- Wrong calculation of insertion position for adjacent/non-adjacent questions
+- Wrong judgment on the applicable conditions of the Partition Law
+
+**Recommended verification method:**
+- Validation of smaller enumerations
+- Complementary set method cross-validation
+- Recursive formula verification
+- Use different classification methods to get the same results
+
+**Output format:**
+ ```
+Question Type: [Permutation/Combination/Mixed]
+Method: [Classification counting/step counting/complement method/bundling method/...]
+Calculation: [Formula Substitution]
+Result: [numeric value]
+Enumeration verification (small scale): [enumeration result]
+```
+
+**Whether to ask the user:** Only if there is ambiguity between "same" and "different" elements
 
 ---
 
 ## probability_statistics
 
-**识别特征：**
-- 涉及概率计算、随机变量、分布、统计量
-- 关键词："概率"、"期望"、"方差"、"分布"、"贝叶斯"、"正态分布"
-- 示例：掷两枚骰子，求点数之和为 7 的概率
-
-**推荐求解策略：**
-- 古典概型：$P(A) = \frac{\text{有利情况数}}{\text{总情况数}}$
-- 条件概率：$P(A|B) = \frac{P(AB)}{P(B)}$，贝叶斯公式
-- 独立事件：$P(AB) = P(A) \cdot P(B)$
-- 二项分布：$P(X=k) = C_n^k p^k (1-p)^{n-k}$
-- 期望：$E(X) = \sum x_i p_i$，方差：$D(X) = E(X^2) - [E(X)]^2$
-
-**必须检查的条件：**
-- 事件是否互斥（$P(A+B) = P(A) + P(B)$）
-- 事件是否独立
-- 全概率公式中划分是否完备
-- 样本空间是否等可能
-- 几何概型中的测度定义
-
-**常见错误：**
-- 加法公式误用（未减去交的概率）
-- 条件概率与无条件概率混淆
-- 期望的线性性质条件忽略
-- 二项分布成功概率 $p$ 恒定条件忽略
-- 各阶矩计算错误
-
-**推荐验证方法：**
-- 概率总和是否等于 1
-- 期望和方差公式验证
-- 用简单枚举验证小规模问题
-- 不同解法得到一致结果
-
-**输出格式：**
-```
-概率类型: [古典概型/条件概率/几何概型]
-样本空间: [描述]
-有利情况: [描述与计数]
-概率: $P = [分数或小数]$
-期望 (如有): $E(X) = [值]$
-```
-
-**是否追问用户：** 仅当"随机"或"等可能"的含义不明确时
+**Identification Features:**
+- Involving probability calculations, random variables, distributions, statistics
+- Keywords: "probability", "expectation", "variance", "distribution", "Bayes", "normal distribution"
+- Example: Roll two dice and find the probability that the sum of the points is 7
+
+**Recommended solution strategy:**
+- Classical profile: $P(A) = \frac{\text{Number of favorable situations}}{\text{Number of total situations}}$
+- Conditional probability: $P(A|B) = \frac{P(AB)}{P(B)}$, Bayesian formula
+- Independent event: $P(AB) = P(A) \cdot P(B)$
+- Binomial distribution: $P(X=k) = C_n^k p^k (1-p)^{n-k}$
+- Expectation: $E(X) = \sum x_i p_i$, Variance: $D(X) = E(X^2) - [E(X)]^2$
+
+**Conditions that must be checked:**
+- Whether events are mutually exclusive ($P(A+B) = P(A) + P(B)$)
+- Whether the events are independent
+- Is the division in the total probability formula complete?
+- Is the sample space equal to possibility?
+- Definition of measures in geometric concepts
+
+**Common mistakes:**
+- Misuse of the addition formula (not subtracting the probability of intersection)
+- Confusion between conditional probability and unconditional probability
+- Expected linearity conditions ignored
+- Binomial distribution probability of success $p$ constant conditions ignored
+- Calculation errors of each order moment
+
+**Recommended verification method:**
+- Whether the sum of probabilities equals 1
+- Validation of expectation and variance formulas
+- Verify small-scale problems with simple enumerations
+- Different solutions get consistent results
+
+**Output format:**
+ ```
+Probability type: [classical concept/conditional probability/geometric concept]
+Sample space: [description]
+Advantages: [Description and Count]
+Probability: $P = [fraction or decimal]$
+Expectation (if any): $E(X) = [value]$
+```
+
+**Whether to ask the user:** Only if the meaning of "random" or "equally possible" is unclear
 
 ---
 
 ## word_problem
 
-**识别特征：**
-- 用自然语言描述的实际问题，需要建立数学模型
-- 关键词：实际问题场景（行程、工程、浓度、利润、年龄等）
-- 示例：甲、乙两人从 A、B 两地相向而行，甲的速度是乙的 1.5 倍，3 小时后相遇，求两人速度
-
-**推荐求解策略：**
-1. 仔细阅读，提取关键信息
-2. 设未知数（明确物理意义）
-3. 根据等量关系列方程/不等式
-4. 求解数学模型
-5. 检验解的合理性（是否符合实际）
-
-**必须检查的条件：**
-- 隐含条件（速度 $>0$，年龄为正整数等）
-- 单位是否统一
-- 解的物理意义是否合理
-- 是否存在多解或模型不唯一
-
-**常见错误：**
-- 设未知数不当导致方程复杂
-- 忽略隐含条件
-- 单位不统一
-- 实际问题中的取整要求忽略
-- 关系式列错（如速度 = 路程/时间）
-
-**推荐验证方法：**
-- 将解代入原问题场景验证
-- 估算合理性
-- 量纲分析
-- 用不同设未知数方式重新建模求解
-
-**输出格式：**
-```
-问题分析: [提取的已知量和未知量]
-设未知数: [变量定义]
-建立模型: [方程/不等式]
-求解: [计算过程]
-结果: [带单位的答案]
-实际合理性检验: [验证]
-```
-
-**是否追问用户：** 是，当条件有歧义或缺少必要信息时
+**Identification Features:**
+- Practical problems described in natural language require the establishment of mathematical models
+- Keywords: Actual problem scenarios (trip, project, concentration, profit, age, etc.)
+- Example: Two people A and B are traveling towards each other from A and B. The speed of A is 1.5 times that of B. They will meet after 3 hours. Find the speed of the two people.
+
+**Recommended solution strategy:**
+1. Read carefully and extract key information
+2. Assume unknown numbers (clear physical meaning)
+3. Series of equations/inequality based on equivalence relations
+4. Solve mathematical models
+5. Test the rationality of the solution (whether it is consistent with reality)
+
+**Conditions that must be checked:**
+- Implied conditions (speed $>0$ , age a positive integer, etc.)
+- Is the unit unified?
+- Is the physical meaning of the solution reasonable?
+- Whether there are multiple solutions or the model is not unique
+
+**Common mistakes:**
+- Improper assumption of unknowns leads to complicated equations
+- Ignore implicit conditions
+- Units are not unified
+- Rounding requirements in actual questions are ignored
+- The relationship is incorrectly listed (e.g. speed = distance/time)
+
+**Recommended verification method:**
+- Substitute the solution into the original problem scenario to verify
+- Reasonableness of estimates
+- Dimensional analysis
+- Re-model and solve using different methods of setting unknowns
+
+**Output format:**
+ ```
+Problem analysis: [Extracted known quantities and unknown quantities]
+Assume unknown number: [variable definition]
+Modeling: [Equations/Inequalities]
+Solution: [Calculation process]
+Result: [Answer with units]
+Actual plausibility test: [Verification]
+```
+
+**Whether to ask the user:** Yes, when the conditions are ambiguous or necessary information is missing
 
 ---
 
 ## limit
 
-**识别特征：**
-- 涉及数列极限或函数极限
-- 关键词："极限"、"$\lim$"、"趋于"、"趋向"、"收敛"
-- 示例：求 $\lim_{x \to 0} \frac{\sin x}{x}$
-
-**推荐求解策略：**
-- 代入法（若能直接代入且无不定式）
-- 不定式类型识别（$\frac{0}{0}, \frac{\infty}{\infty}, 0\cdot\infty, \infty-\infty, 1^\infty, 0^0, \infty^0$）
-- 等价无穷小替换：$\sin x \sim x, \tan x \sim x, \ln(1+x) \sim x, e^x-1 \sim x$（$x \to 0$）
-- 洛必达法则（$0/0$ 或 $\infty/\infty$ 型）
-- 重要极限：$\lim_{x\to 0}\frac{\sin x}{x}=1$，$\lim_{x\to\infty}(1+\frac{1}{x})^x=e$
-- 夹逼定理
-
-**必须检查的条件：**
-- 洛必达法则条件（分子分母可导，极限为 $0/0$ 或 $\infty/\infty$）
-- 等价无穷小替换仅在乘除中使用（加减不直接用）
-- 左右极限是否相等
-- 函数在极限点的定义域
-
-**常见错误：**
-- 等价无穷小在加减法中直接使用
-- 洛必达法则条件不满足时使用
-- 忽略左右极限
-- 极限不存在但错误地得出结论
-- 夹逼定理上下界选择不当
-
-**推荐验证方法：**
-- 数值逼近（代入趋近值计算）
-- 泰勒展开验证
-- 洛必达法则结果验证 $f'(x)/g'(x)$ 的极限
-- 夹逼定理构造验证
-
-**输出格式：**
-```
-极限: $\lim_{x \to [值]} [表达式]$
-类型: [确定型 / 不定式 $\frac{0}{0}$ / ...]
-方法: [等价替换/洛必达/夹逼/...]
-结果: [极限值]
-验证: [数值逼近或替代方法结果]
-```
-
-**是否追问用户：** 否
+**Identification Features:**
+- Involving sequence limits or function limits
+- Keywords: "limit", "$\lim$", "tend to", "tendency", "convergence"
+- Example: Find $\lim_{x \to 0} \frac{\sin x}{x}$
+
+**Recommended solution strategy:**
+- Substitution method (if it can be substituted directly and there is no infinitive)
+- Infinitive type identification ($\frac{0}{0}, \frac{\infty}{\infty}, 0\cdot\infty, \infty-\infty, 1^\infty, 0^0, \infty^0$)
+- Equivalent infinitesimal substitution: $\sin x \sim x, \tan x \sim x, \ln(1+x) \sim x, e^x-1 \sim x$ ( $x \to 0$ )
+- Lópida's Law (Type $0/0$ or $\infty/\infty$)
+- Important limits: $\lim_{x\to 0}\frac{\sin x}{x}=1$, $\lim_{x\to\infty}(1+\frac{1}{x})^x=e$
+- Pinch theorem
+
+**Conditions that must be checked:**
+- Conditions of Lópida's law (the numerator and denominator are differentiable, the limit is $0/0$ or $\infty/\infty$ )
+- Equivalent infinitesimal substitution is only used in multiplication and division (not directly used in addition and subtraction)
+- Are the left and right limits equal?
+- The domain of the function at the limit point
+
+**Common mistakes:**
+- Equivalent infinitesimals are used directly in addition and subtraction
+- Used when the conditions of Lópida's Law are not met
+- Ignore left and right limits
+- Wrong conclusion that limit does not exist
+- Improper selection of upper and lower bounds for the pinch theorem
+
+**Recommended verification method:**
+- Numerical approximation (substitute the approximate value for calculation)
+-Taylor expanded verification
+- The result of L'Block's law verifies the limit of $f'(x)/g'(x)$
+- Construction verification of the pinching theorem
+
+**Output format:**
+ ```
+Limit: $\lim_{x \to [value]} [expression]$
+Type: [deterministic / infinitive $\frac{0}{0}$ / ...]
+Method: [equivalent replacement/Lópida/pinch/...]
+Result: [limit value]
+Verification: [Numerical approximation or alternative method results]
+```
+
+**Whether to ask the user:** No
 
 ---
 
 ## differentiation
 
-**识别特征：**
-- 涉及求导、切线、变化率
-- 关键词："求导"、"导数"、"$f'(x)$"、"$\frac{dy}{dx}$"、"切线方程"、"变化率"
-- 示例：求 $f(x) = x^3\ln x$ 的导数
-
-**推荐求解策略：**
-- 基本求导公式记忆
-- 四则运算法则：$(u \pm v)' = u' \pm v'$，$(uv)' = u'v + uv'$，$(\frac{u}{v})' = \frac{u'v-uv'}{v^2}$
-- 链式法则：$y = f(g(x)) \Rightarrow y' = f'(g(x)) \cdot g'(x)$
-- 隐函数求导
-- 参数方程求导：$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
-- 高阶导数
-
-**必须检查的条件：**
-- 函数在求导点是否可导
-- 链式法则内层函数可导性
-- 隐函数存在定理条件
-- 分式求导分母不为零
-
-**常见错误：**
-- 链式法则遗漏（复合函数求导不完整）
-- 乘积求导公式符号错误
-- 隐函数求导遗漏 $\frac{dy}{dx}$ 项
-- 对数求导法定义域检查遗漏
-- 高阶导数计算中项遗漏
-
-**推荐验证方法：**
-- 积分验证（导数的积分还原为原函数）
-- 数值求导验证（差分近似）
-- 对简单点代入验证
-- 使用求导公式对称性验证
-
-**输出格式：**
-```
-函数: $f(x) = [表达式]$
-导数: $f'(x) = [表达式]$
-方法: [直接求导/链式法则/隐函数求导/...]
-关键步骤: [主要变换过程]
-(可选) 切线方程: $y = f'(x_0)(x - x_0) + f(x_0)$
-```
-
-**是否追问用户：** 仅当对哪个变量求导不明确时
+**Identification Features:**
+- Involves derivation, tangent, rate of change
+- Keywords: "derivative", "derivative", "$f'(x)$", "$\frac{dy}{dx}$", "tangent equation", "rate of change"
+- Example: Find the derivative of $f(x) = x^3\ln x$
+
+**Recommended solution strategy:**
+- Memorize basic derivation formulas
+- Four arithmetic operations: $(u \pm v)' = u' \pm v'$, $(uv)' = u'v + uv'$, $(\frac{u}{v})' = \frac{u'v-uv'}{v^2}$
+- Chain Rule: $y = f(g(x)) \Rightarrow y' = f'(g(x)) \cdot g'(x)$
+- Implicit function derivation
+- Derivative of parametric equations: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
+- Higher order derivatives
+
+**Conditions that must be checked:**
+- Whether the function is differentiable at the derivation point
+- Chain rule inner function differentiability
+- Theorem conditions for the existence of implicit functions
+- The denominator of the derivative of a fraction is not zero.
+
+**Common mistakes:**
+- Missing chain rule (incomplete derivation of composite functions)
+- The sign of the product derivative formula is wrong
+- Implicit function derivation misses $\frac{dy}{dx}$ items
+- Logarithmic derivation domain check missing
+- Missing terms in the calculation of higher-order derivatives
+
+**Recommended verification method:**
+- Integral verification (the integral of the derivative is restored to the original function)
+- Numerical derivation verification (differential approximation)
+- Substitute verification for simple points
+- Symmetry verification using derivation formulas
+
+**Output format:**
+ ```
+Function: $f(x) = [expression]$
+Derivative: $f'(x) = [expression]$
+Method: [direct derivation/chain rule/implicit function derivation/...]
+Key steps: [Main transformation process]
+(Optional) Tangent equation: $y = f'(x_0)(x - x_0) + f(x_0)$
+```
+
+**Whether to ask the user:** Only if it is unclear which variable to derive the derivative from
 
 ---
 
 ## integration
 
-**识别特征：**
-- 涉及不定积分、定积分、面积、体积
-- 关键词："积分"、"$\int$"、"原函数"、"面积"、"体积"、"不定积分"、"定积分"
-- 示例：计算 $\int_0^1 x e^x \,dx$
-
-**推荐求解策略：**
-- 基本积分公式记忆
-- 换元积分法（第一类/凑微分，第二类/三角换元）
-- 分部积分法：$\int u\,dv = uv - \int v\,du$
-- 有理函数积分：部分分式分解
-- 三角有理式积分：万能公式
-- 定积分：牛顿-莱布尼茨公式 $\int_a^b f(x)dx = F(b) - F(a)$
-
-**必须检查的条件：**
-- 被积函数在积分区间上是否连续
-- 反常积分的收敛性
-- 换元后上下限的对应变换
-- 分部积分中 $u$ 和 $dv$ 的选取策略（LIATE 原则）
-
-**常见错误：**
-- 换元后忘记变换积分上下限
-- 分部积分方向选择不当导致更复杂
-- 部分分式分解遗漏项
-- 反常积分不检验收敛性
-- 绝对值函数积分忽略分段
-
-**推荐验证方法：**
-- 求导验证不定积分结果
-- 数值积分近似验证定积分
-- 换元法与原方法交叉验证
-- 对称性利用验证
-
-**输出格式：**
-```
-积分: $\int [表达式] \,dx$ (或定积分上下限)
-方法: [换元法/分部积分/部分分式/...]
-关键步骤: [换元设置 / $u, dv$ 选择 / 分解]
-原函数: $F(x) = [表达式] + C$
-定积分值 (如有): [值]
-```
-
-**是否追问用户：** 仅当积分范围不明确时
+**Identification Features:**
+- Involving indefinite integrals, definite integrals, area, and volume
+- Keywords: "integral", "$\int$", "original function", "area", "volume", "indefinite integral", "definite integral"
+- Example: Calculate $\int_0^1 x e^x \,dx$
+
+**Recommended solution strategy:**
+- Memory of basic integral formulas
+- Integral method with substitution of elements (first type/differentiation, second type/trigonometric substitution)
+- Integration by parts method: $\int u\,dv = uv - \int v\,du$
+- Integral of rational functions: partial fraction decomposition
+- Trigonometric Rational Integration: Universal Formula
+- Definite integral: Newton-Leibniz formula $\int_a^b f(x)dx = F(b) - F(a)$
+
+**Conditions that must be checked:**
+- Whether the integrand is continuous on the integration interval
+- Convergence of anomalous integrals
+- Corresponding transformation of upper and lower limits after substitution
+- Selection strategy of $u$ and $dv$ in partial integration (LIATE principle)
+
+**Common mistakes:**
+- Forgot to change the upper and lower limits of the points after changing yuan
+- Improper selection of the direction of partial integration leads to more complexity
+- Missing items in partial fraction decomposition
+- Abnormal integration does not test convergence
+- Integration of absolute value functions ignores segmentation
+
+**Recommended verification method:**
+- Derivation to verify indefinite integral results
+- Approximate verification of definite integrals by numerical integration
+- Cross-validation between the substitution method and the original method
+- Symmetry exploit verification
+
+**Output format:**
+ ```
+Integral: $\int [expression] \,dx$ (or upper and lower limits of definite integral)
+Method: [Substitution method/integration by parts/partial fractions/...]
+Key steps: [Change setting / $u, dv$ selection / decomposition]
+Original function: $F(x) = [expression] + C$
+Definite integral value (if any): [value]
+```
+
+**Whether to ask the user:** Only when the point range is unclear
 
 ---
 
 ## multivariable_calculus
 
-**识别特征：**
-- 涉及多元函数的偏导数、全微分、重积分
-- 关键词："偏导数"、"$\frac{\partial}{\partial x}$"、"全微分"、"二重积分"、"三重积分"、"梯度"
-- 示例：求 $f(x,y) = x^2y + e^{xy}$ 的偏导数 $\frac{\partial f}{\partial x}$ 和 $\frac{\partial f}{\partial y}$
-
-**推荐求解策略：**
-- 偏导数：对某一变量求导时其他变量视为常数
-- 链式法则：$\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}$
-- 全微分：$dz = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$
-- 方向导数与梯度：$\nabla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$
-- 二重积分：选择积分次序，直角坐标与极坐标互换
-- 三重积分：直角坐标、柱坐标、球坐标
-
-**必须检查的条件：**
-- 混合偏导数的可交换条件（$f_{xy} = f_{yx}$，当二阶偏导连续）
-- 重积分积分区域的正确描述
-- 雅可比行列式的计算（换元时）
-- 极坐标/球坐标的 $r$ 因子
-
-**常见错误：**
-- 偏导数时忘记其他变量视为常数
-- 二重积分交换次序时区域描述错误
-- 极坐标换元漏掉 $r\,dr\,d\theta$ 中的 $r$
-- 雅可比行列式计算错误
-- 格林公式/斯托克斯公式方向判断错误
-
-**推荐验证方法：**
-- 验证 $f_{xy} = f_{yx}$（二阶偏导）
-- 数值积分验证重积分
-- 换元与直接计算交叉验证
-- 对称性利用验证
-
-**输出格式：**
-```
-函数: $f(x,y,\dots) = [表达式]$
-偏导数: $\frac{\partial f}{\partial x} = [表达式], \frac{\partial f}{\partial y} = [表达式]$
-(可选) 梯度: $\nabla f = ([值], [值])$
-(可选) 重积分值: [值]
-方法: [直角坐标/极坐标/柱坐标/球坐标]
-```
-
-**是否追问用户：** 仅当积分区域不明确时
+**Identification Features:**
+- Partial derivatives, total differentials, and heavy integrals involving multivariate functions
+- Keywords: "Partial derivative", "$\frac{\partial}{\partial x}$", "Total differential", "Double integral", "Triple integral", "Gradient"
+- Example: Find the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ of $f(x,y) = x^2y + e^{xy}$
+
+**Recommended solution strategy:**
+- Partial derivatives: When deriving the derivative of a certain variable, other variables are treated as constants
+- Chain Rule: $\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}$
+- Full differential: $dz = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$
+- Directional derivatives and gradients: $\nabla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$
+- Double integral: select the order of integration, interchange rectangular coordinates and polar coordinates
+- Triple integral: rectangular coordinates, cylindrical coordinates, spherical coordinates
+
+**Conditions that must be checked:**
+- Commutative conditions for mixed partial derivatives ($f_{xy} = f_{yx}$, when second-order partial derivatives are continuous)
+- Correct description of the integration area for heavy integration
+- Calculation of Jacobian (when changing elements)
+- $r$ factor for polar/spherical coordinates
+
+**Common mistakes:**
+- Forget about treating other variables as constants when taking partial derivatives
+- Wrong area description when double integral exchange order
+- Polar coordinate substitution misses $r$ in $r\,dr\,d\theta$
+- Jacobian calculation error
+- Green's formula/Stokes' formula direction error
+
+**Recommended verification method:**
+- Verify $f_{xy} = f_{yx}$ (second-order partial derivative)
+- Numerical integration to verify re-integration
+- Substitution and direct calculation cross-validation
+- Symmetry exploit verification
+
+**Output format:**
+ ```
+Function: $f(x,y,\dots) = [expression]$
+Partial derivatives: $\frac{\partial f}{\partial x} = [expression], \frac{\partial f}{\partial y} = [expression]$
+(optional) gradient: $\nabla f = ([value], [value])$
+(optional) reintegration value: [value]
+Method: [Rectangular coordinates/polar coordinates/cylindrical coordinates/spherical coordinates]
+```
+
+**Whether to ask the user:** Only when the points area is unclear
 
 ---
 
 ## linear_algebra
 
-**识别特征：**
-- 涉及矩阵、向量、线性方程组、行列式、特征值
-- 关键词："矩阵"、"行列式"、"特征值"、"特征向量"、"秩"、"线性相关"、"对角化"
-- 示例：求矩阵 $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ 的特征值和特征向量
-
-**推荐求解策略：**
-- 行列式计算：展开法、初等变换、特殊矩阵公式
-- 矩阵运算：加法、乘法、转置、逆矩阵
-- 特征值：解 $|A - \lambda I| = 0$
-- 特征向量：解 $(A - \lambda I)\vec{x} = \vec{0}$
-- 对角化：$A = PDP^{-1}$（$P$ 的列是特征向量，$D$ 是对角矩阵）
-- 秩的计算：初等变换化为行阶梯形
-
-**必须检查的条件：**
-- 矩阵乘法是否可乘（维度匹配）
-- 逆矩阵存在条件（$|A| \neq 0$）
-- 对角化条件（是否有 $n$ 个线性无关的特征向量）
-- 实对称矩阵必然可对角化
-
-**常见错误：**
-- 矩阵乘法不可交换（$AB \neq BA$）
-- 行列式计算时符号错误
-- 特征方程求解错误
-- 特征向量自由变量设值不当
-- 正交化过程计算错误
-
-**推荐验证方法：**
-- 验证 $A\vec{v} = \lambda\vec{v}$ 对每个特征对
-- 验证 $P^{-1}AP = D$
-- 验证 $A \cdot A^{-1} = I$
-- 迹与特征值和的关系验证
-
-**输出格式：**
-```
-矩阵/向量: [表示]
-行列式 (如有): $|A| = [值]$
-特征值: $\lambda_1 = [值], \lambda_2 = [值], \dots$
-特征向量: $\vec{v}_1 = ([], \dots)^T, \dots$
-逆矩阵 (如有): $A^{-1} = [矩阵]$
-秩: $r(A) = [值]$
-```
-
-**是否追问用户：** 否
+**Identification Features:**
+- Involving matrices, vectors, linear equations, determinants, and eigenvalues
+- Keywords: "matrix", "determinant", "eigenvalue", "eigenvector", "rank", "linear correlation", "diagonalization"
+- Example: Find the eigenvalues ​​and eigenvectors of matrix $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$
+
+**Recommended solution strategy:**
+- Determinant calculation: expansion method, elementary transformation, special matrix formula
+- Matrix operations: addition, multiplication, transpose, inverse matrix
+- Eigenvalue: Solution $|A - \lambda I| = 0$
+- Eigenvector: Solution $(A - \lambda I)\vec{x} = \vec{0}$
+- Diagonalization: $A = PDP^{-1}$ (the columns of $P$ are eigenvectors and $D$ is the diagonal matrix)
+- Calculation of rank: elementary transformation into row echelon form
+
+**Conditions that must be checked:**
+- Is matrix multiplication multiplicable (dimension matching)
+- Condition for the existence of inverse matrix ($|A| \neq 0$)
+- Diagonalization condition (whether there are $n$ linearly independent eigenvectors)
+- A real symmetric matrix must be diagonalizable
+
+**Common mistakes:**
+- Matrix multiplication is not commutative ($AB \neq BA$)
+- Wrong sign during determinant calculation
+- Error in solving characteristic equation
+- Improper setting of eigenvector free variables
+- Calculation errors in the orthogonalization process
+
+**Recommended verification method:**
+- Verify $A\vec{v} = \lambda\vec{v}$ for each feature pair
+- Verify $P^{-1}AP = D$
+- Verify $A \cdot A^{-1} = I$
+- Verification of the relationship between trace and eigenvalue sum
+
+**Output format:**
+ ```
+matrix/vector: [representation]
+Determinant (if any): $|A| = [value]$
+Eigenvalues: $\lambda_1 = [value], \lambda_2 = [value], \dots$
+Feature vector: $\vec{v}_1 = ([], \dots)^T, \dots$
+Inverse matrix (if any): $A^{-1} = [matrix]$
+Rank: $r(A) = [value]$
+```
+
+**Whether to ask the user:** No
 
 ---
 
 ## ordinary_differential_equation
 
-**识别特征：**
-- 涉及含有导数的方程
-- 关键词："微分方程"、"通解"、"特解"、"初始条件"、"$y'$"、"$\frac{dy}{dx}$"
-- 示例：解微分方程 $y' + 2xy = x$，满足 $y(0) = 1$
-
-**推荐求解策略：**
-- 识别方程类型：可分离变量、一阶线性、齐次、伯努利、恰当方程等
-- 可分离变量：$\frac{dy}{dx} = f(x)g(y) \Rightarrow \int \frac{dy}{g(y)} = \int f(x)dx$
-- 一阶线性：$y' + P(x)y = Q(x)$，积分因子 $\mu = e^{\int P(x)dx}$
-- 二阶常系数线性：特征方程 $r^2 + pr + q = 0$
-- 常数变易法
-- 初始条件确定特解
-
-**必须检查的条件：**
-- $g(y) = 0$ 是否产生奇解
-- 积分因子是否有定义
-- 特征根类型（实根/重根/复根）对应的解形式
-- 方程是否为线性
-
-**常见错误：**
-- 可分离变量时漏掉 $g(y) = 0$ 的解
-- 积分因子计算错误
-- 特征方程的解形式选错（实根/重根/复根）
-- 线性无关解判断错误
-- 特解形式设错
-
-**推荐验证方法：**
-- 将解代入原方程验证
-- 验证通解包含所有线性无关解
-- 验证特解满足初始条件
-- 数值解法近似验证
-
-**输出格式：**
-```
-方程类型: [可分离变量/一阶线性/二阶常系数/...]
-通解: $y = [表达式] + C$
-(可选) 特解: $y = [表达式]$
-求解方法: [积分因子/特征方程/常数变易法]
-步骤: [主要变换]
-```
-
-**是否追问用户：** 仅当缺少初始条件时
+**Identification Features:**
+- Involves equations containing derivatives
+- Keywords: "differential equation", "general solution", "particular solution", "initial conditions", " $y'$ ", " $\frac{dy}{dx}$ "
+- Example: Solve differential equation $y' + 2xy = x$, satisfying $y(0) = 1$
+
+**Recommended solution strategy:**
+- Identify equation types: separable variables, first-order linear, homogeneous, Bernoulli, proper equation, etc.
+- Detachable variable: $\frac{dy}{dx} = f(x)g(y) \Rightarrow \int \frac{dy}{g(y)} = \int f(x)dx$
+- First-order linear: $y' + P(x)y = Q(x)$, integration factor $\mu = e^{\int P(x)dx}$
+- Second-order linear with constant coefficients: characteristic equation $r^2 + pr + q = 0$
+- Constant variation method
+- Initial conditions determine the specific solution
+
+**Conditions that must be checked:**
+- $g(y) = 0$ Whether a strange solution is generated
+- Is the integration factor defined?
+- The solution form corresponding to the characteristic root type (real root/multiple root/complex root)
+- Is the equation linear?
+
+**Common mistakes:**
+- Missing solution to $g(y) = 0$ when variables are separable
+- Calculation error of integration factor
+- The solution form of the characteristic equation is wrongly selected (real roots/multiple roots/complex roots)
+- Error in judging linearly independent solutions
+- Wrong setting of special solution form
+
+**Recommended verification method:**
+- Substitute the solution into the original equation to verify
+- Verify that the general solution contains all linearly independent solutions
+- Verify that the particular solution satisfies the initial conditions
+- Approximate verification of numerical solutions
+
+**Output format:**
+ ```
+Equation type: [separable variable/first-order linear/second-order constant coefficient/...]
+General solution: $y = [expression] + C$
+(optional) Special solution: $y = [expression]$
+Solution method: [Integral factor/Characteristic equation/Constant variation method]
+Steps: [Main transformation]
+```
+
+**Whether to ask the user:** Only if initial conditions are missing
 
 ---
 
 ## complex_analysis
 
-**识别特征：**
-- 涉及复数函数、留数、围道积分
-- 关键词："复数"、"复变函数"、"留数"、"解析"、"$e^{iz}$"、"围道积分"
-- 示例：计算留数 $\operatorname{Res}\left(\frac{e^z}{z^2}, 0\right)$
-
-**推荐求解策略：**
-- 判断函数是否解析（柯西-黎曼方程）
-- 留数计算：$\operatorname{Res}(f, z_0) = \frac{1}{(m-1)!}\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^m f(z)]$（$m$ 阶极点）
-- 围道积分：$\oint_C f(z)dz = 2\pi i \sum \operatorname{Res}(f, z_k)$
-- 洛朗展开
-- 利用柯西积分公式
-
-**必须检查的条件：**
-- 极点阶数判断
-- 围道内包含的奇点
-- 函数是否在围道内除孤立奇点外解析
-- 分支切割的处理
-
-**常见错误：**
-- 极点阶数判断错误
-- 留数计算公式使用不当
-- 围道内奇点遗漏
-- 围道方向与符号关系搞反
-- 洛朗级数展开错误
-
-**推荐验证方法：**
-- 直接积分与留数定理结果对比
-- 柯西积分公式验证
-- 级数展开交叉验证
-
-**输出格式：**
-```
-函数: $f(z) = [表达式]$
-奇点: $z = [值]$ (类型: [极点/本性奇点/可去奇点])
-留数: $\operatorname{Res}(f, z_0) = [值]$
-(可选) 围道积分值: [值]
-```
-
-**是否追问用户：** 否
+**Identification Features:**
+- Involving complex functions, residues, and circumferential integrals
+- Keywords: "complex number", "complex function", "residue", "analysis", "$e^{iz}$", "circuit integral"
+- Example: Calculate residue $\operatorname{Res}\left(\frac{e^z}{z^2}, 0\right)$
+
+**Recommended solution strategy:**
+- Determine whether the function is analytical (Cauchy-Riemann equation)
+- Residue calculation: $\operatorname{Res}(f, z_0) = \frac{1}{(m-1)!}\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^m f(z)]$ ($m$ order pole)
+-Block points: $\oint_C f(z)dz = 2\pi i \sum \operatorname{Res}(f, z_k)$
+- Laurent Expand
+-Use the Cauchy integral formula
+
+**Conditions that must be checked:**
+- Judgment of pole order
+- Singularities contained within the enclosure
+- Whether the function is analytic in the enclosure except for isolated singularities
+- Processing of branch cutting
+
+**Common mistakes:**
+- Wrong judgment of pole order
+- Improper use of the residue calculation formula
+- Omission of singular points in the enclosure
+- The direction of the enclosure and the relationship between symbols are reversed
+- Laurent series expansion error
+
+**Recommended verification method:**
+- Comparison of direct integration and residue theorem results
+- Cauchy integral formula verification
+- Series expansion cross validation
+
+**Output format:**
+ ```
+Function: $f(z) = [expression]$
+Singularity: $z = [value]$ (type: [pole/nature singularity/removable singularity])
+Residue: $\operatorname{Res}(f, z_0) = [value]$
+(optional) perimeter integral value: [value]
+```
+
+**Whether to ask the user:** No
 
 ---
 
 ## real_analysis
 
-**识别特征：**
-- 涉及严格的实分析概念证明
-- 关键词："$\epsilon$-$\delta$ 定义"、"一致连续"、"一致收敛"、"勒贝格"、"测度"、"完备性"
-- 示例：用 $\epsilon$-$\delta$ 语言证明 $\lim_{x\to 2} x^2 = 4$
-
-**推荐求解策略：**
-- 理解并应用 $\epsilon$-$\delta$（$\epsilon$-$N$）定义
-- 从 $|f(x)-L|<\epsilon$ 出发反推 $|x-a|<\delta$
-- 利用放缩技巧
-- 确界原理、单调有界定理等实数完备性定理
-- 一致连续性的判定：闭区间上连续则一致连续
-
-**必须检查的条件：**
-- $\epsilon$ 的任意性
-- $\delta$ 对 $\epsilon$ 的依赖关系
-- 定义域是否为闭区间
-- 函数连续性的严格定义
-
-**常见错误：**
-- $\delta$ 只与 $\epsilon$ 有关，不应依赖于 $x$
-- 放缩过度导致 $\delta$ 不满足要求
-- 混淆一致连续与逐点连续
-- 实数完备性定理的逻辑推理链断裂
-- 反证法的假设不当
-
-**推荐验证方法：**
-- 逻辑推导每一步检查
-- 构造具体 $\epsilon$ 值验证
-- 反例检验
-
-**输出格式：**
-```
-命题: [陈述]
-证明方法: [$\epsilon$-$\delta$ / 反证法 / 确界 / ...]
-证明: [严格推导过程]
-结论: [命题成立/不成立]
-```
-
-**是否追问用户：** 是，需要确认所需严格程度
+**Identification Features:**
+- Involves rigorous real-analytic proof of concept
+- Keywords: "$\epsilon$ - $\delta$ definition", "consistent continuity", "consistent convergence", "Lebesgue", "measure", "completeness"
+- Example: Prove $\lim_{x\to 2} x^2 = 4$ in $\epsilon$ - $\delta$ language
+
+**Recommended solution strategy:**
+- Understand and apply $\epsilon$ - $\delta$ ( $\epsilon$ - $N$ ) definitions
+- Starting from $|f(x)-L|<\epsilon$ and working backward to $|x-a|<\delta$
+- Utilize zooming techniques
+- Real number completeness theorems such as the certainty principle and the monotonic bounded theorem
+- Judgment of consistent continuity: If it is continuous on a closed interval, it is consistent and continuous.
+
+**Conditions that must be checked:**
+- Arbitrariness of $\epsilon$
+- Dependency of $\delta$ on $\epsilon$
+- Whether the domain is a closed interval
+- Strict definition of function continuity
+
+**Common mistakes:**
+- $\delta$ is only related to $\epsilon$ and should not depend on $x$
+- Excessive scaling causes $\delta$ to not meet the requirements
+- Confusing uniform continuity with point-wise continuity
+- The logical reasoning chain of the real number completeness theorem is broken
+- Improper assumptions in proof by contradiction
+
+**Recommended verification method:**
+- Logical derivation for each step check
+-Construct concrete $\epsilon$ value validation
+- Counterexample test
+
+**Output format:**
+ ```
+proposition: [statement]
+Proof method: [$\epsilon$-$\delta$ / proof by contradiction / certainty / ...]
+Proof: [Strict derivation process]
+Conclusion: [The proposition is true/not true]
+```
+
+**Whether to ask the user:** Yes, you need to confirm the required strictness
 
 ---
 
 ## abstract_algebra
 
-**识别特征：**
-- 涉及群、环、域等代数结构的定义和性质
-- 关键词："群"、"环"、"域"、"同态"、"同构"、"子群"、"正规子群"、"理想"
-- 示例：证明 $G$ 的子群 $H$ 是正规子群当且仅当 $\forall g \in G, gHg^{-1} = H$
-
-**推荐求解策略：**
-- 验证群公理（封闭性、结合律、单位元、逆元）
-- 子群判定：$H \leq G \iff \forall a,b \in H, ab^{-1} \in H$
-- 正规子群：$\forall g\in G, gH = Hg$ 或 $gHg^{-1} \subseteq H$
-- 同态基本定理：$G/\ker\phi \cong \operatorname{Im}\phi$
-- 拉格朗日定理：子群的阶整除群的阶
-- 环的理想判断
-
-**必须检查的条件：**
-- 运算是否封闭
-- 群元素的阶
-- 子群与正规子群的区别
-- 同态映射是否良定义
-
-**常见错误：**
-- 群公理验证遗漏
-- 正规子群与子群混淆
-- 商群构造时正规子群条件缺失
-- 同态映射定义的良定性忽略
-- 拉格朗日定理逆命题误用
-
-**推荐验证方法：**
-- 逐条验证公理
-- 构造反例验证非性质
-- 用具体群（$\mathbb{Z}_n, S_3$ 等）测试一般结论
-- 同构映射双向验证
-
-**输出格式：**
-```
-代数结构: [群/环/域/模]
-已知: [给定条件]
-求证: [要证明的性质]
-证明: [逻辑推导]
-关键定理: [使用的定理名称]
-```
-
-**是否追问用户：** 是，确认代数结构的精确定义
+**Identification Features:**
+- Involving the definition and properties of algebraic structures such as groups, rings, fields, etc.
+- Keywords: "group", "ring", "domain", "homomorphism", "isomorphism", "subgroup", "normal subgroup", "ideal"
+- Example: Prove that the subgroup $H$ of $G$ is a normal subgroup if and only if $\forall g \in G, gHg^{-1} = H$
+
+**Recommended solution strategy:**
+- Verify group axioms (closure, associativity, identity element, inverse element)
+- Subgroup determination: $H \leq G \iff \forall a,b \in H, ab^{-1} \in H$
+- Regular subgroup: $\forall g\in G, gH = Hg$ or $gHg^{-1} \subseteq H$
+- Fundamental theorem of homomorphism: $G/\ker\phi \cong \operatorname{Im}\phi$
+- Lagrange's theorem: The order of a subgroup divides the order of the group
+- Ideal judgment of the ring
+
+**Conditions that must be checked:**
+- Whether the operation is closed
+- order of group elements
+- The difference between subgroups and regular subgroups
+- Is homomorphic mapping well defined?
+
+**Common mistakes:**
+- Missing group axiom verification
+- Confusion between normal subgroups and subgroups
+- The normal subgroup condition is missing when constructing the quotient group
+- Well-qualified neglect of homomorphic mapping definitions
+- Misuse of the converse of Lagrange's theorem
+
+**Recommended verification method:**
+- Verify the axioms one by one
+- Construct counterexamples to verify non-properties
+- Test general conclusions with specific groups ($\mathbb{Z}_n, S_3$, etc.)
+- Two-way verification of isomorphic mapping
+
+**Output format:**
+ ```
+Algebraic structure: [group/ring/field/module]
+Known: [given conditions]
+Proof: [Property to be proved]
+Proof: [logical deduction]
+Key theorem: [name of theorem used]
+```
+
+**Whether to ask the user:** Yes, confirm the precise definition of the algebraic structure
 
 ---
 
 ## topology
 
-**识别特征：**
-- 涉及开集、闭集、连续性、紧致性、连通性等拓扑概念
-- 关键词："开集"、"闭集"、"紧致"、"连通"、"同胚"、"拓扑空间"、"豪斯多夫"
-- 示例：证明紧致空间的连续像是紧致的
-
-**推荐求解策略：**
-- 理解拓扑空间的定义（开集公理）
-- 连续性：开集的原像是开集（或闭集的原像是闭集）
-- 紧致性：任意开覆盖存在有限子覆盖
-- 连通性：不能表示为两个不相交非空开集的并
-- 豪斯多夫空间：任意两点存在不相交的开邻域
-
-**必须检查的条件：**
-- 空间的拓扑是否明确给定
-- 是否为度量空间诱导的拓扑
-- 映射的连续性定义是否正确
-- 子空间拓扑的继承
-
-**常见错误：**
-- 混淆度量空间与一般拓扑空间的性质
-- 紧致性判定中开覆盖遗漏
-- 连通与道路连通混淆
-- 闭集在连续映射下的像不一定闭
-- 紧致空间与有界闭集的混淆
-
-**推荐验证方法：**
-- 在具体拓扑空间中验证一般定理
-- 构造反例验证非性质
-- 逐条检查公理和定义
-
-**输出格式：**
-```
-空间: $(X, \tau)$ / 具体空间
-性质/命题: [陈述]
-证明思路: [核心策略]
-证明: [严格推导]
-反例 (如有): [说明]
-```
-
-**是否追问用户：** 是，确认拓扑空间定义
+**Identification Features:**
+- Involving topological concepts such as open sets, closed sets, continuity, compactness, and connectivity
+- Keywords: "open set", "closed set", "compact", "connected", "homeomorphism", "topological space", "Hausdorff"
+- Example: Prove that the continuity of a compact space is compact
+
+**Recommended solution strategy:**
+- Understand the definition of topological space (open set axiom)
+- Continuity: The original shape of an open set is an open set (or the original shape of a closed set is a closed set)
+- Compactness: Any open cover has finite subcovers
+- Connectivity: cannot be expressed as the union of two disjoint non-empty open sets
+- Hausdorff space: any two points have disjoint open neighborhoods
+
+**Conditions that must be checked:**
+- Is the topology of the space clearly given?
+- Is it a metric space induced topology?
+- Is the mapping continuity defined correctly?
+- Inheritance of subspace topology
+
+**Common mistakes:**
+- Confusing the properties of metric spaces and general topological spaces
+- Omission of open coverage in tightness judgment
+- Confusion between connectivity and road connectivity
+- The image of a closed set under continuous mapping is not necessarily closed
+- Confusion between compact spaces and bounded closed sets
+
+**Recommended verification method:**
+- Verify general theorems in specific topological spaces
+- Construct counterexamples to verify non-properties
+- Check the axioms and definitions one by one
+
+**Output format:**
+ ```
+Space: $(X, \tau)$ / specific space
+Property/proposition: [statement]
+Proof idea: [Core strategy]
+Proof: [strict derivation]
+Counterexample (if any): [Explanation]
+```
+
+**Whether to ask the user:** Yes, confirm the topological space definition
 
 ---
 
 ## number_theory
 
-**识别特征：**
-- 涉及整数的整除性、素数、同余
-- 关键词："整除"、"素数"、"同余"、"$\pmod{n}$"、"费马小定理"、"欧拉定理"、"中国剩余定理"
-- 示例：证明不存在最大的素数
-
-**推荐求解策略：**
-- 带余除法：$a = bq + r$，$0 \leq r < |b|$
-- 最大公因数：欧几里得算法
-- 同余运算和性质
-- 费马小定理：$a^{p-1} \equiv 1 \pmod{p}$（$p$ 为素数，$p \nmid a$）
-- 欧拉定理：$a^{\varphi(n)} \equiv 1 \pmod{n}$（$\gcd(a,n)=1$）
-- 中国剩余定理求解同余方程组
-
-**必须检查的条件：**
-- 模运算中模数是否为正
-- 费马小定理中 $p$ 是否为素数
-- 欧拉定理中 $\gcd(a,n) = 1$
-- 中国剩余定理中模数两两互素
-
-**常见错误：**
-- 同余运算时模运算忘记取模
-- 费马小定理条件不满足时误用
-- 中国剩余定理使用时模数不互素
-- 整除符号混淆（$a \mid b$ vs $b \mid a$）
-- 素数判定忽略 $2$ 的特殊性
-
-**推荐验证方法：**
-- 代入具体数值验证
-- 不同方法交叉求解同余方程
-- 逆元验证（$a \cdot a^{-1} \equiv 1 \pmod{n}$）
-
-**输出格式：**
-```
-问题类型: [整除/同余/素数/丢番图方程]
-定理引用: [费马小定理/欧拉定理/中国剩余定理/...]
-求解/证明: [步骤]
-结果: [数值或结论]
-```
-
-**是否追问用户：** 否
+**Identification Features:**
+- Involving divisibility of integers, prime numbers, and congruence
+- Keywords: "Divisibility", "Prime Numbers", "Congruence", "$\pmod{n}$", "Fermat's Little Theorem", "Euler's Theorem", "Chinese Remainder Theorem"
+- Example: Prove that there is no largest prime number
+
+**Recommended solution strategy:**
+- Division with remainder: $a = bq + r$, $0 \leq r < |b|$
+- Greatest common factor: Euclidean algorithm
+- Congruence operations and properties
+- Fermat's Little Theorem: $a^{p-1} \equiv 1 \pmod{p}$ ($p$ is a prime number, $p \nmid a$)
+- Euler's Theorem: $a^{\varphi(n)} \equiv 1 \pmod{n}$ ( $\gcd(a,n)=1$ )
+- Chinese Remainder Theorem to solve the system of congruence equations
+
+**Conditions that must be checked:**
+- Whether the modulus in modular operation is positive
+- Whether $p$ in Fermat's Little Theorem is a prime number
+- Euler's theorem $\gcd(a,n) = 1$
+- In the Chinese Remainder Theorem, the modulus are pairwise prime
+
+**Common mistakes:**
+- Forgetting to take the modulus in the modulo operation during congruence operation
+- Misuse when the conditions of Fermat’s Little Theorem are not met
+- Modules are not mutually prime when using the Chinese Remainder Theorem
+- Divisor confusion ($a \mid b$ vs $b \mid a$)
+- Prime number determination ignores the specialness of $2$
+
+**Recommended verification method:**
+- Substitute specific numerical values ​​for verification
+- Cross-solve congruence equations using different methods
+- Inverse element verification ($a \cdot a^{-1} \equiv 1 \pmod{n}$)
+
+**Output format:**
+ ```
+Question Type: [Divisibility/Congruence/Prime Numbers/Diophantine Equation]
+Theorem reference: [Fermat's Little Theorem/Euler's Theorem/Chinese Remainder Theorem/...]
+Solution/Proof: [Step]
+Result: [numeric value or conclusion]
+```
+
+**Whether to ask the user:** No
 
 ---
 
 ## discrete_math
 
-**识别特征：**
-- 涉及图论、逻辑、集合论、布尔代数、关系
-- 关键词："图"、"顶点"、"边"、"逻辑"、"命题"、"集合"、"关系"、"格"、"布尔代数"
-- 示例：证明在任意 6 个人中，必存在 3 个人互相认识或互相不认识
-
-**推荐求解策略：**
-- 图论：握手定理、欧拉路径条件、平面图欧拉公式、着色
-- 逻辑：真值表、逻辑等价、范式转换
-- 集合论：文氏图、集合运算、基数
-- 鸽巢原理
-- 布尔代数的化简（卡诺图）
-
-**必须检查的条件：**
-- 图是否有重边/自环
-- 逻辑命题的真值定义
-- 集合运算的定义域（全集）
-- 鸽巢原理中鸽子和鸽巢的对应
-
-**常见错误：**
-- 握手定理的度数计算错误
-- 哈密顿图/欧拉图条件混淆
-- 逻辑等价变换遗漏步骤
-- 集合包含关系方向搞反
-- 鸽巢原理应用不当
-
-**推荐验证方法：**
-- 小规模枚举
-- 真值表验证逻辑等价
-- 具体图实例测试定理
-- 反例构造
-
-**输出格式：**
-```
-分支: [图论/逻辑/集合论/关系]
-命题: [陈述]
-方法: [握手定理/真值表/鸽巢原理/...]
-证明/求解: [步骤]
-结论: [结果]
-```
-
-**是否追问用户：** 仅当图的定义有歧义时
+**Identification Features:**
+- Involving graph theory, logic, set theory, Boolean algebra, relations
+- Keywords: "graph", "vertex", "edge", "logic", "proposition", "set", "relation", "lattice", "Boolean algebra"
+- Example: Prove that among any 6 people, there must be 3 people who know each other or do not know each other
+
+**Recommended solution strategy:**
+- Graph theory: handshake theorem, Euler path conditions, Euler formula for plane graphs, coloring
+- Logic: truth table, logical equivalence, paradigm shift
+- Set theory: Venn diagram, set operations, cardinality
+- Pigeonhole principle
+- Simplification of Boolean algebra (Karnaugh map)
+
+**Conditions that must be checked:**
+- Whether the graph has multiple edges/self-loops
+- Definition of truth value of logical propositions
+- Domain of set operations (complete set)
+- The correspondence between pigeons and pigeon nests in the pigeon nest principle
+
+**Common mistakes:**
+- Degree calculation error for handshake theorem
+- Hamiltonian graph/Eulerian graph condition confusion
+- Missing steps in logical equivalent transformation
+- The direction of the collection inclusion relationship is reversed
+- Improper application of the pigeonhole principle
+
+**Recommended verification method:**
+- Small-scale enumeration
+- Truth table to verify logical equivalence
+- Test theorems with specific graph examples
+- Counterexample construction
+
+**Output format:**
+ ```
+Branch: [Graph Theory/Logic/Set Theory/Relationship]
+proposition: [statement]
+Method: [Handshake theorem/truth table/pigeon nest principle/...]
+Proof/Solution: [Step]
+Conclusion: [result]
+```
+
+**Whether to ask the user:** Only if the definition of the graph is ambiguous
 
 ---
 
 ## optimization
 
-**识别特征：**
-- 涉及在约束条件下求最大值或最小值
-- 关键词："最大化"、"最小化"、"最优"、"约束条件"、"线性规划"、"非线性规划"
-- 示例：在 $x+y \leq 10, x \geq 0, y \geq 0$ 条件下，求 $z = 3x + 2y$ 的最大值
-
-**推荐求解策略：**
-- 线性规划：单纯形法或图解法（二维）
-- 等式约束极值：拉格朗日乘数法
-- 不等式约束：KKT 条件
-- 无约束极值：梯度为零 + Hessian 矩阵判断
-- 动态规划（多阶段决策）
-
-**必须检查的条件：**
-- 可行域是否非空
-- 线性规划是否有界
-- KKT 条件是否充分
-- 约束条件是否线性无关
-- 目标函数的凸性
-
-**常见错误：**
-- 可行域画错或描述错误
-- 忘记检查边界和顶点
-- 拉格朗日乘数法解错
-- KKT 条件遗漏
-- 局部最优与全局最优混淆
-
-**推荐验证方法：**
-- 二维问题画图验证
-- 代入可行域边界点检验
-- 对偶问题交叉验证
-- 敏感度分析
-
-**输出格式：**
-```
-优化类型: [线性规划/非线性规划/整数规划/动态规划]
-目标函数: [表达式]
-约束条件: [不等式组/方程组]
-方法: [单纯形法/拉格朗日法/KKT/图解法]
-最优解: $(x^*, y^*) = ([值], [值])$
-最优值: $z^* = [值]$
-```
-
-**是否追问用户：** 仅当约束不明确时
+**Identification Features:**
+- Involves finding the maximum or minimum value under constraints
+- Keywords: "maximization", "minimization", "optimality", "constraints", "linear programming", "nonlinear programming"
+- Example: Under the condition of $x+y \leq 10, x \geq 0, y \geq 0$, find the maximum value of $z = 3x + 2y$
+
+**Recommended solution strategy:**
+- Linear programming: simplex method or graphical method (2D)
+- Equality constrained extreme values: Lagrange multiplier method
+- Inequality constraints: KKT conditions
+- Unconstrained extreme value: gradient is zero + Hessian matrix judgment
+- Dynamic programming (multi-stage decision-making)
+
+**Conditions that must be checked:**
+- Whether the feasible region is non-empty
+- Is linear programming bounded?
+- Are the KKT conditions sufficient?
+- Whether the constraints are linearly independent
+- Convexity of the objective function
+
+**Common mistakes:**
+- The feasible region is incorrectly drawn or described incorrectly
+- Forgot to check bounds and vertices
+- Lagrange multiplier method solution error
+- KKT condition missing
+- Confusion between local optimal and global optimal
+
+**Recommended verification method:**
+- Two-dimensional problem drawing verification
+- Substitute into feasible region boundary point test
+- Dual problem cross-validation
+- Sensitivity analysis
+
+**Output format:**
+ ```
+Optimization type: [Linear programming/nonlinear programming/integer programming/dynamic programming]
+Objective function: [expression]
+Constraints: [Inequalities/Equations]
+Method: [Simplex method/Lagrangian method/KKT/Graphic method]
+Optimal solution: $(x^*, y^*) = ([value], [value])$
+Optimal value: $z^* = [value]$
+```
+
+**Whether to ask the user:** Only if the constraint is unclear
 
 ---
 
 ## mathematical_modeling
 
-**识别特征：**
-- 要求根据描述构建数学模型
-- 关键词："建立模型"、"数学模型"、"建模"、无现成公式套用的开放问题
-- 示例：建立传染病传播的数学模型
-
-**推荐求解策略：**
-1. 理解问题背景，识别关键变量
-2. 做出合理假设（简化现实）
-3. 建立变量间的关系（微分方程、差分方程、概率模型等）
-4. 确定参数
-5. 求解和分析模型
-6. 验证模型是否合理
-
-**必须检查的条件：**
-- 假设是否合理且被明确陈述
-- 量纲是否一致
-- 模型是否有解
-- 参数是否有实际意义
-- 模型的适用范围
-
-**常见错误：**
-- 假设过于简化导致模型无意义
-- 关键变量遗漏
-- 因果关系搞反
-- 未考虑量纲一致性
-- 参数取值无依据
-
-**推荐验证方法：**
-- 敏感性分析（参数微小变化对结果的影响）
-- 极端情况测试
-- 与实际数据对比
-- 量纲分析
-
-**输出格式：**
-```
-问题描述: [概述]
-假设: [列出所有假设]
-变量定义: [每个变量的含义和单位]
-模型: [数学表达式]
-求解: [方法]
-结果分析: [模型预测]
-局限性: [模型局限]
-```
-
-**是否追问用户：** 是，需要确认建模目标和可接受的简化程度
+**Identification Features:**
+- Required to construct a mathematical model based on the description
+- Keywords: "model building", "mathematical model", "modeling", open problems without ready-made formulas
+- Example: Building a mathematical model of the spread of infectious diseases
+
+**Recommended solution strategy:**
+1. Understand the problem background and identify key variables
+2. Make reasonable assumptions (simplifying reality)
+3. Establish relationships between variables (differential equations, difference equations, probability models, etc.)
+4. Determine parameters
+5. Solve and analyze the model
+6. Verify whether the model is reasonable
+
+**Conditions that must be checked:**
+- Are the assumptions reasonable and clearly stated?
+- Are the dimensions consistent?
+- Does the model have a solution?
+-Whether the parameters have actual meaning
+- Scope of application of the model
+
+**Common mistakes:**
+- Assumptions are oversimplified and make the model meaningless
+- Key variables missing
+- The causal relationship is reversed
+- Dimensional consistency is not considered
+- There is no basis for parameter values
+
+**Recommended verification method:**
+- Sensitivity analysis (the impact of small changes in parameters on the results)
+- Extreme case testing
+- Compare with actual data
+- Dimensional analysis
+
+**Output format:**
+ ```
+Problem Description: [Overview]
+Assumptions: [List all assumptions]
+Variable definition: [Meaning and unit of each variable]
+model: [mathematical expression]
+Solution: [Method]
+Result analysis: [Model prediction]
+Limitations: [Model limitations]
+```
+
+**Do you want to ask the user:** Yes, you need to confirm the modeling goals and the acceptable level of simplification
 
 ---
 
 ## proof
 
-**识别特征：**
-- 要求证明某个数学命题
-- 关键词："证明"、"求证"、"试证"、"证明或否定"
-- 示例：证明 $\sqrt{2}$ 是无理数
-
-**推荐求解策略：**
-- 确定证明方法：直接证明、反证法、数学归纳法、构造法、等价变换
-- 数学归纳法：验证 $n=1$ 成立，假设 $n=k$ 成立推出 $n=k+1$ 成立
-- 反证法：假设结论不成立，推导矛盾
-- 构造法：构造满足条件的具体对象
-- 分析命题结构，确定前提和结论
-
-**必须检查的条件：**
-- 前提条件是否完整
-- 推理每一步是否有依据
-- 反证法的矛盾是否确实与已知矛盾
-- 数学归纳法的基础步骤和归纳步骤是否完备
-- 是否隐含使用了未验证的假设
-
-**常见错误：**
-- 循环论证（用结论证明前提）
-- 反证法中假设的反面不完整（忘掉边界情况）
-- 数学归纳法省略基础步骤
-- 推理跳跃，缺少中间步骤
-- 使用未证明的定理
-
-**推荐验证方法：**
-- 每一步推理的逻辑检查
-- 特殊值代入验证结论
-- 尝试寻找反例
-- 用不同方法再次证明
-
-**输出格式：**
-```
-命题: [陈述]
-证明方法: [直接证明/反证法/数学归纳法/构造法/...]
-假设 (如反证法): [反证假设]
-证明: [逐步推导]
-结论: 命题成立 $\square$
-```
-
-**是否追问用户：** 仅当命题陈述有歧义时
+**Identification Features:**
+- Ask to prove a certain mathematical proposition
+- Keywords: "proof", "verification", "test", "prove or deny"
+- Example: Prove that $\sqrt{2}$ is an irrational number
+
+**Recommended solution strategy:**
+- Determine the proof method: direct proof, proof by contradiction, mathematical induction, construction method, equivalent transformation
+- Mathematical induction: verify that $n=1$ is established, assume that $n=k$ is established and conclude that $n=k+1$ is established
+- Evidence by contradiction: Assume that the conclusion is not valid and derive contradictions
+- Construction method: Construct specific objects that meet the conditions
+- Analyze the structure of propositions and determine premises and conclusions
+
+**Conditions that must be checked:**
+- Are the prerequisites complete?
+- Reasoning whether there is a basis for each step
+- Whether the contradiction in proof by contradiction is indeed contradictory to what is known
+- Whether the basic steps and induction steps of mathematical induction are complete
+- Whether untested assumptions are implicitly used
+
+**Common mistakes:**
+- Circular argument (use conclusion to prove premise)
+- The opposite of the hypothesis in proof by contradiction is incomplete (forget about edge cases)
+- Mathematical induction omits basic steps
+- Jumps in reasoning, missing intermediate steps
+- Use of unproven theorems
+
+**Recommended verification method:**
+- Logic check for each step of reasoning
+-Special value substitution verification conclusion
+- Try to find counterexamples
+- Prove it again in different ways
+
+**Output format:**
+ ```
+proposition: [statement]
+Proof method: [direct proof/contradiction/mathematical induction/construction method/...]
+Hypothesis (such as proof by contradiction): [hypothesis by contradiction]
+Proof: [Step-by-step derivation]
+Conclusion: The proposition holds $\square$
+```
+
+**Whether to ask the user:** Only if the propositional statement is ambiguous
 
 ---
 
 ## counterexample
 
-**识别特征：**
-- 要求找出反例来反驳某个命题
-- 关键词："反例"、"不成立"、"举出反例"、"是否成立"、"举出...反驳"
-- 示例：命题"若 $f'(x_0)=0$，则 $x_0$ 是 $f(x)$ 的极值点"是否成立？若不成立请举反例。
-
-**推荐求解策略：**
-1. 分析命题的条件和结论
-2. 找出条件满足但结论不成立的情形
-3. 从边界情况、退化情况开始寻找
-4. 选择尽可能简单的反例
-5. 验证反例满足条件但不满足结论
-
-**必须检查的条件：**
-- 反例是否满足命题的所有条件
-- 反例是否确实违反结论
-- 反例是否在合理的定义域内
-- 反例是否是最简形式
-
-**常见错误：**
-- 反例不满足命题条件
-- 反例虽然违反结论但定义域不符
-- 反例过于复杂
-- 把"命题成立"错误判断为"命题不成立"
-- 遗漏命题隐含条件
-
-**推荐验证方法：**
-- 逐条检查条件是否满足
-- 逐一检查结论是否违反
-- 简化反例是否能保持反例性质
-
-**输出格式：**
-```
-命题: [原始命题陈述]
-分析: [为什么这个命题可能不成立]
-反例: [具体的反例构造]
-验证条件: [逐条检查条件是否满足]
-验证结论: [说明结论不成立]
-最小反例: [是否还能更简单]
-```
-
-**是否追问用户：** 否
+**Identification Features:**
+- Ask for counterexamples to refute a certain proposition
+- Keywords: "counterexample", "not established", "cite counterexamples", "whether it is established", "cite...refutation"
+- Example: Is the proposition "If $f'(x_0)=0$ , then $x_0$ is the extreme point of $f(x)$" true? If this is not true, please give a counterexample.
+
+**Recommended solution strategy:**
+1. Analyze the conditions and conclusions of propositions
+2. Find situations where the conditions are met but the conclusion is not true
+3. Start looking for boundary conditions and degradation conditions
+4. Choose the simplest possible counterexample
+5. Verify that the counterexample satisfies the condition but does not satisfy the conclusion
+
+**Conditions that must be checked:**
+- Does the counterexample satisfy all the conditions of the proposition?
+-Whether the counterexample actually violates the conclusion
+- Whether the counterexample is within a reasonable definition domain
+- Whether the counterexample is in its simplest form
+
+**Common mistakes:**
+- The counterexample does not satisfy the propositional condition
+- Although the counterexample violates the conclusion, the definition domain does not match
+- Counterexamples are too complex
+- Misjudge "the proposition is true" as "the proposition is not true"
+- Missing propositional implicit conditions
+
+**Recommended verification method:**
+- Check whether the conditions are met one by one
+- Check whether the conclusions are violated one by one
+- Whether simplifying counterexamples can maintain the properties of counterexamples
+
+**Output format:**
+ ```
+proposition: [original proposition statement]
+Analysis: [Why this proposition may not be true]
+Counterexample: [Specific counterexample construction]
+Verification conditions: [Check whether the conditions are met one by one]
+Verification conclusion: [Explanation that the conclusion is not valid]
+Minimal counterexample: [Can it be simpler]
+```
+
+**Whether to ask the user:** No
 
 ---
 
 ## solution_checking
 
-**识别特征：**
-- 给定一个解答（可能是学生答案），要求检查对错
-- 关键词："对吗"、"是否正确"、"检查"、"批改"、"有没有错误"、"这个解法"
-- 示例：检查下面解法的正误：解 $x^2=4$ 得 $x=2$
-
-**推荐求解策略：**
-1. 重新独立求解该问题
-2. 将给定解答与标准解逐行对比
-3. 检查逻辑推导的每一步
-4. 验证最终结果
-5. 分类错误类型（计算错误、逻辑错误、概念错误、遗漏）
-
-**必须检查的条件：**
-- 最终结果是否正确
-- 推导过程是否有逻辑跳跃
-- 定义域/约束条件是否被满足
-- 有无遗漏解
-- 计算过程是否正确
-
-**常见错误（批改者错误）：**
-- 只看结果不看过程
-- 等价解（不同形式）判为错误
-- 忽略合理的不同解法
-- 末步错误忽略前面的正确推导
-
-**推荐验证方法：**
-- 代入验证最终结果
-- 用不同方法得到同一问题答案
-- 检查每一步变换的等价性
-- 对数值结果估算合理性
-
-**输出格式：**
-```
-问题: [原问题]
-给定解答: [待检查的解答]
-独立求解结果: [正确答案]
-判断: [正确 / 部分正确 / 错误]
-错误定位: [第 X 步错误，具体错误描述]
-错误类型: [计算错误/逻辑错误/概念错误/遗漏/表述不当]
-修正建议: [应当如何改正]
-```
-
-**是否追问用户：** 仅当解答模糊不清或笔迹难辨时
+**Identification Features:**
+- Given a solution (possibly a student answer), ask to check if it is correct or incorrect
+- Keywords: "is it right", "is it correct", "check", "correct", "are there any errors", "this solution"
+- Example: Check whether the following solution is correct or incorrect: solve $x^2=4$ to get $x=2$
+
+**Recommended solution strategy:**
+1. Solve the problem independently again
+2. Compare the given solution line by line with the standard solution.
+3. Check every step of logical derivation
+4. Verify final results
+5. Classification error types (calculation errors, logical errors, conceptual errors, omissions)
+
+**Conditions that must be checked:**
+- Is the final result correct?
+- Are there any logical jumps in the derivation process?
+- Whether the domain/constraint conditions are satisfied
+- Are there any missing solutions?
+- Is the calculation process correct?
+
+**Common mistakes (corrector errors):**
+- Only look at the results and not the process
+- Equivalent solutions (different forms) are judged as errors
+- Ignore legitimate alternative solutions
+- The final step error ignores the previous correct derivation
+
+**Recommended verification method:**
+- Substitute the final result of verification
+- Get the answer to the same question in different ways
+- Check the equivalence of each transformation step
+- Estimating plausibility of numerical results
+
+**Output format:**
+ ```
+Question: [original question]
+Given solution: [Solution to be checked]
+Independent solution result: [Correct answer]
+Judgment: [Correct/Partially Correct/Incorrect]
+Error location: [Error in step X, specific error description]
+Error type: [Calculation error/Logical error/Conceptual error/Omission/Inappropriate presentation]
+Correction suggestions: [How to correct it]
+```
+
+**Whether to question the user:** Only if the answer is unclear or the handwriting is difficult to read
 
 ---
 
 ## problem_generation
 
-**识别特征：**
-- 要求生成新的数学问题
-- 关键词："出题"、"生成题目"、"帮我出一题"、"编一道题"、"设计题目"
-- 示例：生成一道二次函数最值问题的应用题，难度为高考中等
-
-**推荐求解策略：**
-1. 确定题目主题和知识点
-2. 设计题目结构和已知条件
-3. 构造数值（确保数值合理且计算简洁）
-4. 自己先求解一遍验证可解性
-5. 提供参考答案
-
-**必须检查的条件：**
-- 题目条件是否完整且自洽
-- 是否有解且解唯一（如需唯一）
-- 数值是否合理（不会出现奇怪的小数）
-- 难度是否符合要求
-- 条件之间无矛盾
-
-**常见错误：**
-- 条件不足导致多解或无解
-- 条件矛盾
-- 数值构造不合理导致计算复杂
-- 超出指定难度范围
-- 题目表述有歧义
-
-**推荐验证方法：**
-- 独立求解生成的题目
-- 检查是否有非预期解法
-- 验证数值的合理性
-- 试做估计用时
-
-**输出格式：**
-```
-知识点: [涵盖的知识点]
-题目: [完整题目描述]
-难度: [简单/中等/困难]
-参考答案: [完整解答过程]
-答案: [最终结果]
-考察要点: [题目主要考察什么]
-```
-
-**是否追问用户：** 是，确认难度、知识点范围、题型偏好
+**Identification Features:**
+- Ask to generate new math problems
+- Keywords: "give a question", "generate a question", "help me give a question", "make up a question", "design a question"
+- Example: Generate an application question of the maximum value of a quadratic function, with a difficulty level of College Entrance Examination Medium
+
+**Recommended solution strategy:**
+1. Determine the topic and knowledge points of the question
+2. Design the question structure and known conditions
+3. Construct the numerical value (make sure the numerical value is reasonable and the calculation is simple)
+4. Solve it yourself first to verify the solvability
+5. Provide reference answers
+
+**Conditions that must be checked:**
+- Are the question conditions complete and self-consistent?
+- Whether there is a solution and the solution is unique (if unique)
+- Are the values ​​reasonable (no strange decimals will appear)
+- Whether the difficulty meets the requirements
+- There is no contradiction between the conditions
+
+**Common mistakes:**
+- Insufficient conditions lead to multiple or no solutions
+- Contradictory conditions
+- Unreasonable numerical construction leads to complicated calculations
+- Beyond the specified difficulty range
+- There is ambiguity in the title description
+
+**Recommended verification method:**
+- Solve generated questions independently
+- Check for unexpected solutions
+- Verify the rationality of the values
+- Estimated trial time
+
+**Output format:**
+ ```
+Knowledge points: [Knowledge points covered]
+Title: [Full title description]
+Difficulty: [Easy/Medium/Hard]
+Reference answer: [Complete solution process]
+Answer: [final result]
+Examination points: [What is the main examination of the question]
+```
+
+**Whether to ask the user:** Yes, confirm the difficulty, knowledge point range, and question type preference
 
 ---
 
 ## research_level_problem
 
-**识别特征：**
-- 极高难度，可能接近竞赛决赛、数学研究或开放问题
-- 关键词：涉及费马大定理、黎曼猜想级别的引用，或明显超出标准课程
-- 示例：证明黎曼 $\zeta$ 函数的非平凡零点都位于 $\operatorname{Re}(s) = 1/2$ 上
-
-**推荐求解策略：**
-- 首先判断问题是否已知被解决
-- 若为已知结果，提供该结果的概述和相关文献
-- 若为开放问题，诚实告知
-- 提供可能的思路或相关子结果
-- 引用权威来源
-
-**必须检查的条件：**
-- 问题是否等价于已知的开放问题
-- 问题是否有已知的部分进展
-- 用户是否理解问题的难度
-- 是否有必要降级为近似或特殊情形
-
-**常见错误：**
-- 对开放问题给出错误证明
-- 低估问题难度
-- 忽略已知的进展
-- 使用不成熟的推理
-
-**推荐验证方法：**
-- 查阅文献验证结论
-- 检查证明中是否有漏洞
-- 与已知结果对比
-
-**输出格式：**
-```
-问题评估: [已知结果 / 部分解决 / 开放问题]
-已知进展: [相关定理和部分结果]
-参考文献: [关键引用]
-建议: [可能的思路或简化方向]
-注意事项: [问题难度说明]
-```
-
-**是否追问用户：** 是，确认用户期望的处理方式和深度
+**Identification Features:**
+- Very high level of difficulty, possibly close to competition finals, mathematical studies or open questions
+- Keywords: References involving Fermat’s Last Theorem and Riemann Hypothesis, or that are obviously beyond the standard curriculum
+- Example: Prove that the non-trivial zeros of Riemann's $\zeta$ function are located on $\operatorname{Re}(s) = 1/2$
+
+**Recommended solution strategy:**
+- First determine whether the problem is known to have been solved
+- If the result is known, provide a summary of the result and relevant literature
+- If it’s an open question, tell it honestly
+- Provide possible ideas or related sub-results
+- Citing authoritative sources
+
+**Conditions that must be checked:**
+- Is the problem equivalent to a known open problem?
+- Is there any known partial progress on the issue?
+- Whether the user understands the difficulty of the question
+- Whether it is necessary to downgrade to approximate or special circumstances
+
+**Common mistakes:**
+- Provide error proofs for open questions
+- Underestimating the difficulty of the problem
+- Ignore known progress
+- Use premature reasoning
+
+**Recommended verification method:**
+- Review the literature to verify the conclusion
+- Check the proof for holes
+- Compare with known results
+
+**Output format:**
+ ```
+Problem Assessment: [Known Result/Partially Solved/Open Issue]
+Known progress: [Related theorems and partial results]
+References: [Key Citation]
+Suggestions: [Possible ideas or simplification directions]
+Note: [Question Difficulty Description]
+```
+
+**Whether to ask the user:** Yes, confirm the processing method and depth expected by the user
 
 ---
 
 ## ambiguous_or_incomplete
 
-**识别特征：**
-- 条件缺失、表述有歧义、信息不足
-- 关键词：模糊的指代、未定义的符号、自相矛盾的条件
-- 示例："求这个函数的最大值"（未给出函数）
-
-**推荐求解策略：**
-- 指出缺失/模糊的具体内容
-- 给出合理假设下的可能解释
-- 请求用户补充信息
-- 在用户回复前不擅自作答
-
-**必须检查的条件：**
-- 问题中所有符号是否已定义
-- 数值范围是否明确
-- 假设是否自洽
-- 是否有多种合理解释
-
-**常见错误：**
-- 自行脑补条件并给出错误答案
-- 忽略歧义直接作答
-- 未将所有歧义列出
-- 补充了不合理的假设
-
-**推荐验证方法：**
-- 列出所有可能的解释
-- 检查每种解释下的答案是否一致
-
-**输出格式：**
-```
-问题: [用户原始输入]
-歧义/缺失点: [逐条列举]
-可能的解释:
-  解释1: [描述] → 在此解释下答案为 [值]
-  解释2: [描述] → 在此解释下答案为 [值]
-需要确认: [具体需要用户明确的内容]
-```
-
-**是否追问用户：** 是，必须追问
+**Identification Features:**
+- Missing conditions, ambiguous expressions, and insufficient information
+- Keywords: vague reference, undefined symbols, contradictory conditions
+- Example: "Find the maximum value of this function" (no function given)
+
+**Recommended solution strategy:**
+- Point out missing/ambiguous specific content
+- Give possible explanations under reasonable assumptions
+- Request additional information from the user
+- Do not answer without permission before the user replies
+
+**Conditions that must be checked:**
+- Are all symbols in the question defined?
+- Is the numerical range clear?
+- Is the hypothesis self-consistent?
+- Are there multiple reasonable explanations?
+
+**Common mistakes:**
+- Make up the conditions on your own and give wrong answers
+- Ignore the ambiguity and answer directly
+- Not all ambiguities are listed
+- Added unreasonable assumptions
+
+**Recommended verification method:**
+- List all possible explanations
+- Check whether the answers under each explanation are consistent
+
+**Output format:**
+ ```
+Question: [Original user input]
+Ambiguities/missing points: [itemize]
+Possible explanation:
+Explanation 1: [Description] → In this explanation the answer is [Value]
+Explanation 2: [Description] → In this explanation the answer is [Value]
+Confirmation required: [Specific content needs to be clarified by the user]
+```
+
+**Whether to ask the user:** Yes, must ask
 
 ---
 
 ## out_of_scope
 
-**识别特征：**
-- 输入不是数学问题
-- 涉及非数学学科、日常对话、无法归类的内容
-- 示例："今天天气怎么样"、"帮我写一封邮件"
+**Identification Features:**
+- Input is not a math problem
+- Involving non-mathematical subjects, daily conversations, and content that cannot be classified
+- Examples: "How is the weather today", "Write an email for me"
 
-**推荐求解策略：**
-- 礼貌说明 Math.skill 的职责范围
-- 如果知道答案可简要回复
-- 建议用户使用其他更合适的工具
+**Recommended solution strategy:**
+- Polite description of Math.skill’s scope of responsibilities
+- If you know the answer, please give a brief reply
+- Recommend users to use other more suitable tools
 
-**必须检查的条件：**
-- 是否包含任何数学内容
-- 是否可被重新表述为数学问题
-- 用户是否有意测试系统边界
+**Conditions that must be checked:**
+- Does it contain any mathematical content?
+- Can it be reformulated as a mathematical problem
+- Whether the user is interested in testing system boundaries
 
-**常见错误：**
-- 强行将非数学问题当作数学问题处理
-- 误判边缘问题（如物理中的数学推导）
+**Common mistakes:**
+- Forcing non-mathematical problems to be treated as mathematical problems
+- Misjudgment of edge problems (such as mathematical derivation in physics)
 
-**推荐验证方法：**
-- 确认输入确实不包含可处理的数学内容
+**Recommended verification method:**
+- Confirm that the input does not contain processable mathematical content
 
-**输出格式：**
-```
-判断: 该输入不属于数学问题范畴
-具体原因: [说明]
-建议: [如果适用，建议用户使用其他工具或重新表述]
+**Output format:**
+ ```
+Judgment: The input does not belong to the category of mathematical problems
+Specific reason: [Explanation]
+Recommendation: [Recommend user to use other tools or rephrase if applicable]
 ```
 
-**是否追问用户：** 是，建议重新表述或确认意图
+**Whether to ask the user:** Yes, it is recommended to rephrase or confirm the intention
 
 ---
 
-最后更新: 2026-05-30
+Last updated: 2026-05-30
diff --git a/modules/error_prevention.md b/modules/error_prevention.md
index 0c87989..3e259de 100644
--- a/modules/error_prevention.md
+++ b/modules/error_prevention.md
@@ -4,6 +4,17 @@ This module defines concrete error prevention rules, common pitfalls, and preven
 
 ---
 
+## 0. General Anti-Hallucination and Anti-BS Rules
+
+Before applying any domain-specific rules, the following strict principles MUST be observed to prevent "Plausible BS" and mathematical hallucinations:
+
+1. **Zero-Tolerance for Fabricated Logic**: Never invent theorems, properties, or algebraic rules that do not exist to force a derivation to work.
+2. **Honest Admissions of Ignorance**: If you do not know a formula or cannot compute an intermediate step, explicitly state "I am unable to compute this step" and STOP. Do not guess or approximate and present it as exact.
+3. **No "Fake" Corrections**: If a verification method fails, do not invent a trivial error (like a "sign error") to justify outputting the same wrong answer. Actually perform the recalculation.
+4. **Do Not Pretend to Verify**: If a verification step requires complex computation that you cannot confidently perform, do NOT say "By calculation, this holds true". State that you cannot perform the verification.
+
+---
+
 ## 1. Algebraic Error Prevention
 
 ### Rules
diff --git a/modules/hard_problem_protocol.md b/modules/hard_problem_protocol.md
index a7ddfaf..5350ce1 100644
--- a/modules/hard_problem_protocol.md
+++ b/modules/hard_problem_protocol.md
@@ -1,281 +1,281 @@
-# Math.skill 难题处理协议 (Hard Problem Protocol)
+# Math.skill Hard Problem Protocol
 
-当遇到超出标准解题范围的问题时，按以下协议流程处理。
+When encountering problems beyond the scope of standard problem solving, follow the following protocol process.
 
 ---
 
-## 1. 问题分类 (Problem Classification)
+## 1. Problem Classification
 
-在开始解题之前，首先对问题进行分类：
+Before starting to solve the problem, first classify the problem:
 
-| 分类 | 特征 | 示例 |
+| Classification | Features | Examples |
 |------|------|------|
-| **标准教材题** | 有明确的标准解法，可直接应用已知定理 | 计算导数、解常系数微分方程、计算行列式 |
-| **竞赛题** | 需要技巧但不超出初等/高等数学范围，已知有解 | IMO 题目、Putnam 题目、各国家集训队题目 |
-| **研究级别问题** | 涉及前沿数学，可能需要原创性思路 | 某篇论文中提出的猜想、研究生级别的研究问题 |
-| **已知定理** | 是某个已知定理的特例或直接推论 | 用户可能不知道的已知结论 |
-| **已知开放问题** | 数学界尚未解决的问题 | Riemann 猜想、Goldbach 猜想、Collatz 猜想 |
-| **条件缺失** | 问题表述不完整，缺少必要条件 | 缺少定义域、缺少初始条件、缺少收敛性假设 |
+| **Standard textbook questions** | There are clear standard solutions and known theorems can be directly applied | Calculate derivatives, solve differential equations with constant coefficients, and calculate determinants |
+| **Competition questions** | Requires skills but does not exceed the scope of elementary/advanced mathematics, and is known to have solutions | IMO questions, Putnam questions, and national training team questions |
+| **Research level questions** | Involving cutting-edge mathematics, original ideas may be required | Conjectures raised in a certain paper, graduate-level research questions |
+| **Known theorem** | Is a special case or direct corollary of a known theorem | A known conclusion that the user may not know |
+| **Known open problems** | Unsolved problems in mathematics | Riemann conjecture, Goldbach conjecture, Collatz conjecture |
+| **Missing conditions** | The problem statement is incomplete and necessary conditions are missing | The domain is missing, the initial conditions are missing, and the convergence assumption is missing |
 
-### 分类决策流程
+### Classification decision-making process
 
-1. 先检查问题表述是否完整 → 如不完整，归入"条件缺失"
-2. 检查问题是否为已知开放问题的等价形式 → 如是，归入"已知开放问题"
-3. 检查是否存在标准解法或已知定理的直接应用 → 如是，归入"标准教材题"或"已知定理"
-4. 评估问题所需的技术深度 → 判断为"竞赛题"或"研究级别问题"
+1. First check whether the problem statement is complete → If it is incomplete, it will be classified as "missing condition"
+2. Check whether the problem is an equivalent form of a known open problem → If so, classify it as a "known open problem"
+3. Check whether there are standard solutions or direct applications of known theorems → If so, classify them into "standard textbook questions" or "known theorems"
+4. Assess the technical depth required for the problem → Determine it as a "competition question" or "research level question"
 
 ---
 
-## 2. 搜索阶段 (Search Phase)
+## 2. Search Phase
 
-对非标准教材题，执行以下搜索：
+For non-standard textbook questions, perform the following search:
 
-### 搜索关键词构造
-1. 提取问题中的核心数学对象
-2. 构造 2-3 组不同的搜索查询
-3. 使用英文关键词 + LaTeX 语法
+### Search keyword structure
+1. Extract the core mathematical objects in the problem
+2. Construct 2-3 different sets of search queries
+3. Use English keywords + LaTeX syntax
 
-### 搜索目标清单
-- [ ] 相关定理及其适用范围
-- [ ] 该问题是否为已知开放问题
-- [ ] 是否有类似结构的已知问题
-- [ ] 是否存在已知反例
-- [ ] 是否有已知的标准解法模式
+### Search target list
+- [ ] Relevant theorems and their scope of application
+- [ ] Is the issue a known open issue?
+- [ ] Are there any known issues with similar structures?
+- [ ] Whether there are known counterexamples
+- [ ] Is there a known standard solution pattern?
 
 ---
 
-## 3. 找到类似问题时的处理 (If Similar Problems Found)
-
-### 处理原则
-1. **参考思路，不复制解答**：理解网上解答的逻辑结构，但不逐字复制
-2. **独立重新推导**：关闭搜索页面，从问题本身的定义和条件出发推导
-3. **逐步验算**：每一步都必须通过验算检查
-4. **用自己的语言重新组织**：最终呈现的解答必须是自己的理解
-
-### 输出格式
-在解答中明确标注搜索结果的影响：
-```
-## 类似问题参考
-- 在 [来源] 中找到结构类似的问题 [简述]
-- 本题采用的方法与参考方案在 [具体方面] 有所不同
-- 以下为独立推导的完整解答
+## 3. Processing when similar problems are found (If Similar Problems Found)
+
+### Processing principles
+1. **Reference ideas, do not copy answers**: Understand the logical structure of online answers, but do not copy verbatim
+2. **Independent re-derivation**: Close the search page and deduce based on the definition and conditions of the problem itself
+3. **Step-by-step verification**: Each step must pass the verification check
+4. **Reorganize in your own words**: The final solution must be your own understanding
+
+### Output format
+Clearly label the impact of search results in your answer:
+ ```
+## Reference to similar questions
+- Find similar structured questions in [Source] [Brief Description]
+- The method used in this question is different from the reference plan in [specific aspects]
+- The following is the complete solution of independent derivation
 ```
 
 ---
 
-## 4. 找不到类似问题的处理（第一原理模式）
+## 4. No solution to similar problems can be found (first principles mode)
 
-如果搜索后未找到任何类似问题，进入**第一原理模式 (First Principles Mode)**。
+If the search does not find any similar problems, enter **First Principles Mode**.
 
-### 第一原理模式流程
+### First principles model process
 
-#### 步骤 1：写下全部相关定义
-```
-对象 X 的定义：……
-属性 P 的定义：……
-问题的精确定义：[用数学语言重新表述]
+#### Step 1: Write down all relevant definitions
+ ```
+Definition of object X:…
+Definition of property P:…
+Precise definition of the problem: [Restated in mathematical language]
 ```
 
-#### 步骤 2：将目标分解为子目标
-```
-目标：证明 P(X)
-子目标 1：证明 Q(X)
-子目标 2：证明 R(X)
-子目标 3：由 Q 和 R 推出 P
+#### Step 2: Break down the goal into sub-goals
+ ```
+Goal: Prove P(X)
+Sub-goal 1: Prove Q(X)
+Subgoal 2: Prove R(X)
+Subgoal 3: deduce P from Q and R
 ```
 
-#### 步骤 3：尝试简单情形
-- n=1, n=2 等小参数情形
-- 退化情形（如矩阵退化为标量）
-- 对称情形
-- 齐次情形
-
-#### 步骤 4：尝试构造反例
-- 尝试找到使结论不成立的特殊情形
-- 如果找到反例 → 结论不成立，问题到此为止
-- 如果找不到反例 → 增强了对结论为真的信心
-
-#### 步骤 5：尝试归纳法
-- 检查是否可对某个参数（维数、度数、阶数等）做归纳
-- 检查归纳步骤是否需要加强归纳假设
-
-#### 步骤 6：尝试数学变换
-- 变量替换
-- 坐标变换
-- 对偶变换
-- 生成函数
-- Fourier/Laplace 变换
-- 对称化
-
-#### 步骤 7：尝试建立引理
-- 将困难证明拆分为多个独立引理
-- 先证引理，再用引理组合证明主结论
-
-#### 步骤 8：尝试等价重述
-- 逆否命题
-- 对偶形式
-- 几何解释
-- 代数重新表述
-- 用随机变量的语言重新表述
-
-#### 步骤 9：数值实验
-- 对具体参数值做数值计算
-- 观察数值模式
-- 用数值结果指导理论推导方向
-
-#### 步骤 10：尝试特殊情形推广
-- 如果证明了 n=1、n=2、n=3 等情形
-- 检查能否将证明方法推广到一般 n
-
-#### 步骤 11：尝试弱化目标
-- 能否证明更弱的结论？
-- 在额外条件下能否证明？
-- 部分结果也有价值
-
-#### 步骤 12：逐步验证
-- 每一步推导后立即验证逻辑正确性
-- 检查除零、分母为零、不等式方向等细节
+#### Step 3: Try a simple scenario
+- n=1, n=2 and other small parameter cases
+- Degenerate situations (such as matrices degenerating into scalars)
+- Symmetrical situation
+- homogeneous situation
+
+#### Step 4: Try to construct a counterexample
+- Try to find special circumstances that make the conclusion invalid
+- If a counterexample is found → the conclusion is not valid and the problem ends here
+- If no counterexample can be found → Increases confidence that the conclusion is true
+
+#### Step 5: Try induction
+- Check whether a parameter (dimension, degree, order, etc.) can be generalized
+- Check whether the induction step needs to strengthen the inductive hypothesis
+
+#### Step 6: Try mathematical transformations
+- Variable substitution
+- Coordinate transformation
+- Dual transformation
+- Generate function
+- Fourier/Laplace transformation
+- Symmetrization
+
+#### Step 7: Try to establish the lemma
+- Split the difficult proof into multiple independent lemmas
+- Prove the lemma first, and then use the combination of lemmas to prove the main conclusion
+
+#### Step 8: Try an equivalent restatement
+- Converse proposition
+- Dual form
+- Geometric explanation
+- Algebraic reformulation
+- Restated in the language of random variables
+
+#### Step 9: Numerical Experiment
+- Perform numerical calculations on specific parameter values
+- Observe numerical mode
+- Use numerical results to guide the direction of theoretical derivation
+
+#### Step 10: Try Special Situation Promotions
+- If it is proved that n=1, n=2, n=3, etc.
+- Check whether the proof method can be generalized to general n
+
+#### Step 11: Try weakening the target
+- Can a weaker conclusion be proved?
+- Can it be proven under additional conditions?
+- Some results are also valuable
+
+#### Step 12: Step-by-Step Verification
+- Immediately verify the logical correctness after each step of derivation
+- Check details such as division by zero, zero denominator, direction of inequality, etc.
 
 ---
 
-## 5. 疑似开放问题的处理 (Possible Open Problem)
+## 5. Handling of suspected open problems (Possible Open Problem)
 
-### 识别特征
-- 问题形式极其简单但证明异常困难
-- 涉及数论的素数性质
-- 涉及图论的极值猜想
-- 涉及混沌系统或动力系统
-- 问题陈述类似于已知的 Millenium 问题
+### Identifying features
+- The problem form is extremely simple but the proof is extremely difficult
+- Involving the properties of prime numbers in number theory
+- Extreme value conjecture involving graph theory
+- Involving chaotic systems or dynamical systems
+-Problem statement similar to known Millenium issues
 
-### 处理流程
+### Processing process
 
-#### 步骤 1：搜索确认
-用问题关键词 + "open problem" 或 "unsolved" 或 "conjecture" 搜索。
+#### Step 1: Search and confirm
+Search with question keyword + "open problem" or "unsolved" or "conjecture".
 
-#### 步骤 2：如确认为开放问题
-**MUST 执行的声明**：
-```
-## 问题状态
+#### Step 2: If confirmed as an open issue
+**MUST enforce the statement**:
+ ```
+## Problem status
 
-此问题是已知的数学开放问题：[标准名称]
+This problem is a known open problem in mathematics: [standard name]
 
-### 已知事实（已验证）
-- [事实 1]
-- [事实 2]
+### Known facts (verified)
+- [Fact 1]
+- [Fact 2]
 
-### 猜想内容
-- 该猜想声称：[精确陈述]
+### Guess content
+- The conjecture claims: [precise statement]
 
-### 已知的部分结果
-- [已知结果 1]
-- [已知结果 2]
+### Known partial results
+- [Known Result 1]
+- [Known Result 2]
 
-### 已知的失败尝试及原因
-- [方向 1]：[失败原因]
-- [方向 2]：[失败原因]
+### Known failed attempts and reasons
+- [Direction 1]: [Reason for failure]
+- [Direction 2]: [Reason for failure]
 
-### 数值证据
-- 已验证到 n ≤ [范围]
-- 未发现反例
+### Numerical evidence
+- verified to n ≤ [range]
+- No counterexamples found
 
-### 声明
-目前不存在被数学界公认的完整证明。以下是基于现有理论的探索性分析，不代表完整证明。
+### Statement
+There is currently no complete proof recognized by the mathematical community. The following is an exploratory analysis based on existing theories and does not represent a complete proof.
 ```
 
-#### 步骤 3：探索性分析
-可以提供：
-- 部分结果（在某些特殊条件下成立）
-- 数值实验的结果
-- 可能的攻克方向（但明确标注为推测）
-- 与已知结论的联系
+#### Step 3: Exploratory Analysis
+Can provide:
+- Partial results (held under certain special conditions)
+- Results of numerical experiments
+- Possible directions of attack (but clearly marked as speculation)
+- Connection to known conclusions
 
-#### 步骤 4：严禁行为
-- **绝对禁止声称解决了开放问题**（除非提供完整的、可验证的、无间隙的证明）
-- **绝对禁止虚构突破性结论**
-- **绝对禁止声称"我找到了简单证明"**——如果真那么简单，Fermat's Last Theorem 和 Poincaré 猜想就不会流传数百年
+#### Step 4: Strictly Prohibited Behavior
+- **Absolutely prohibited from claiming to solve open problems** (unless complete, verifiable, gap-free proof is provided)
+- **It is absolutely forbidden to make up any groundbreaking conclusions**
+- **Absolutely forbidden to claim "I found a simple proof"** - If it were that simple, Fermat's Last Theorem and Poincaré's conjecture wouldn't have been around for hundreds of years
 
 ---
 
-## 6. 迭代推进策略 (Iterative Advancement)
-
-### 持续向前推进
-遇到困难时不停止，尝试：
-1. 改变视角（几何 ↔ 代数 ↔ 分析）
-2. 增加或放松条件
-3. 退回到"如果……成立"的假设推演
-4. 从反方向思考：如果要结果不成立，需要什么条件？
-
-### 达到可验证阶段的标志
-- 得到了一些明确的结论（即使不是最终目标）
-- 能够进行数值检验
-- 能够举出特例验证
-
-### 明确解释障碍
-遇到无法突破的步骤时：
-```
-## 当前障碍
-在第 [X] 步遇到障碍：
-- 我们需要证明 [具体陈述]
-- 问题出在 [具体困难]
-- 已知的 [定理/方法] 不能直接应用，因为 [原因]
+## 6. Iterative Advancement Strategy
+
+### Keep moving forward
+Don't stop when you encounter difficulties, try:
+1. Change perspective (geometry ↔ algebra ↔ analysis)
+2. Increase or relax conditions
+3. Return to the hypothetical deduction of "if... is true"
+4. Think from the opposite direction: If the result does not hold, what conditions are needed?
+
+### Sign of reaching the verifiable stage
+- Got some clear conclusions (if not the ultimate goal)
+- Ability to perform numerical tests
+- Be able to cite special cases for verification
+
+### Clearly explain barriers
+When you encounter a step that you cannot break through:
+ ```
+## Current obstacle
+Encountered an obstacle at step [X]:
+- We need to prove [specific statement]
+- The problem lies in [specific difficulty]
+- Known [theorems/methods] cannot be applied directly because of [reason]
 ```
 
-### 建议后续方向
-```
-## 可能的突破方向
-1. [方向 1]：[具体原因和预期]
-2. [方向 2]：[具体原因和预期]
-3. [方向 3]：[需要验证的前提条件]
+### Suggest follow-up directions
+ ```
+## Possible breakthrough direction
+1. [Direction 1]: [Specific reasons and expectations]
+2. [Direction 2]: [Specific reasons and expectations]
+3. [Direction 3]: [Prerequisites requiring verification]
 ```
 
 ---
 
-## 7. 决策树 (Decision Tree)
+## 7. Decision Tree
 
-```
-收到复杂数学问题
+ ```
+receive complex math questions
 │
-├─ 问题是否完整？
-│  ├─ 否 → 请求用户补充条件 → 回到判断
-│  └─ 是 ↓
+├─ Are the questions complete?
+│ ├─ No → Request the user to add conditions → Return to judgment
+│ └─ Yes ↓
 │
-├─ 是否为已知开放问题？
-│  ├─ 是 → 执行开放问题协议 → 输出部分结果，声明未解决
-│  └─ 否 ↓
+├─ Is it a known open issue?
+│ ├─ Yes → Execute the open problem protocol → Output partial results and declare unsolved
+│ └─ No ↓
 │
-├─ 是否有标准解法？
-│  ├─ 是 → 执行标准解题流程 → 输出标准解答
-│  └─ 否 ↓
+├─ Is there a standard solution?
+│ ├─ Yes → Execute the standard problem-solving process → Output the standard solution
+│ └─ No ↓
 │
-├─ 搜索是否有类似问题？
-│  ├─ 找到类似问题 → 参考思路，独立推导 → 输出解答
-│  └─ 未找到 ↓
+├─ Search for similar questions?
+│ ├─ Find similar questions → Reference ideas and independent derivation → Output the solution
+│ └─ Not found ↓
 │
-├─ 进入第一原理模式
-│  ├─ 是否能在合理时间内推进？
-│  │  ├─ 能 → 推进至可验证阶段 → 输出部分结果或完整解答
-│  │  └─ 不能 ↓
+├─Enter first principles mode
+│ ├─ Can it be advanced within a reasonable time?
+│ │ ├─ Can → Advance to the verifiable stage → Output partial results or complete answers
+│ │ └─ Can’t ↓
 │  │
-│  └─ 输出当前进展 + 障碍说明 + 后续方向
+│ └─ Output current progress + obstacle description + follow-up direction
 │
-└─ 在任何阶段发现条件缺失 → 返回请求补充
+└─ If conditions are found to be missing at any stage → return to request for supplementation
 ```
 
 ---
 
-## 8. 记录与追溯
-
-### 难题记录
-对于每个难题，记录以下信息：
-- 问题分类
-- 搜索的关键词和结果
-- 尝试过的方法及其结果（成功或失败）
-- 当前进展状态
-- 未解决的障碍
-
-### 给用户的建议
-在适当时机，向用户建议：
-- 咨询领域专家（教授、研究员）
-- 在 MathOverflow 上以适当方式提问
-- 查阅特定方向的文献
-- 考虑是否有简化版本可以首先解决
+## 8. Recording and Traceability
+
+### Problem record
+For each puzzle, record the following information:
+- Problem classification
+- Search keywords and results
+- Methods tried and their results (success or failure)
+- Current progress status
+- unresolved obstacles
+
+### Advice to users
+At the appropriate time, suggest to users:
+- Consulting experts (professors, researchers)
+- Ask questions in an appropriate manner on MathOverflow
+- Review literature in a specific direction
+- Consider if there is a simplified version that can be solved first
diff --git a/modules/higher_math_modules.md b/modules/higher_math_modules.md
index 40c2bd7..75cdc96 100644
--- a/modules/higher_math_modules.md
+++ b/modules/higher_math_modules.md
@@ -1,335 +1,335 @@
-# Math.skill 高等数学模块 (Higher Math Modules)
+# Math.skill Higher Math Modules
 
-以下定义 8 个高等数学领域的详细解题与验算要求。
+The following defines detailed problem solving and calculation requirements for 8 areas of advanced mathematics.
 
 ---
 
-## 模块 1：极限 (Limits)
+## Module 1: Limits
 
-### 覆盖范围
-- 数列极限：`lim_{n→∞} a_n` 的存在性与计算
-- 函数极限：`lim_{x→a} f(x)`（含 `x→∞`）
-- 单侧极限：`lim_{x→a+}`、`lim_{x→a-}`
-- 无穷小比较：等价无穷小、高阶/低阶/同阶/等价
-- 泰勒展开方法求极限
-- 夹逼定理（Squeeze Theorem）
-- 单调有界定理（单调有界必收敛）
-- Stolz 定理（数列 L'Hôpital）
-- L'Hôpital 法则（函数极限）
-- 渐近展开
+### Coverage
+- Sequence limit: Existence and calculation of `lim_{n→∞} a_n`
+- Function limit: `lim_{x→a} f(x)` (including `x→∞`)
+- One-sided limits: `lim_{x→a+}`, `lim_{x→a-}`
+- Comparison of infinitesimals: equivalent infinitesimals, higher order/lower order/same order/equivalent
+-Taylor expansion method to find the limit
+- Squeeze Theorem
+- Monotone bounded theorem (monotone bounded must converge)
+- Stolz's theorem (sequence L'Hôpital)
+- L'Hôpital's law (function limits)
+- asymptotic expansion
 
-### 核心定理及适用条件
+### Core theorem and applicable conditions
 
-| 定理 | 适用条件 |
+| Theorem | Applicable conditions |
 |------|----------|
-| 夹逼定理 | `a_n ≤ b_n ≤ c_n` 且 `lim a_n = lim c_n = L` |
-| 单调有界定理 | 数列单调递增且有上界；或单调递减且有下界 |
-| Stolz 定理 | 分母严格单调趋于 `∞`（`∞/∞`型）或分子分母同趋于 `0`（`0/0`型）；需满足极限存在或为 `∞` |
-| L'Hôpital 法则 | 不定式 `0/0` 或 `∞/∞`；导数比值的极限存在或为 `∞` |
-| 泰勒展开 | 函数在展开点附近有足够阶的连续导数 |
-
-### 常见陷阱
-1. **Stolz 定理条件不满足**：分母必须是单调无穷大，不能只趋于无穷
-2. **L'Hôpital 循环使用**：反复使用后无法消除不定式，需改用其他方法
-3. **泰勒展开阶数不足**：导致等价无穷小抵消，需要更高阶展开
-4. **忽视了极限不存在的情形**：左右极限不等
-5. **对数列使用 L'Hôpital**：L'Hôpital 仅适用于函数极限
-
-### 验算策略
-- **定理条件检查**：逐一验证所用定理的条件是否满足
-- **左右一致性检查**：对于 `x→a` 的极限，检查左右极限是否一致
-- **展开阶数检查**：确认泰勒展开阶数足够，余项可控
-- **数值验证**：取具体数值近似计算，与解析结果比较
-- **夹逼验证**：检查夹逼的上下界是否正确且相等
+| Pinch Theorem | `a_n ≤ b_n ≤ c_n` and `lim a_n = lim c_n = L` |
+| Monotone bounded theorem | The sequence is monotonically increasing and has an upper bound; or is monotonically decreasing and has a lower bound |
+| Stolz's theorem | The denominator is strictly monotonic to `∞` (type `∞/∞`) or the numerator and denominator tend to `0` (type `0/0`); the limit must exist or be `∞` |
+| L'Hôpital's Law | Infinitive `0/0` or `∞/∞` ; the limit of the derivative ratio exists or is `∞` |
+| Taylor expansion | The function has continuous derivatives of sufficient order near the expansion point |
+
+### Common pitfalls
+1. **Stolz’s theorem conditions are not met**: the denominator must be monotonic infinity and cannot just tend to infinity.
+2. **L'Hôpital recycling**: The infinitive cannot be eliminated after repeated use, and other methods need to be used.
+3. **Insufficient order of Taylor expansion**: It leads to equivalent infinitesimal cancellation and requires higher order expansion.
+4. **Ignoring the situation where the limit does not exist**: The left and right limits are not equal
+5. **Use L'Hôpital for a sequence**: L'Hôpital only works with function limits
+
+### Verification strategy
+- **Theorem Condition Check**: Verify whether the conditions of the theorem used are met one by one.
+- **Left and right consistency check**: For the limit of `x→a`, check whether the left and right limits are consistent
+- **Expansion order check**: Confirm that the Taylor expansion order is sufficient and the remainder is controllable
+- **Numerical verification**: Approximate calculation using specific numerical values ​​and compare with analytical results
+- **Clamp verification**: Check whether the upper and lower bounds of the clamp are correct and equal
 
 ---
 
-## 模块 2：微分学 (Differentiation)
+## Module 2: Differentiation
 
-### 覆盖范围
-- 一元函数导数
-- 偏导数（多元函数）
-- 隐函数求导
-- 参数方程求导
-- 高阶导数
-- 极值与最值（一元和多元）
-- 凹凸性与拐点
-- 泰勒公式（含余项）
-- 方向导数与梯度（基础）
+### Coverage
+- Derivatives of functions of one variable
+- Partial derivatives (multivariate functions)
+- Implicit function derivation
+- Derivation of parametric equations
+- Higher order derivatives
+- Extreme value and maximum value (univariate and multivariate)
+- Concave-convexity and inflection points
+- Taylor formula (including remainder)
+- Directional Derivatives and Gradient (Basic)
 
-### 核心定理及适用条件
+### Core theorem and applicable conditions
 
-| 定理 | 适用条件 |
+| Theorem | Applicable conditions |
 |------|----------|
-| Fermat 定理 | 函数在极值点可导 → 导数为零 |
-| Rolle 定理 | `f(a)=f(b)` 且在 `[a,b]` 连续，`(a,b)` 可导 |
-| Lagrange 中值定理 | 在 `[a,b]` 连续，`(a,b)` 可导 |
-| 隐函数存在定理 | 偏导数连续且 Jacobian 非零 |
-| 多元极值判定 | Hessian 矩阵正定（极小）/负定（极大）/不定（鞍点） |
-
-### 常见陷阱
-1. **隐函数求导时忘记使用链式法则**：`d/dx[f(g(x),h(x))]`
-2. **忘记检查极值的充分条件**：一阶导为零只是必要条件
-3. **边界点遗漏**：闭区间最值可能在端点取到
-4. **Hessian 矩阵半定性**：不能确定极值类型，需要更高阶分析
-5. **线性近似误差估计不足**：泰勒余项的阶数判断错误
-
-### 验算策略
-- **逆运算验证**：对求得的不定积分求导，应返回原函数
-- **数值差分验证**：用有限差分 `(f(x+h)-f(x))/h` 与解析导数比较
-- **一阶/二阶条件验证**：检查极值点是否同时满足一阶和二阶条件
-- **边界点穷举**：检查所有候选点（驻点、边界点、不可导点）
-- **Hessian 验证**：对于多元极值，计算 Hessian 并验证正定/负定性
+| Fermat's theorem | The function is differentiable at the extreme point → the derivative is zero |
+| Rolle's Theorem | `f(a)=f(b)` and continuous at `[a,b]`, `(a,b)` is differentiable |
+| Lagrange's Mean Value Theorem | Continuous at `[a,b]`, Differentiable at `(a,b)` |
+| Existence theorem of implicit functions | Partial derivatives are continuous and Jacobian is non-zero |
+| Multivariate extreme value determination | Hessian matrix positive definite (minimum)/negative definite (maximum)/indefinite (saddle point) |
+
+### Common pitfalls
+1. **Forgot to use the chain rule when deriving implicit functions**: `d/dx[f(g(x),h(x))]`
+2. **Forgot to check the sufficient conditions for extreme values**: It is only a necessary condition that the first derivative is zero
+3. **Boundary point missing**: The maximum value of a closed interval may be obtained at the endpoint
+4. **Hessian matrix semi-qualitative**: The extreme value type cannot be determined and higher-order analysis is required.
+5. **Linear approximation error underestimation**: wrong order judgment of Taylor remainder
+
+### Verification strategy
+- **Inverse operation verification**: Derivative of the obtained indefinite integral should return the original function
+- **Numerical Difference Verification**: Compare with analytic derivative using finite difference `(f(x+h)-f(x))/h`
+- **First-order/second-order condition verification**: Check whether the extreme point satisfies both first-order and second-order conditions
+- **Boundary point exhaustion**: Check all candidate points (stationary points, boundary points, non-differentiable points)
+- **Hessian verification**: For multivariate extrema, calculate the Hessian and verify positive/negative definiteness
 
 ---
 
-## 模块 3：积分学 (Integration)
+## Module 3: Integration
 
-### 覆盖范围
-- 不定积分：换元法、分部积分、有理函数积分、三角积分
-- 定积分：Newton-Leibniz 公式
-- 反常积分（广义积分）：收敛性判定
-- 多重积分：累次积分、坐标变换（极坐标、柱坐标、球坐标）
-- 换元积分（含 Jacobian）
-- 分部积分
-- 曲线积分与曲面积分（基础级别）
+### Coverage
+- Indefinite integrals: substitution method, integral by parts, integral of rational functions, trigonometric integral
+- Definite integral: Newton-Leibniz formula
+- Abnormal integral (generalized integral): Convergence determination
+- Multiple integration: cumulative integration, coordinate transformation (polar coordinates, cylindrical coordinates, spherical coordinates)
+- Exchange points (including Jacobian)
+- Division points
+- Curve integral and surface integral (basic level)
 
-### 核心定理及适用条件
+### Core theorem and applicable conditions
 
-| 定理 | 适用条件 |
+| Theorem | Applicable conditions |
 |------|----------|
-| Newton-Leibniz 公式 | 被积函数在积分区间连续，原函数在区间上连续可导 |
-| 换元积分法 | 换元函数连续可导，积分区间变换正确 |
-| 分部积分法 | 两部分分别可导/可积 |
-| 积分中值定理 | 函数在闭区间连续 |
-| 反常积分比较判别法 | 非负被积函数，比较函数已知收敛性 |
-
-### 常见陷阱
-1. **换元时忘记变换积分限**：定积分换元后上下限需对应新变量
-2. **分部积分方向选择不当**：应选择能简化的拆分方向
-3. **反常积分收敛条件遗漏**：先判断是否收敛，再计算
-4. **多重积分次序交换条件未检查**：Fubini 定理需要绝对可积
-5. **坐标变换时 Jacobian 遗漏或算错**：
-6. **有理函数部分分式分解错误**：分母因式分解不全
-
-### 验算策略
-- **不定积分求导验证**：对结果求导，应等于被积函数
-- **定积分数值近似**：使用 Simpson 或梯形法估算，与解析结果比较
-- **换元变换验证**：检查积分区间变换是否正确
-- **收敛性验证**：对反常积分，先验收敛性再算值
-- **分部积分逆运算**：可用分部积分的另一方向验证
+| Newton-Leibniz formula | The integrand is continuous in the integration interval, and the original function is continuously differentiable in the interval |
+| Substitution integral method | The substitution function is continuously differentiable, and the integration interval transformation is correct |
+| Integration by parts method | The two parts can be differentiated/integrated respectively |
+| Integral mean value theorem | Functions are continuous in closed intervals |
+| Abnormal integral comparison and discrimination method | Non-negative integrand, comparison function with known convergence |
+
+### Common pitfalls
+1. **Forgot to transform the integral limit when changing elements**: After changing the element of the definite integral, the upper and lower limits need to correspond to the new variables
+2. **Inappropriate selection of division integration direction**: The splitting direction that can simplify should be selected
+3. **Abnormal integral convergence condition is missing**: first determine whether convergence occurs, and then calculate
+4. **Multiple integration order commutative conditions not checked**: Fubini’s theorem requires absolute integrability
+5. **Jacobian omission or miscalculation during coordinate transformation**:
+6. **Partial fraction decomposition error of rational function**: Incomplete factorization of the denominator
+
+### Verification strategy
+- **Indefinite integral derivation verification**: Derivative of the result should be equal to the integrand
+- **Numerical approximation of definite integrals**: estimate using Simpson or trapezoidal method and compare with analytical results
+- **Change transformation verification**: Check whether the integration interval transformation is correct
+- **Convergence Verification**: For anomalous integrals, a priori convergence is required before calculating the value.
+- **Inverse operation of integral by parts**: can be verified by the other direction of integral by parts
 
 ---
 
-## 模块 4：线性代数 (Linear Algebra)
+## Module 4: Linear Algebra (Linear Algebra)
 
-### 覆盖范围
-- 矩阵运算（加、乘、转置、逆）
-- 行列式计算与性质
-- 矩阵的秩
-- 线性方程组求解（齐次/非齐次）
-- 向量空间与子空间
-- 线性相关与线性无关
-- 特征值与特征向量
-- 矩阵对角化
-- Gram-Schmidt 正交化
-- 二次型与正定性
+### Coverage
+- Matrix operations (addition, multiplication, transpose, inversion)
+- Calculation and properties of determinants
+- rank of matrix
+- Solving linear equations (homogeneous/non-homogeneous)
+- Vector spaces and subspaces
+-Linear dependence and linear independence
+- Eigenvalues ​​and eigenvectors
+-Matrix diagonalization
+- Gram-Schmidt orthogonalization
+- Quadratic form and positive certainty
 
-### 核心定理及适用条件
+### Core theorem and applicable conditions
 
-| 定理 | 适用条件 |
+| Theorem | Applicable conditions |
 |------|----------|
-| 秩-零化度定理 | 对任何矩阵 `A`：`rank(A) + nullity(A) = n` |
-| Cramer 法则 | 系数矩阵为方阵且行列式非零 |
-| Cayley-Hamilton 定理 | 任何方阵满足其特征多项式 |
-| 谱定理（实对称） | 实对称矩阵可正交对角化 |
-| 正定二次型判定 | 所有顺序主子式 > 0 |
-
-### 常见陷阱
-1. **特征向量计算错误**：`(A-λI)v=0`，不是 `(A-λ)v=0`
-2. **对角化条件不满足**：矩阵必须可对角化（几何重数 = 代数重数）
-3. **行列式为零时求逆**：应先检查可逆性
-4. **秩的误判**：行秩 = 列秩，但执行行和列操作需谨慎
-5. **线性方程组解的表示不完整**：忘记包含齐次解的通解部分
-6. **Gram-Schmidt 数值不稳定**：在大规模计算中可能失去正交性
-
-### 验算策略
-- **维数验证**：`rank + nullity = n`
-- **回代验证**：解代入原方程，验证 `Ax = b`
-- **特征值验证**：验证 `det(A - λI) = 0` 对每个特征值
-- **特征向量验证**：验证 `Av = λv`
-- **对角化验证**：验证 `A = PDP^{-1}` 或 `P^T A P = D`
-- **逆矩阵验证**：验证 `A A^{-1} = I`
-- **正交性验证**：验证 `Q^T Q = I`
+| Rank-zero degree theorem | For any matrix `A`: `rank(A) + nullity(A) = n` |
+| Cramer's rule | The coefficient matrix is ​​a square matrix and the determinant is non-zero |
+| Cayley-Hamilton Theorem | Any square matrix satisfies its characteristic polynomial |
+| Spectral theorem (real symmetry) | Real symmetric matrices can be orthogonally diagonalized |
+| Determination of positive definite quadratic form | All sequential principal formulas > 0 |
+
+### Common pitfalls
+1. **Feature vector calculation error**: `(A-λI)v=0`, not `(A-λ)v=0`
+2. **Diagonalization condition is not met**: The matrix must be diagonalizable (geometric multiplicity = algebraic multiplicity)
+3. **Inverse when the determinant is zero**: Reversibility should be checked first
+4. **Misjudgment of rank**: row rank = column rank, but you need to be careful when performing row and column operations
+5. **Incomplete representation of solutions to linear equations**: Forgot to include the general solution part of the homogeneous solution
+6. **Gram-Schmidt numerical instability**: possible loss of orthogonality in large-scale calculations
+
+### Verification strategy
+- **Dimension Validation**: `rank + nullity = n`
+- **Back substitution verification**: Substitute the solution into the original equation and verify `Ax = b`
+- **Eigenvalue Validation**: Verify `det(A - λI) = 0` for each eigenvalue
+- **Feature Vector Validation**: Validation `Av = λv`
+- **Diagonal Verification**: Verify `A = PDP^{-1}` or `P^T A P = D`
+- **Inverse Matrix Verification**: Verification `A A^{-1} = I`
+- **Orthogonality Verification**: Verify `Q^T Q = I`
 
 ---
 
-## 模块 5：常微分方程 (ODEs)
+## Module 5: Ordinary Differential Equations (ODEs)
 
-### 覆盖范围
-- 可分离变量方程
-- 一阶线性微分方程
-- 齐次方程
-- 常系数线性微分方程（齐次/非齐次）
-- 初值问题 (IVP)
-- 简单微分方程组
+### Coverage
+- Equations of separable variables
+- First order linear differential equation
+- Homogeneous equations
+- Linear differential equations with constant coefficients (homogeneous/non-homogeneous)
+- Initial value problem (IVP)
+- System of simple differential equations
 
-### 核心定理及适用条件
+### Core theorem and applicable conditions
 
-| 定理 | 适用条件 |
+| Theorem | Applicable conditions |
 |------|----------|
-| 存在唯一性定理 (Picard-Lindelöf) | `f(t,y)` 关于 y 满足 Lipschitz 条件 |
-| 叠加原理（齐次线性） | 解的线性组合仍是解 |
-| 特征方程法 | 常系数齐次线性微分方程 |
-| 待定系数法 | 非齐次项为多项式、指数、三角函数的积/和 |
-| 常数变异法 | 适用于任何非齐次线性 ODE（如果能求齐次解） |
-
-### 常见陷阱
-1. **遗忘积分常数**：不定积分后必须加 `+ C`
-2. **通解与特解混淆**：初值问题需要确定常数
-3. **常数的数量与阶数不匹配**：n 阶 ODE 的通解应有 n 个独立常数
-4. **齐次与非齐次解混淆**：通解 = 齐次通解 + 非齐次特解
-5. **重根情形处理错误**：特征方程有重根时解的构造特殊
-6. **待定系数法形式猜测不全**：非齐次项与齐次解有重叠时需乘以 `t^k`
-
-### 验算策略
-- **代入原方程验证**：将求得的解代入 ODE，检查等式是否恒成立
-- **初值条件检查**：将 `t=t_0` 代入解，验证是否满足初值
-- **解的独立性检查**：Wronski 行列式非零
-- **通解完备性**：独立解的数量 = 方程阶数
-- **常数的数量检查**：通解中独立常数的数量 = 方程阶数
+| Existence and uniqueness theorem (Picard-Lindelöf) | `f(t,y)` satisfies the Lipschitz condition with respect to y |
+| Superposition principle (homogeneous linearity) | A linear combination of solutions is still a solution |
+| Characteristic equation method | Homogeneous linear differential equations with constant coefficients |
+| Undetermined coefficient method | Non-homogeneous terms are the products/sums of polynomials, exponentials, and trigonometric functions |
+| Constant mutation method | Applicable to any non-homogeneous linear ODE (if a homogeneous solution can be found) |
+
+### Common pitfalls
+1. **Forgot the integral constant**: `+ C` must be added after the indefinite integral
+2. **Confusion between general solution and special solution**: Initial value problem requires determination of constants
+3. **The number of constants does not match the order**: The general solution of an n-order ODE should have n independent constants
+4. **Confusion between homogeneous and non-homogeneous solutions**: General solution = Homogeneous general solution + Non-homogeneous special solution
+5. **Error in handling multiple root cases**: The solution is specially constructed when the characteristic equation has multiple roots.
+6. **Incomplete guess on the form of the undetermined coefficient method**: When the non-homogeneous term overlaps with the homogeneous solution, it needs to be multiplied by `t^k`
+
+### Verification strategy
+- **Substitute into the original equation to verify**: Substitute the obtained solution into ODE and check whether the equation is always true.
+- **Initial value condition check**: Substitute `t=t_0` into the solution to verify whether the initial value is met
+- **Independence check of solutions**: Wronski determinant is non-zero
+- **General solution completeness**: number of independent solutions = order of equation
+- **Check for the number of constants**: Number of independent constants in the general solution = Order of the equation
 
 ---
 
-## 模块 6：实分析基础 (Real Analysis)
+## Module 6: Basics of Real Analysis (Real Analysis)
 
-### 覆盖范围
-- 数列收敛与发散判定
-- 连续性与一致连续性
-- 可微性与导数
-- Riemann 积分的基本性质
-- 级数收敛性（正项、交错、任意项）
-- 函数序列与函数项级数的一致收敛性
+### Coverage
+- Judgment of convergence and divergence of sequence
+- Continuity and consistent continuity
+- Differentiability and derivatives
+- Basic properties of Riemann integrals
+- Series convergence (positive terms, staggered terms, arbitrary terms)
+- Consistent convergence of function sequences and function term series
 
-### 核心定理及适用条件
+### Core theorem and applicable conditions
 
-| 定理 | 适用条件 |
+| Theorem | Applicable conditions |
 |------|----------|
-| Cauchy 收敛准则 | 完备度量空间（实数、复数） |
-| Bolzano-Weierstrass 定理 | 有界数列必有收敛子列 |
-| 连续函数的介值定理 | 函数在闭区间连续 |
-| 一致连续性定理 | 闭区间上的连续函数必一致连续 |
-| Weierstrass M 判别法 | 函数项级数有收敛的控制级数 |
-| 一致收敛极限的连续性 | 一致收敛的连续函数序列，极限函数连续 |
-
-### 常见陷阱
-1. **量化词顺序颠倒**：一致连续的 `∀ε ∃δ ∀x,y` 和点态连续的 `∀ε ∀x ∃δ` 不同
-2. **ε-δ 与 ε-N 混淆**：数列用 ε-N，函数用 ε-δ
-3. **点态收敛与一致收敛混淆**：一致收敛意味着点态收敛，反之不然
-4. **条件缺乏时使用结论**：如在不连续的闭区间上使用介值定理
-5. **级数级数重排的条件**：绝对收敛才能任意重排
-6. **极限与积分交换**：需要一致收敛或控制收敛定理
-
-### 验算策略
-- **量化词顺序验证**：检查 ε-δ 或 ε-N 的量化词结构
-- **反例构造**：如果对某个推广有怀疑，构造反例测试
-- **逐点收敛 vs 一致收敛**：检查上确界是否趋于零
-- **连续性验证**：检查极限定义和左右极限是否一致
+| Cauchy Convergence Criterion | Complete Metric Space (Real Numbers, Complex Numbers) |
+| Bolzano-Weierstrass Theorem | A bounded sequence must have a convergent subsequence |
+| Intermediate value theorem of continuous functions | Functions are continuous in closed intervals |
+| Uniform continuity theorem | Continuous functions on closed intervals must be uniformly continuous |
+| Weierstrass M criterion | Control series with convergent function term series |
+| Continuity of uniform convergence limit | Consistently convergent continuous function sequence, limit function continuity |
+
+### Common pitfalls
+1. **The order of quantifiers is reversed**: The consistent and continuous `∀ε ∃δ ∀x,y` is different from the point-state continuous `∀ε ∀x ∃δ`
+2. **ε-δ and ε-N are confused**: use ε-N for sequences and ε-δ for functions.
+3. **Confusion between point-state convergence and uniform convergence**: Uniform convergence means point-state convergence, but not vice versa.
+4. **Use conclusions when conditions are lacking**: such as using the intermediate value theorem on discontinuous closed intervals
+5. **Conditions for series rearrangement**: Absolute convergence can lead to arbitrary rearrangement
+6. **Exchange of limits and integrals**: Requires consistent convergence or controlled convergence theorem
+
+### Verification strategy
+- **Quantifier order verification**: Check the quantifier structure of ε-δ or ε-N
+- **Counterexample construction**: If you have doubts about a certain promotion, construct a counterexample test
+- **Pointwise Convergence vs Uniform Convergence**: Check if the supremum approaches zero
+- **Continuity Verification**: Check whether the limit definition and the left and right limits are consistent
 
 ---
 
-## 模块 7：抽象代数基础 (Abstract Algebra)
+## Module 7: Abstract Algebra
 
-### 覆盖范围
-- 群：子群、正规子群、商群、同态基本定理
-- 环：子环、理想、商环
-- 域的基本概念
-- 多项式环基础
+### Coverage
+- Groups: subgroups, normal subgroups, quotient groups, fundamental theorem of homomorphism
+- Ring: sub-ring, ideal, business ring
+- Basic concepts of domain
+- Basics of Polynomial Rings
 
-### 核心定理及适用条件
+### Core theorem and applicable conditions
 
-| 定理 | 适用条件 |
+| Theorem | Applicable conditions |
 |------|----------|
-| Lagrange 定理 | 有限群，子群的阶整除群的阶 |
-| Cayley 定理 | 任何群同构于某个置换群的子群 |
-| 同态基本定理 | `G/ker(φ) ≅ Im(φ)` |
-| 第一同构定理 | 子群与正规子群的对应关系 |
-| 中国剩余定理（环论版） | 理想两两互素 |
-
-### 常见陷阱
-1. **子群验证遗漏**：需要验证封闭性、含单位元、含逆元（或等价地 `ab⁻¹ ∈ H`）
-2. **正规子群 vs 子群**：所有子群都是正规子群的说法仅在 Abel 群中成立
-3. **同态验证不完整**：需要验证保持运算且保持单位元
-4. **商群良定性**：需要先确认正规子群
-5. **群的阶与元素阶关系**：元素的阶整除群的阶，逆命题不总是成立
-6. **以偏概全**：在有少量小阶群的结论后直接推广到所有群
-
-### 验算策略
-- **封闭性检查**：生成所有元素对，验证乘积在集合内
-- **逆元检查**：每个元素有唯一逆元
-- **结合律检查**：对特定元素三重积检查
-- **同态结构保持验证**：`φ(ab) = φ(a)φ(b)`
-- **商群良定性验证**：左右陪集相等
-- **优先使用小阶反例**：用小阶群（如 S₃、D₄、Z₆）测试每个猜想
+| Lagrange's theorem | Finite group, the order of the subgroup is divided by the order of the group |
+| Cayley's theorem | Any group is isomorphic to a subgroup of a permutation group |
+| Fundamental Theorem of Homomorphism | `G/ker(φ) ≅ Im(φ)` |
+| First isomorphism theorem | Correspondence between subgroups and normal subgroups |
+| Chinese Remainder Theorem (Ring Theory Edition) | Ideal pairs are relatively prime |
+
+### Common pitfalls
+1. **Subgroup verification missing**: Need to verify closure, identity element, and inverse element (or equivalently `ab⁻¹ ∈ H` )
+2. **Normal subgroup vs subgroup**: The statement that all subgroups are normal subgroups is only true in Abelian groups
+3. **Incomplete homomorphism verification**: It is necessary to verify that the operation is maintained and the unit is maintained
+4. **Goodness of business group**: It is necessary to confirm the formal subgroup first
+5. **Relationship between the order of the group and the order of the elements**: The order of the element divides the order of the group, and the converse proposition is not always true.
+6. **Generalize from partial to complete**: After having a few conclusions about small-order groups, directly generalize to all groups.
+
+### Verification strategy
+- **Closedness Check**: Generate all pairs of elements and verify that the product is within the set
+- **Inverse check**: Each element has a unique inverse
+- **Associative Law Check**: Check the triple product of specific elements
+- **Homomorphic structure preservation verification**: `φ(ab) = φ(a)φ(b)`
+- **Good Qualitative Verification of Business Groups**: The left and right cosets are equal
+- **Prefer small-order counterexamples**: test each conjecture with small-order groups (such as S₃, D₄, Z₆)
 
 ---
 
-## 模块 8：拓扑学基础 (Topology Basics)
+## Module 8: Topology Basics
 
-### 覆盖范围
-- 开集、闭集、拓扑空间定义
-- 连续映射的拓扑定义
-- 紧致性（compactness）的基本性质
-- 连通性（connectedness）的基本性质
-- 度量空间基本概念
+### Coverage
+- Definition of open sets, closed sets, and topological spaces
+- Topological definition of continuous mapping
+- Basic properties of compactness
+- Basic properties of connectedness
+- Basic concepts of metric space
 
-### 核心定理及适用条件
+### Core theorem and applicable conditions
 
-| 定理 | 适用条件 |
+| Theorem | Applicable conditions |
 |------|----------|
-| Heine-Borel 定理 | `R^n` 中子集紧致 iff 有界且闭 |
-| 连续函数保持紧致性 | 连续像将紧集映为紧集 |
-| 介值定理（连通性版本） | 连续像将连通集映为连通集 |
-| Tietze 扩张定理 | 正规空间中的闭子集 |
-| Bolzano-Weierstrass 定理（度量空间版） | 列紧与紧致的等价性 |
-
-### 常见陷阱
-1. **将欧氏空间的直觉外推到一般拓扑空间**：不是所有拓扑空间都是 Hausdorff 空间
-2. **紧致性等价条件混淆**：
-   - 在度量空间中：紧致 ↔ 列紧 ↔ 完备且全有界
-   - 在一般拓扑空间中：这些等价不一定成立
-3. **开集与闭集的对偶性**：开集的补是闭集，但存在既开又闭、不既开也不闭的集合
-4. **连续映射的序列定义不适用于一般拓扑空间**：需要网的收敛概念
-5. **连通空间不一定是道路连通的**：拓扑学中著名的反例（如拓扑学家的正弦曲线）
-6. **紧致空间的闭子集是紧的**，但需要紧致空间是 Hausdorff 的
-
-### 验算策略
-- **定义一致性检查**：逐条对照拓扑空间定义中的开集公理
-- **反例搜索**：对每个疑似的推广，在 `R^n` 中或有限拓扑中找反例
-- **度量空间与一般拓扑空间的区分**：注意哪些性质仅适用于度量空间
-- **Hausdorff 条件检查**：某些结论需要 Hausdorff 条件（如紧子集的闭性）
-- **紧致性验证**：检查每个开覆盖是否有有限子覆盖
+| Heine-Borel Theorem | `R^n` The neutron set compact iff is bounded and closed |
+| Continuous functions maintain compactness | Continuous images map compact sets into compact sets |
+| Intermediate value theorem (connectivity version) | Continuous image maps connected sets to connected sets |
+| Tietze's expansion theorem | Closed subsets in normal spaces |
+| Bolzano-Weierstrass theorem (metric space version) | Equivalence of compaction and compactness |
+
+### Common pitfalls
+1. **Extrapolate the intuition of Euclidean spaces to general topological spaces**: Not all topological spaces are Hausdorff spaces
+2. **Confusion of compactness equivalence conditions**:
+- In metric space: compact ↔ compact ↔ complete and fully bounded
+- In general topological spaces: these equivalences do not necessarily hold
+3. **Duality of open sets and closed sets**: The complement of an open set is a closed set, but there are sets that are both open and closed, and neither open nor closed.
+4. **The sequence definition of continuous mapping is not suitable for general topological spaces**: the convergence concept of the network is required
+5. **Connected spaces are not necessarily road-connected**: Famous counterexamples in topology (such as topologists’ sinusoids)
+6. **The closed subset of a compact space is compact**, but the compact space needs to be Hausdorff's
+
+### Verification strategy
+- **Definition consistency check**: Check the open set axioms in the definition of topological space item by item
+- **Counterexample search**: For each suspected generalization, find counterexamples in `R^n` or limited topology
+- **The difference between metric spaces and general topological spaces**: Note which properties only apply to metric spaces
+- **Hausdorff condition check**: Some conclusions require Hausdorff conditions (such as closure of compact subsets)
+- **Compactness Verification**: Check whether each open cover has finite subcovers
 
 ---
 
-## 跨模块通用注意事项
-
-### 定义域与条件检查清单
-在解题前强制检查：
-- [ ] 函数的定义域是什么？
-- [ ] 所用定理的条件是否全部满足？
-- [ ] 是否存在退化边界情况（零、无穷、奇点）？
-- [ ] 使用的所有变量是否有明确的范围？
-
-### 跨领域常见错误
-1. **除法前未检查非零**
-2. **开方前未检查非负**
-3. **在不可导点使用求导**
-4. **在不连续点使用积分**
-5. **在非 Riemann 可积函数上使用 Riemann 积分**
-6. **未检查收敛性就计算级数和**
-
-### 验算的跨领域一致性
-- 每个模块的结论应与其他模块的结论自洽
-- 如果有多种方法求解同一问题，用另一种方法验算
-- 特别注意符号、单位、量纲的一致性
+## Common cross-module considerations
+
+### Domain and Condition Checklist
+It is mandatory to check before solving the problem:
+- [ ] What is the domain of a function?
+- [ ] Are all the conditions of the theorem used met?
+- [ ] Are there degenerate boundary cases (zero, infinity, singularity)?
+- [ ] Do all variables used have clear scopes?
+
+### Common cross-domain errors
+1. **Non-zero is not checked before division**
+2. **Non-negative is not checked before formulating**
+3. **Use derivation at non-differentiable points**
+4. **Use integrals at discontinuous points**
+5. **Using Riemann integrals on non-Riemann integrable functions**
+6. **Calculating series sums without checking convergence**
+
+### Cross-domain consistency of verification
+- The conclusions of each module should be consistent with the conclusions of other modules
+- If there are multiple ways to solve the same problem, use another method to check the calculation
+- Pay special attention to the consistency of symbols, units and dimensions
diff --git a/modules/interaction_policy.md b/modules/interaction_policy.md
index 1b68ef6..57e1450 100644
--- a/modules/interaction_policy.md
+++ b/modules/interaction_policy.md
@@ -1,611 +1,635 @@
-# Math.skill 交互策略 (Interaction Policy)
+# Math.skill Interaction Policy
 
-每种交互场景需包含：(a) 检测标准，(b) 响应策略，(c) 必要验证，(d) 示例回复。
+Each interaction scenario needs to include: (a) detection criteria, (b) response strategy, (c) necessary verification, (d) sample responses.
 
 ---
 
-## 场景 1：用户问题不完整
+## Scenario 0: Unable to verify or model confidence is extremely low (Anti-Hallucination)
 
-### (a) 检测标准
-- 缺少定义域（如未指定 `x` 的取值范围）
-- 缺少初始条件或边界条件
-- 缺少函数的明确定义
-- 缺少方程中的参数取值范围
-- 量纲不一致或符号未定义
+### (a) Testing standards
+- The internal confidence of the model is low and no clear steps can be derived.
+- At least two verification methods fail consecutively and cannot be repaired on their own.
+- Requires complex calculations beyond the current computational capabilities of the model.
 
-### (b) 响应策略
-1. 指出问题中不完整的具体位置
-2. 列出缺失的必要信息
-3. 针对每种可能的补充条件，说明对解答的影响
-4. 等待用户补充后再继续
+### (b) Response strategy
+1. It is absolutely prohibited to output "Plausible BS".
+2. It is absolutely prohibited to fabricate illusions or fictitious theorems that have passed verification.
+3. It is necessary to directly admit that the solution cannot be solved and explain at which step an insurmountable obstacle or verification failure was encountered.
 
-### (c) 必要验证
-- 确认补充后的信息与原始问题无矛盾
+### (c) Necessary verification
+- Self-censorship: Are you trying to concoct a plausible answer? If so, stop immediately and output a failure statement.
 
-### (d) 示例回复
+### (d) Sample reply
+ ```
+I can't give you a final answer to this question.
+During the calculation, my verification mechanism failed repeatedly, which meant that the intermediate results I derived were unreliable.
+In order to maintain mathematical rigor, I cannot provide you with unverified or possibly incorrect answers. The step I'm stuck on is: [Specify where the validation fails or the calculation cannot continue].
 ```
-您的问题中缺少以下信息：
-- 当前未指定 x 的取值范围（定义域）。x 是实数还是正整数？
-  这会影响级数是否收敛以及函数的可导性。
-- 缺少初始条件。如果是解微分方程，需要 y(0) 的值。
 
-请补充这些信息后，我可以给出完整解答。
+---
+
+## Scenario 1: User question is incomplete
+
+### (a) Testing standards
+- Missing domain (e.g. the value range of `x` is not specified)
+- Missing initial or boundary conditions
+- Lack of clear definition of functions
+- Missing parameter ranges in equations
+- Inconsistent dimensions or undefined symbols
+
+### (b) Response strategy
+1. Point out the specific location of the incompleteness in the question
+2. List missing necessary information
+3. For each possible supplementary condition, explain the impact on the solution
+4. Wait for the user to add before continuing
+
+### (c) Necessary verification
+- Confirm that the supplemented information does not conflict with the original question
+
+### (d) Sample reply
+ ```
+The following information is missing from your question:
+- The range (domain) of x is currently unspecified. Is x a real number or a positive integer?
+This affects whether the series converges and the differentiability of the function.
+- Missing initial conditions. If you are solving a differential equation, you need the value of y(0).
+
+Please add this information so I can give a complete answer.
 ```
 
 ---
 
-## 场景 2：图片题目文字信息不足
+## Scenario 2: Insufficient text information in the picture title
 
-### (a) 检测标准
-- 用户上传了图片但文字描述不完整
-- 图片中关键符号、公式无法准确识别
-- 图片质量低导致部分内容模糊
+### (a) Testing standards
+- User uploaded a picture but the text description is incomplete
+- Key symbols and formulas in pictures cannot be accurately identified
+- Low picture quality causes some content to be blurry
 
-### (b) 响应策略
-1. 说明已尝试识别但不确定的部分
-2. 请求用户转录关键公式和条件
-3. 指出需要清晰化的具体位置
+### (b) Response strategy
+1. Describe the parts that you have tried to identify but are not sure about
+2. Ask users to transcribe key formulas and conditions
+3. Point out specific locations that need clarification.
 
-### (c) 必要验证
-- 等待用户确认关键信息后再开始解题
+### (c) Necessary verification
+- Wait for the user to confirm key information before starting to solve the problem
 
-### (d) 示例回复
-```
-感谢上传图片。我已尝试识别题目，但对以下部分不确定：
-- 第 2 行的积分下限是否是 0？
-- 第 3 行求和符号的上限是 n 还是 ∞？
+### (d) Sample reply
+ ```
+Thanks for uploading the picture. I have tried identifying the question but am not sure about the following parts:
+- Is the lower limit of points in line 2 0?
+- Is the upper limit on the sign of the summation in line 3 n or ∞?
 
-能否请您确认或转录这几处？这将确保我给出的答案完全正确。
+Could you please confirm or transcribe these points? This will ensure that the answer I give is completely correct.
 ```
 
 ---
 
-## 场景 3：用户要求"只给答案"
+## Scenario 3: User requests "just give answers"
 
-### (a) 检测标准
-- 明确说"只要答案"、"直接给结果"、"不需要过程"
-- "简答"、"答案是什么"、"最后结果多少"
+### (a) Testing standards
+- Clearly say "just the answer", "give the result directly", "no process required"
+- "Short answer", "What is the answer", "What is the final result"
 
-### (b) 响应策略
-1. 直接给答案
-2. 附加极简的验算（一两行）
-3. 不展开推导过程
+### (b) Response strategy
+1. Give the answer directly
+2. Add minimalist verification (one or two lines)
+3. Do not start the derivation process
 
-### (c) 必要验证
-- 答案必须经过验算（即使不展示过程）
-- 检查是否有多个解（如二次方程有两个根）
+### (c) Necessary verification
+- The answer must be verified (even if the process is not shown)
+- Check if there are multiple solutions (e.g. a quadratic equation has two roots)
 
-### (d) 示例回复
-```
-答案为：x = 3 或 x = -1
-简单验算：代入原方程，3² - 2×3 - 3 = 0 ✓；(-1)² - 2×(-1) - 3 = 0 ✓
+### (d) Sample reply
+ ```
+The answer is: x = 3 or x = -1
+Simple calculation: Substitute into the original equation, 3² - 2×3 - 3 = 0 ✓; (-1)² - 2×(-1) - 3 = 0 ✓
 ```
 
 ---
 
-## 场景 4：用户要求"不要用导数"
+## Scenario 4: User request "Don't use derivatives"
 
-### (a) 检测标准
-- "不要用微积分"
-- "用初等方法"
-- "不用求导"
-- "不要用导数"
+### (a) Testing standards
+- "Don't use calculus"
+- "Use elementary methods"
+- "No need to ask for derivation"
+- "Don't use derivatives"
 
-### (b) 响应策略
-1. 完全遵守约束，不使用导数相关方法
-2. 使用替代方法（配方法、判别式、不等式、对称性、几何法等）
-3. 如实在无法避免使用导数，提前说明并征得同意
+### (b) Response strategy
+1. Fully comply with constraints and do not use derivative-related methods
+2. Use alternative methods (combination method, discriminant, inequality, symmetry, geometric method, etc.)
+3. If it is really unavoidable to use derivatives, explain it in advance and obtain consent.
 
-### (c) 必要验证
-- 确认解答中不包含任何求导步骤
-- 检查替代方法的等价性
+### (c) Necessary verification
+- Confirm that the solution does not contain any derivation steps
+- Check equivalence of alternative methods
 
-### (d) 示例回复
-```
-好的，我将使用配方法求极值，不涉及导数。
+### (d) Sample reply
+ ```
+Okay, I'm going to use the collocation method to find the extremum, no derivatives involved.
 
 f(x) = x² - 4x + 7 = (x-2)² + 3
 
-因为 (x-2)² ≥ 0 恒成立，当 x=2 时取最小值 0，所以 f(x) 的最小值为 3，在 x=2 处取得。
+Since (x-2)² ≥ 0 always holds, taking the minimum value 0 when x=2, the minimum value of f(x) is 3, obtained at x=2.
 ```
 
 ---
 
-## 场景 5：用户要求"用初中/高中方法"
+## Scenario 5: User request "Use junior high school/high school method"
 
-### (a) 检测标准
-- "用初中方法"、"用高中知识"
-- "不要用高等数学"
-- "用中学范围内的方法"
+### (a) Testing standards
+- "Use junior high school methods", "Use high school knowledge"
+- "Don't use advanced mathematics"
+- "Use middle school-wide methods"
 
-### (b) 响应策略
-1. 确认选用的方法在指定学段的知识范围内
-2. 必要时指出高级方法的简洁性，但不使用
-3. 如果问题确实超出指定学段的能力范围，如实告知
+### (b) Response strategy
+1. Confirm that the chosen method is within the knowledge range of the designated academic period
+2. Point out the simplicity of high-level methods when necessary, but don’t use them
+3. If the problem is indeed beyond the ability of the designated academic period, tell the truth.
 
-### (c) 必要验证
-- 确认所用方法不超出指定学段的教学大纲
-- 验证初等方法是否严格证明正确
+### (c) Necessary verification
+- Confirm that the methods used do not exceed the syllabus for the designated academic period
+- Verify that elementary methods are strictly proven correct
 
-### (d) 示例回复
-```
-使用高中方法（基本不等式）：
-设 a, b > 0，由均值不等式：a + b ≥ 2√(ab)，等号当且仅当 a = b。
+### (d) Sample reply
+ ```
+Using the high school method (basic inequalities):
+Assume a, b > 0, from the mean inequality: a + b ≥ 2√(ab), the equal sign is if and only if a = b.
 
-这里 a = x, b = 1/x，所以 x + 1/x ≥ 2，最小值在 x = 1 处取得。
+Here a = x, b = 1/x, so x + 1/x ≥ 2, and the minimum is taken at x = 1.
 ```
 
 ---
 
-## 场景 6：用户要求"用高等数学方法"
+## Scenario 6: User request "use advanced mathematical methods"
 
-### (a) 检测标准
-- "用高等数学方法"
-- "用大学数学"、"用更高级的工具"
-- "用分析/代数/拓扑方法"
-- "用拉格朗日乘数法"等指定高级方法
+### (a) Testing standards
+- "Use advanced mathematical methods"
+- "Use college mathematics", "Use more advanced tools"
+- "Use analytical/algebraic/topological methods"
+- Specify advanced methods such as "Use Lagrange Multiplier Method"
 
-### (b) 响应策略
-1. 使用指定领域的高级工具
-2. 选择合适的定理和框架
-3. 对方法的选择做简要说明
-4. 可同时展示高等方法和初等方法供对比（如果有助于理解）
+### (b) Response strategy
+1. Use advanced tools in specific areas
+2. Choose appropriate theorems and frameworks
+3. Briefly explain the choice of method
+4. Advanced methods and elementary methods can be displayed simultaneously for comparison (if it helps to understand)
 
-### (c) 必要验证
-- 确认高级定理的条件完全满足
-- 验算与初等方法的结果一致（如果初等方法适用）
+### (c) Necessary verification
+- Confirm that the conditions of the advanced theorem are fully satisfied
+- Check that the calculation is consistent with the results of the elementary method (if the elementary method is applicable)
 
-### (d) 示例回复
-```
-使用 Lagrange 乘数法：
+### (d) Sample reply
+ ```
+Use the Lagrange multiplier method:
 
-定义 Lagrange 函数：L(x,y,λ) = f(x,y) - λ g(x,y)
-求 ∂L/∂x = 0，∂L/∂y = 0，∂L/∂λ = 0
-解得：...
+Define the Lagrange function: L(x,y,λ) = f(x,y) - λ g(x,y)
+Find ∂L/∂x = 0, ∂L/∂y = 0, ∂L/∂λ = 0
+Solution:...
 ```
 
 ---
 
-## 场景 7：用户给的答案是错误的，让帮忙检查
+## Scenario 7: The answer given by the user is wrong, please help to check
 
-### (a) 检测标准
-- "我这样算对吗？"
-- "帮我看看哪里错了"
-- "我的答案和参考答案不一样"
-- 用户展示了含有错误的推导过程
+### (a) Testing standards
+- "Am I right?"
+- "Help me see what's wrong"
+- "My answer is different from the reference answer"
+- The user demonstrated the derivation process containing errors
 
-### (b) 响应策略
-1. 首先尊重用户的努力，用建设性的语气
-2. 明确指出错误出在哪一步
-3. 解释为什么错了（数学原理）
-4. 给出正确的做法
-5. 避免使用负面评价词汇（如"太差了"、"明显错了"）
+### (b) Response strategy
+1. First respect the user’s efforts and use a constructive tone
+2. Clearly indicate where the error occurred
+3. Explain why it is wrong (mathematical principles)
+4. Give the correct approach
+5. Avoid using negative words (such as "too bad", "obviously wrong")
 
-### (c) 必要验证
-- 多角度验证正确解答
-- 对用户的错误推导，解释为何特定步骤不成立
+### (c) Necessary verification
+- Verify correct answers from multiple angles
+- Error derivation to the user, explaining why a specific step does not work
 
-### (d) 示例回复
-```
-您的思路大部分是对的，但在第 3 步出现了一个问题：
+### (d) Sample reply
+ ```
+Most of your ideas are correct, but there is a problem in step 3:
 
-第 3 步：√(a²) 不等于 a，而应等于 |a|。
-因为平方根函数返回的是非负值。在这个问题中 a 可能为负，所以省略绝对值会导致符号错误。
+Step 3: √(a²) is not equal to a, but equal to |a|.
+Because the square root function returns a non-negative value. In this problem a may be negative, so omitting the absolute value results in a sign error.
 
-正确的做法是：
+The correct approach is:
 √(x² - 2x + 1) = √((x-1)²) = |x-1|
 
-在 x < 1 时等于 1-x，在 x ≥ 1 时等于 x-1。
+Equals 1-x when x < 1 and equals x-1 when x ≥ 1.
 ```
 
 ---
 
-## 场景 8：用户要求生成练习题
+## Scenario 8: User requests to generate practice questions
 
-### (a) 检测标准
-- "给我出几道题"
-- "有没有类似的题目可以练习"
-- "帮我编一道题"
-- "生成练习题"
+### (a) Testing standards
+- "Ask me some questions"
+- "Are there any similar questions that I can practice on?"
+- "Make up a question for me"
+- "Generate exercises"
 
-### (b) 响应策略
-1. 生成在指定范围内的题目
-2. 提供完整答案
-3. 给出评分标准/解题要点
-4. 标注难度等级
-5. 题目应至少有 3 道，覆盖不同难度
+### (b) Response strategy
+1. Generate questions within the specified range
+2. Provide complete answers
+3. Provide scoring criteria/problem-solving points
+4. Mark the difficulty level
+5. There should be at least 3 questions covering different levels of difficulty
 
-### (c) 必要验证
-- 确保生成的题目条件完整且可解
-- 验证参考答案的正确性
-- 确保题目不是已知竞赛题的直接复制
+### (c) Necessary verification
+- Ensure that the generated questions are complete and solvable
+- Verify the correctness of the reference answers
+- Make sure the question is not a direct copy of a known competition question
 
-### (d) 示例回复
-```
-以下是为您生成的 3 道练习（难度递增）：
+### (d) Sample reply
+ ```
+Here are 3 exercises (in increasing difficulty) generated for you:
 
-**题 1** (基础, ★☆☆): 计算 lim_{x→0} (sin 3x)/(tan 2x)
-答案：3/2。要点：利用等价无穷小 sin u ~ u, tan u ~ u。
+**Question 1** (Basic, ★☆☆): Calculate lim_{x→0} (sin 3x)/(tan 2x)
+Answer: 3/2. Key point: Use the equivalent infinitesimal sin u ~ u, tan u ~ u.
 
-**题 2** (中等, ★★☆): 判断 lim_{x→0} (ln(1+x) - x)/x² 的值
-答案：-1/2。要点：泰勒展开 ln(1+x) = x - x²/2 + O(x³)。
+**Question 2** (Medium, ★★☆): Determine the value of lim_{x→0} (ln(1+x) - x)/x²
+Answer: -1/2. Key points: Taylor expansion ln(1+x) = x - x²/2 + O(x³).
 
-**题 3** (较难, ★★★): 计算 lim_{x→0} (arcsin x - arctan x)/x³
-答案：1/3。要点：两个函数的泰勒展开均需展开到 3 阶。
+**Question 3** (harder, ★★★): Calculate lim_{x→0} (arcsin x - arctan x)/x³
+Answer: 1/3. Important: The Taylor expansion of both functions needs to be expanded to order 3.
 ```
 
 ---
 
-## 场景 9：用户要求证明的错误命题
+## Scenario 9: Wrong proposition asked by the user to prove
 
-### (a) 检测标准
-- 用户提出的命题在数学上是错误的
-- 题目要求证明一个实际上不成立的结论
+### (a) Testing standards
+- The proposition proposed by the user is mathematically incorrect
+- The question requires proving a conclusion that is actually untenable
 
-### (b) 响应策略
-1. 明确说明该命题不正确
-2. 提供反例（最简单、最直观的可能反例）
-3. 解释为什么反例推翻了命题
-4. 如果可能的条件下命题可以成立，指出需要添加什么额外条件
+### (b) Response strategy
+1. Clearly state that the proposition is incorrect
+2. Provide counterexamples (the simplest and most intuitive possible counterexamples)
+3. Explain why the counterexample overturns the proposition
+4. If the proposition can be true under possible conditions, indicate what additional conditions need to be added
 
-### (c) 必要验证
-- 反例必须严格满足原命题的所有条件
-- 反例必须明确违背原命题的结论
+### (c) Necessary verification
+- The counterexample must strictly satisfy all the conditions of the original proposition
+- The counterexample must clearly violate the conclusion of the original proposition
 
-### (d) 示例回复
-```
-这个命题是不正确的。反例如下：
+### (d) Sample reply
+ ```
+This proposition is incorrect. Counter-example is as follows:
 
-设 f(x) = x³，则 f'(x) = 3x²。
-在 x=0 处，f'(0) = 0，但 x=0 不是 f 的极值点（因为 f 在 0 附近严格单调递增）。
+Let f(x) = x³, then f'(x) = 3x².
+At x=0, f'(0) = 0, but x=0 is not an extreme point of f (because f is strictly monotonically increasing near 0).
 
-f'(x)=0 是极值点的必要条件而非充分条件。
-要成为极值点，f' 还需在 x₀ 两侧改变符号，
-或者满足高阶导数的条件（f''(x₀) ≠ 0）。
+f'(x)=0 is a necessary condition but not a sufficient condition for the extreme point.
+To become an extreme point, f' also needs to change signs on both sides of x₀,
+Or satisfy the conditions of higher-order derivatives (f''(x₀) ≠ 0).
 ```
 
 ---
 
-## 场景 10：用户问"这一步为什么可以这样做"
+## Scenario 10: The user asks "Why can this step be done like this?"
 
-### (a) 检测标准
-- "为什么可以交换求和与极限？"
-- "这里为什么能直接求导？"
-- "这步的依据是什么？"
-- "为什么要乘这个因子？"
+### (a) Testing standards
+- "Why can sums and limits be exchanged?"
+- "Why can the derivation be obtained directly here?"
+- "What is the basis for this step?"
+- "Why multiply by this factor?"
 
-### (b) 响应策略
-1. 给出该操作所依据的定理或定义
-2. 验证定理的条件在此处成立
-3. 如果是暗含的数学约定，解释约定的原因
+### (b) Response strategy
+1. Give the theorem or definition on which the operation is based
+2. The conditions for verifying the theorem are established here
+3. If it is an implicit mathematical convention, explain the reason for the convention.
 
-### (c) 必要验证
-- 确认引用的定理条件在当前上下文中确实满足
+### (c) Necessary verification
+- Verify that the cited theorem conditions are indeed satisfied in the current context
 
-### (d) 示例回复
-```
-这一步可以这样做是因为 **一致收敛的函数项级数的逐项求导定理**：
+### (d) Sample reply
+ ```
+This step can be done because of the term-by-term differentiation theorem for uniformly convergent functional term series:
 
-如果 ∑f_n(x) 在某区间上点态收敛于 S(x)，
-且 ∑f'_n(x) 在该区间上**一致收敛**，
-则 S'(x) = ∑f'_n(x)。
+If ∑f_n(x) converges to S(x) in a certain interval,
+And ∑f'_n(x) **consistently converges** in this interval,
+Then S'(x) = ∑f'_n(x).
 
-在这个问题中，通过 Weierstrass M 判别法可以证明 ∑f'_n(x) 在 [a,b] 上一致收敛，
-因此逐项求导是合法的。
+In this problem, it can be proved by the Weierstrass M criterion that ∑f'_n(x) converges uniformly on [a,b],
+Therefore it is legal to take the derivative term by term.
 ```
 
 ---
 
-## 场景 11：用户要求多种解法
+## Scenario 11: Users require multiple solutions
 
-### (a) 检测标准
-- "有没有其他方法？"
-- "给 2-3 种解法"
-- "用不同的方法解"
-- "多种解法"
+### (a) Testing standards
+- "Is there any other way?"
+- "Give 2-3 solutions"
+- "Solve it in different ways"
+- "Multiple solutions"
 
-### (b) 响应策略
-1. 提供 2-3 种本质上不同的解法
-2. 每种解法独立完整
-3. 对比各方法的适用场景和优缺点
-4. 确保各种方法得到一致的结果
+### (b) Response strategy
+1. Provide 2-3 essentially different solutions
+2. Each solution is independent and complete
+3. Compare the applicable scenarios, advantages and disadvantages of each method
+4. Ensure consistent results across methods
 
-### (c) 必要验证
-- 每种方法的最终答案必须一致
-- 每种方法独立验算
+### (c) Necessary verification
+- The final answer for each method must be consistent
+- Each method is independently verified
 
-### (d) 示例回复
-```
-以下是三种不同方法：
+### (d) Sample reply
+ ```
+Here are three different methods:
 
-**方法一：配方法（最直接）**
-f(x) = x² - 4x + 7 = (x-2)² + 3 ≥ 3，最小值在 x=2
+**Method 1: Preparation method (most direct)**
+f(x) = x² - 4x + 7 = (x-2)² + 3 ≥ 3, minimum at x=2
 
-**方法二：导数法（通用）**
-f'(x) = 2x - 4 = 0 → x = 2，f''(2) = 2 > 0，极小值 f(2) = 3
+**Method 2: Derivative method (general)**
+f'(x) = 2x - 4 = 0 → x = 2, f''(2) = 2 > 0, minimum value f(2) = 3
 
-**方法三：对称性＋判别式法**
-令 y = x² - 4x + 7，则 x² - 4x + (7-y) = 0
+**Method 3: Symmetry + Discriminant Method**
+Let y = x² - 4x + 7, then x² - 4x + (7-y) = 0
 Δ = 16 - 4(7-y) = 4y - 12 ≥ 0 → y ≥ 3
 
-三种方法得到一致答案：最小值为 3。
-方法一最简洁，方法二最通用，方法三常用于含参讨论。
+All three methods yield a consistent answer: the minimum value is 3.
+Method 1 is the simplest, Method 2 is the most versatile, and Method 3 is often used in discussions involving parameters.
 ```
 
 ---
 
-## 场景 12：用户要求验算
+## Scenario 12: User requests verification
 
-### (a) 检测标准
-- "帮我验算一下"
-- "检查有没有错"
-- "验证一下这个答案"
-- "确认这个结果是否正确"
+### (a) Testing standards
+- "Help me check the calculations"
+- "Check for errors"
+- "Verify this answer"
+- "Confirm whether this result is correct"
 
-### (b) 响应策略
-1. 运行完整验算引擎
-2. 逐条汇报验算结果
-3. 对每个有问题/有问题的部分详细说明
-4. 如果结果有误，指出错误并提供正确答案
+### (b) Response strategy
+1. Run the complete verification engine
+2. Report the verification results item by item
+3. Detailed explanation of each problematic/problematic part
+4. If the result is wrong, point out the error and provide the correct answer
 
-### (c) 必要验证
-- 代入验证
-- 逆运算验证
-- 边界检查
-- 量纲/单位检查
-- 特殊值测试
+### (c) Necessary verification
+- Substitute verification
+- Inverse operation verification
+- Boundary checking
+- Dimension/unit check
+- Special value testing
 
-### (d) 示例回复
-```
-## 验算报告
+### (d) Sample reply
+ ```
+## Calculation report
 
-**代入原方程验证**：
-令 x = -2：(-2)³ + (-2)² - (-2) = -8 + 4 + 2 = -2 ≠ 0 ❌
+**Substitute into the original equation to verify**:
+Let x = -2: (-2)³ + (-2)² - (-2) = -8 + 4 + 2 = -2 ≠ 0 ❌
 
-**结论**：x = -2 不是原方程的根。
+**Conclusion**: x = -2 is not a root of the original equation.
 
-**寻找正确根**：
-尝试 x = 1：1 + 1 - 1 = 1 ≠ 0 ❌
-尝试 x = -1：-1 + 1 + 1 = 1 ≠ 0 ❌
-尝试 x = 0：0 + 0 + 0 = 0 ✓
+**Find the correct root**:
+Try x = 1: 1 + 1 - 1 = 1 ≠ 0 ❌
+Try x = -1: -1 + 1 + 1 = 1 ≠ 0 ❌
+Try x = 0: 0 + 0 + 0 = 0 ✓
 
-继续因式分解得到完整解集，建议重新检查求解过程。
+Continue factoring to obtain a complete solution set, and it is recommended to recheck the solution process.
 ```
 
 ---
 
-## 场景 13：用户问"我是否漏解了？"
+## Scenario 13: User asks "Did I miss a solution?"
 
-### (a) 检测标准
-- "有没有遗漏的解？"
-- "还有其他解吗？"
-- "解全了吗？"
+### (a) Testing standards
+- "Are there any missing solutions?"
+- "Are there any other solutions?"
+- "Is everything solved?"
 
-### (b) 响应策略
-1. 完整检查定义域
-2. 检查是否存在多分支（如平方根的正负号）
-3. 检查边界情况
-4. 检查方程变换是否丢失解（如两边平方、除以可能为零的因子）
-5. 报告完整解集
+### (b) Response strategy
+1. Complete inspection of the domain
+2. Check whether there are multiple branches (such as the sign of the square root)
+3. Check boundary conditions
+4. Check whether the equation transformation loses solutions (such as squaring both sides, dividing by factors that may be zero)
+5. Report the complete solution set
 
-### (c) 必要验证
-- 对称性：如果方程对称，检查对应的对称解
-- 奇点：检查不在定义域内的值
-- 参数：如果解依赖于参数，给出参数划分
+### (c) Necessary verification
+- Symmetry: If the equation is symmetric, check the corresponding symmetric solution
+- Singularity: Check for values ​​that are not in the domain
+- Parameters: If the solution depends on parameters, give the parameter partition
 
-### (d) 示例回复
-```
-您的当前解答 x = 0, x = 5 不完整。检查后发现缺失以下解：
+### (d) Sample reply
+ ```
+Your current solution x = 0, x = 5 is incomplete. After checking, it was found that the following solutions were missing:
 
-**遗漏原因**：在第 2 步两边除以 x 时，隐含假定了 x ≠ 0，
-但实际上 x = 0 也是一个解（0² = 0 × 5）。
+**Reason for omission**: When dividing both sides by x in step 2, it is implicitly assumed that x ≠ 0,
+But in fact x = 0 is also a solution (0² = 0 × 5).
 
-**完整解集**：x = 0, x = 5
+**Complete solution set**: x = 0, x = 5
 
-**验算**：
+**Check**:
 x = 0：0² = 0，0 × 5 = 0 ✓
 x = 5：5² = 25，5 × 5 = 25 ✓
 
-建议：在除以任何可能为零的表达式之前，先单独讨论该表达式为零的情况。
+Recommendation: Before dividing by any expression that may be zero, discuss the zero case in its own right.
 ```
 
 ---
 
-## 场景 14：用户问"这个是不是无解？"
+## Scenario 14: The user asks "Is this unsolvable?"
 
-### (a) 检测标准
-- "无解吗？"
-- "有没有解？"
-- "是不是没有答案"
+### (a) Testing standards
+- "No solution?"
+- "Is there any solution?"
+- "Is there no answer?"
 
-### (b) 响应策略
-1. 彻底检查解的存在条件
-2. 分析导致无解的原因（不可满足的条件、矛盾）
-3. 如果确实无解，明确说明原因
-4. 如果实际上有解只是在特殊情形，解释具体条件
-5. 避免过早下结论
+### (b) Response strategy
+1. Thoroughly check the existence conditions of the solution
+2. Analyze the reasons leading to no solution (unsatisfiable conditions, contradictions)
+3. If there is really no solution, clearly explain the reason
+4. If there is actually a solution only in special circumstances, explain the specific conditions
+5. Avoid jumping to premature conclusions
 
-### (c) 必要验证
-- 尝试直接构造一个解
-- 尝试证明无解（反证法）
-- 检查特殊情况和边界
+### (c) Necessary verification
+- Try to construct a solution directly
+- Try to prove that there is no solution (proof by contradiction)
+- Check for special cases and boundaries
 
-### (d) 示例回复
-```
-经分析，此方程确实**无实数解**。
+### (d) Sample reply
+ ```
+After analysis, this equation indeed has no real solution.
 
-原因是：
-方程 e^x = -1 在实数范围内无解，因为 e^x > 0 对所有实数 x 成立。
+The reason is:
+The equation e^x = -1 has no solution in the real range because e^x > 0 holds for all real x.
 
-但是如果您考虑复数范围，解为：x = iπ + 2πik，k ∈ Z。
+But if you consider the complex range, the solution is: x = iπ + 2πik, k ∈ Z.
 
-请确认您需要在实数范围还是复数范围内求解。
+Please confirm whether you need to solve in the real range or the complex range.
 ```
 
 ---
 
-## 场景 15：用户提出了疑似开放问题
+## Scenario 15: User raised a suspected open question
 
-### (a) 检测标准
-- 问题与已知开放问题高度相似
-- 问题形式简单但难度极大
-- 涉及素数分布、NP vs P、Riemann 猜想等
+### (a) Testing standards
+- The issue is highly similar to a known open issue
+- Question form is simple but extremely difficult
+- Involving prime number distribution, NP vs P, Riemann conjecture, etc.
 
-### (b) 响应策略
-1. 按难题处理协议（hard_problem_protocol.md）执行
-2. 搜索确认是否为已知开放问题
-3. 如实告知问题的状态
-4. 提供已知部分结果
-5. 绝不声称已解决
+### (b) Response strategy
+1. Execute according to the problem processing protocol (hard_problem_protocol.md)
+2. Search to confirm whether it is a known open issue
+3. Tell the status of the problem truthfully
+4. Provide known partial results
+5. Never claim to have solved it
 
-### (c) 必要验证
-- 搜索已知结果
-- 确认问题分类
+### (c) Necessary verification
+- Search known results
+- Confirm problem classification
 
-### (d) 示例回复
-```
-根据目前的数学知识，这个问题与已知的 **[猜想名称]** 密切相关，
-该猜想目前尚未被数学界完全证明。
+### (d) Sample reply
+ ```
+According to current mathematical knowledge, this problem is closely related to the known **[guess name]**,
+This conjecture has not yet been fully proven by the mathematical community.
 
-**已知的事实**：
-- [事实 1]
-- [事实 2]
+**Known Facts**:
+- [Fact 1]
+- [Fact 2]
 
-**已证的特殊情形**：
-- 已证 n ≤ 10⁶ 成立
-- [其他特殊情形]
+**Certified special circumstances**:
+- It has been proved that n ≤ 10⁶
+-[Other special circumstances]
 
-如果您希望，我可以帮您探索该问题在特定条件下的部分结果，
-但需要明确说明目前不存在完整的证明。
+If you wish, I can help you explore some of the results of this problem under specific conditions,
+But it needs to be made clear that no complete proof currently exists.
 ```
 
 ---
 
-## 场景 16：用户要求联网验证
+## Scenario 16: User requires online verification
 
-### (a) 检测标准
-- "你网上查一下"
-- "帮我验证这个定理"
-- "查一下文献"
-- "找一下参考资料"
+### (a) Testing standards
+- "Check it online"
+- "Help me verify this theorem"
+- "Check the literature"
+- "Look for reference materials"
 
-### (b) 响应策略
-1. 按搜索策略（search_policy.md）执行搜索
-2. 报告搜索结果和来源
-3. 根据搜索结果验证或修正
+### (b) Response strategy
+1. Perform search by search policy (search_policy.md)
+2. Report search results and sources
+3. Verify or correct based on search results
 
-### (c) 必要验证
-- 搜索结果与自身推导的一致性
-- 来源的可信度评估
+### (c) Necessary verification
+- Consistency between search results and own derivation
+- Credibility assessment of the source
 
-### (d) 示例回复
-```
-## 搜索结果
+### (d) Sample reply
+ ```
+## Search results
 
-在 Math StackExchange 找到一个相关问题（高赞，2018 年 3 月）：
-该方法通过 Cauchy-Schwarz 不等式得到了相同结论。
+Found a related question on Math StackExchange (upvoted, March 2018):
+This method leads to the same conclusion via the Cauchy-Schwarz inequality.
 
-在 OEIS 中查找到该数列（A000045），确认了生成函数形式。
+The sequence (A000045) was found in OEIS and the generating function form was confirmed.
 
-## 验证结论
+## Verify conclusion
 
-搜索结果与我的推导一致。但需要指出：
-- StackExchange 的解答中第 3 步假设了 a₁ ≠ 0，与本题条件略有不同
-- 本题独立推导的结果不受此差异影响
+The search results are consistent with my derivation. But it needs to be pointed out:
+- Step 3 of StackExchange's answer assumes a₁ ≠ 0, which is slightly different from the conditions of this question.
+- The results of independent derivation of this question are not affected by this difference.
 ```
 
 ---
 
-## 场景 17：用户要求极其详细的推导
+## Scenario 17: User requests extremely detailed derivation
 
-### (a) 检测标准
-- "每一步都详细写出来"
-- "越详细越好"
-- "一步一步推导"
-- "不要跳步"
+### (a) Testing standards
+- "Write out every step in detail"
+- "The more detailed the better"
+- "Step-by-step derivation"
+- "Don't skip steps"
 
-### (b) 响应策略
-1. 展开每一个步骤
-2. 解释每一步的数学原因和依据
-3. 即使是"显然"的化简也要写明
-4. 中间不跳过任何代数变换
+### (b) Response strategy
+1. Expand each step
+2. Explain the mathematical reasons and basis for each step
+3. Even if it is an "obvious" simplification, it must be stated
+4. No algebraic transformations are skipped in the middle
 
-### (c) 必要验证
-- 每一步的逻辑连续性
-- 不遗漏任何中间步骤
+### (c) Necessary verification
+- Logical continuity at every step
+- No intermediate steps are missed
 
-### (d) 示例回复
-```
-## 详细推导
+### (d) Sample reply
+ ```
+## Detailed derivation
 
-**第 1 步**：写出原表达式
+**Step 1**: Write the original expression
 f(x) = x² - 4x + 7
 
-**第 2 步**：提取 x² 的系数，准备配方
-观察前两项 x² - 4x，加入并减去 (4/2)² = 4：
+**Step 2**: Extract the coefficients of x² and prepare the recipe
+Observing the first two terms x² - 4x, add and subtract (4/2)² = 4:
 f(x) = x² - 4x + 4 - 4 + 7
       = (x² - 4x + 4) + (7 - 4)
 
-**第 3 步**：将前三项写成完全平方
-因为 x² - 4x + 4 = (x - 2)²（验证：(x-2)² = x² - 4x + 4）
-所以：f(x) = (x - 2)² + 3
+**Step 3**: Write the first three terms as perfect squares
+Because x² - 4x + 4 = (x - 2)² (verify: (x-2)² = x² - 4x + 4)
+So: f(x) = (x - 2)² + 3
 
-**第 4 步**：利用平方的非负性
-因为任何实数的平方 ≥ 0，即 (x-2)² ≥ 0，
-所以 f(x) = (x-2)² + 3 ≥ 0 + 3 = 3
+**Step 4**: Exploit the non-negativity of squares
+Since the square of any real number ≥ 0, that is (x-2)² ≥ 0,
+So f(x) = (x-2)² + 3 ≥ 0 + 3 = 3
 
-**第 5 步**：确定最小值及取最值的条件
-当 (x-2)² = 0，即 x - 2 = 0，x = 2 时取等号。
-所以最小值 f(2) = 3。
+**Step 5**: Determine the minimum value and the conditions for taking the maximum value
+Take the equal sign when (x-2)² = 0, i.e. x - 2 = 0, x = 2.
+So the minimum value f(2) = 3.
 ```
 
 ---
 
-## 场景 18：用户要求严格证明
+## Scenario 18: User requires strict certification
 
-### (a) 检测标准
-- "严密的证明"
-- "严格的数学证明"
+### (a) Testing standards
+- "Strong proof"
+- "Strict mathematical proof"
 - "rigorous proof"
-- "不要直观解释，要证明"
-
-### (b) 响应策略
-1. 使用形式化的数学语言
-2. 完整写出所有量化词（∀、∃）
-3. 每个逻辑步骤明确标注依据
-4. 区分"已知条件"、"中间结论"和"最终结论"
-5. 包含所有的前提验证
-
-### (c) 必要验证
-- 量化词顺序和逻辑结构的正确性
-- 每个定理条件是否被满足
-- 反例检查（尝试寻找逻辑漏洞）
-
-### (d) 示例回复
-```
-## 命题
-设 f: ℝ → ℝ 满足：∀x∈ℝ, f(x+1) = f(x)。若 f 在 [0,1] 上可积，则 f 在任意有限区间 [a,b] 上可积。
-
-## 证明
-
-**已知**：
-(1) f 的周期为 1：∀x∈ℝ, f(x+1) = f(x)
-(2) f 在 [0,1] 上 Riemann 可积
-
-**推导**：
-步骤 1（划分）：将 [a,b] 划分为整数个完整周期和两个部分周期。
-设 n = ⌊a⌋，m = ⌈b⌉，则 [a,b] = [a, n+1] ∪ [n+1, m] ∪ [m, b]
-
-步骤 2（完整周期的可积性）：
-由 (2) 知 f 在 [0,1] 可积，由 (1) 周期平移知 f 在任意 [k, k+1] 可积。
-因此 f 在 [n+1, m] 上可积（有限个可积区间的并可积）。
-
-步骤 3（部分周期）：
-[a, n+1] ⊂ [n, n+1]，闭区间上可积函数的限制在子区间上可积。
-因此 f 在 [a, n+1] 上可积。
-同理 f 在 [m, b] 上可积。
-
-步骤 4（整体可积性）：
-f 在 [a,b] = [a, n+1] ∪ [n+1, m] ∪ [m, b] 的每个子区间可积，
-三个可积区间的并即 [a,b]，因此 f 在 [a,b] 上可积。∎
+- "Don't explain intuitively, prove it"
+
+### (b) Response strategy
+1. Use formal mathematical language
+2. Completely write out all quantification words (∀, ∃)
+3. Clearly label the basis for each logical step
+4. Distinguish between "known conditions", "intermediate conclusions" and "final conclusions"
+5. Includes all prerequisite verifications
+
+### (c) Necessary verification
+- The correctness of the order and logical structure of quantifiers
+- Whether the conditions of each theorem are satisfied
+- Counterexample checking (try to find logical holes)
+
+### (d) Sample reply
+ ```
+## Proposition
+Let f: ℝ → ℝ satisfy: ∀x∈ℝ, f(x+1) = f(x). If f is integrable on [0,1], then f is integrable on any finite interval [a,b].
+
+## prove
+
+**Known**:
+(1) The period of f is 1: ∀x∈ℝ, f(x+1) = f(x)
+(2) f is Riemann integrable on [0,1]
+
+**Derivation**:
+Step 1 (Partition): Divide [a,b] into an integer number of complete cycles and two partial cycles.
+Assume n = ⌊a⌋, m = ⌈b⌉, then [a,b] = [a, n+1] ∪ [n+1, m] ∪ [m, b]
+
+Step 2 (integrability of complete periods):
+From (2), we know that f is integrable in [0,1], and from (1) periodic translation, we know that f is integrable at any [k, k+1].
+Therefore f is integrable on [n+1, m] (the union of a finite number of integrable intervals is integrable).
+
+Step 3 (partial cycle):
+[a, n+1] ⊂ [n, n+1], the restriction of integrable functions on closed intervals is integrable on subintervals.
+Therefore f is integrable on [a, n+1].
+In the same way, f is integrable on [m, b].
+
+Step 4 (global integrability):
+f is integrable in every subinterval of [a,b] = [a, n+1] ∪ [n+1, m] ∪ [m, b],
+The union of three integrable intervals is [a,b], so f is integrable on [a,b]. ∎
 ```
diff --git a/modules/output_templates.md b/modules/output_templates.md
index 4a1fafc..edb3120 100644
--- a/modules/output_templates.md
+++ b/modules/output_templates.md
@@ -1,241 +1,241 @@
-# Math.skill 输出模板 (Output Templates)
+# Math.skill Output Templates
 
-以下定义 6 种输出模板，每种包含：适用场景、标准结构、完整示例、特殊注意事项。
+The following defines 6 output templates, each including: applicable scenarios, standard structure, complete examples, and special considerations.
 
 ---
 
-## 模板 A：标准解答 (Standard Solution)
+## Template A: Standard Solution
 
-### 何时使用
-- 常规数学问题的解答
-- 包含计算、推导、应用定理的标准题型
-- 用户未指定特殊输出格式时，默认使用此模板
+### When to use
+- Answers to general math problems
+- Standard question types including calculation, derivation, and application of theorems
+- This template is used by default when the user does not specify a special output format.
 
-### 标准结构
+### Standard structure
 
-```markdown
-## 题意解析
-[用数学语言重新表述问题，明确已知条件和求解目标]
+ ```markdown
+## Question meaning analysis
+[Reformulate the problem in mathematical language and clarify the known conditions and solution goals]
 
-## 方法选择
-[解释选择该方法的理由，与其他可能方法的对比]
+## Method selection
+[Explain the rationale for choosing this method and compare it with other possible methods]
 
-## 解题过程
-[完整的逐步推导]
+## Problem solving process
+[Full step-by-step derivation]
 
-## 验算
-[代入验证 / 逆运算验证 / 数值近似验证]
+## Check calculation
+[Substitution Verification/Inverse Operation Verification/Numerical Approximation Verification]
 
-## 最终答案
-[明确标注最终结果的格式]
+## Final answer
+[Clearly mark the format of the final result]
 
-## 易错点
-[列出常见的错误思路和避免方法]
+## Easy to make mistakes
+[List common mistakes and how to avoid them]
 ```
 
-### 完整示例
+### Complete example
 
-```markdown
-## 题意解析
-已知函数 f(x) = x³ - 3x² + 2，求 f 在 [-1, 3] 上的最大值和最小值。
+ ```markdown
+## Question meaning analysis
+Given the function f(x) = x³ - 3x² + 2, find the maximum and minimum values ​​of f on [-1, 3].
 
-## 方法选择
-一元闭区间连续函数的最值问题：先求驻点，再结合端点比较。
-选择求导法是因为 f 在 [-1,3] 上处处可导，临界点可以通过 f'(x) = 0 找到。
+## Method selection
+The optimal value problem of a one-variable closed interval continuous function: first find the stationary point, and then compare it with the endpoints.
+The derivation method was chosen because f is differentiable everywhere on [-1,3] and the critical point can be found by f'(x) = 0.
 
-## 解题过程
-**步骤 1**：求导
+## Problem solving process
+**Step 1**: Derivative
 f'(x) = 3x² - 6x = 3x(x - 2)
 
-**步骤 2**：求驻点
-令 f'(x) = 0，解得 x = 0 或 x = 2
-两个驻点均在区间 [-1, 3] 内。
+**Step 2**: Find the stationary point
+Let f'(x) = 0, solve for x = 0 or x = 2
+Both stationary points are in the interval [-1, 3].
 
-**步骤 3**：计算函数值
+**Step 3**: Calculate the function value
 f(-1) = (-1)³ - 3(-1)² + 2 = -1 - 3 + 2 = -2
 f(0) = 0³ - 3·0² + 2 = 2
 f(2) = 2³ - 3·2² + 2 = 8 - 12 + 2 = -2
 f(3) = 3³ - 3·3² + 2 = 27 - 27 + 2 = 2
 
-**步骤 4**：比较得出最值
-最大值 = max{-2, 2, -2, 2} = 2
-最小值 = min{-2, 2, -2, 2} = -2
+**Step 4**: Compare to get the best value
+maximum value = max{-2, 2, -2, 2} = 2
+Minimum value = min{-2, 2, -2, 2} = -2
 
-## 验算
+## Check calculation
 - f(-1) = -2：(-1)³ = -1, -3(-1)² = -3, -1-3+2 = -2 ✓
 - f(0) = 2：0-0+2 = 2 ✓
 - f(2) = -2：8-12+2 = -2 ✓
 - f(3) = 2：27-27+2 = 2 ✓
-- 端点 -1 和 3 已检查
-- 驻点处 f''(0) = -6 < 0（极大值点），f''(2) = 6 > 0（极小值点），与计算结果一致
+- Endpoints -1 and 3 checked
+- At the stationary point, f''(0) = -6 < 0 (maximum point), f''(2) = 6 > 0 (minimum point), consistent with the calculation results
 
-## 最终答案
-最大值为 2，在 x = 0 和 x = 3 处取得。
-最小值为 -2，在 x = -1 和 x = 2 处取得。
+## Final answer
+The maximum value is 2, taken at x = 0 and x = 3.
+The minimum value is -2, taken at x = -1 and x = 2.
 
-## 易错点
-1. 忘记检查端点值——闭区间最值可能在端点取到
-2. 漏掉 f(3) 等于 f(0)，导致认为最大值只在 x=0 取到
-3. 将 f'(x) = 0 的解直接当作最值点，未比较大小
+## Easy to make mistakes
+1. Forget to check the endpoint value - the maximum value of a closed interval may be taken at the endpoint
+2. Missing out that f(3) is equal to f(0) leads to the belief that the maximum value can only be obtained at x=0
+3. The solution of f'(x) = 0 is directly regarded as the maximum point without comparing the size.
 ```
 
-### 特殊注意事项
-- "题意解析"部分避免直接复制用户原话，应展示对问题的理解
-- "验算"部分至少包含一种验证方式
-- "易错点"应基于真实常见错误，不要虚构
+### Special Notes
+- In the "Analysis of Question Meaning" part, avoid directly copying the user's original words, and should show your understanding of the question.
+- The "Verification" section contains at least one verification method
+- "Error prone points" should be based on real common mistakes, not fictional ones
 
 ---
 
-## 模板 B：仅答案 (Answer Only)
+## Template B: Answer Only
 
-### 何时使用
-- 用户明确要求"只要答案"
-- 用户说"直接给结果"、"不需要过程"
-- 用户在对话中已经展示过完整思路，只需最终确认
+### When to use
+- User explicitly asked for "just the answer"
+- Users say "give results directly" and "no process required"
+- The user has already shown the complete idea in the conversation and only needs final confirmation
 
-### 标准结构
+### Standard structure
 
-```markdown
-答案是：[公式或数值]
+ ```markdown
+The answer is: [formula or numerical value]
 
-简单验算：[一行验证]
+Simple verification: [One line verification]
 ```
 
-### 完整示例
+### Complete example
 
-```markdown
-答案是：∬_D (x² + y²) dxdy = 8π/3
+ ```markdown
+The answer is: ∬_D (x² + y²) dxdy = 8π/3
 
-简单验算：利用极坐标，∫₀²∫₀²ᵖⁱ r²·r dθ dr = 2π·[r⁴/4]₀² = 2π·4 = 8π ✓
+Simple calculation: using polar coordinates, ∫₀²∫₀²ᵖⁱ r²·r dθ dr = 2π·[r⁴/4]₀² = 2π·4 = 8π ✓
 ```
 
-### 特殊注意事项
-- 验算行必须存在，保证答案的可信度
-- 如果问题有多解，必须全部列出
-- 如果是近似值，需注明精度
+### Special Notes
+- The verification row must exist to ensure the credibility of the answer
+- If the problem has multiple solutions, all must be listed
+- If it is an approximate value, please indicate the accuracy
 
 ---
 
-## 模板 C：证明 (Proof)
+## Template C: Proof
 
-### 何时使用
-- 用户要求证明某个命题
-- "请证明……"、"求证……"
-- 需要展示逻辑推导链的数学论证
+### When to use
+- The user asks to prove a certain proposition
+- "Please prove...", "Please prove..."
+- Mathematical arguments that need to show logical chains of derivation
 
-### 标准结构
+### Standard structure
 
-```markdown
-## 命题
-[被证明命题的精确陈述]
+ ```markdown
+## Proposition
+[precise statement of the proposition being proved]
 
-## 证明思路
-[证明的整体逻辑结构、关键引理、核心技巧]
+## Proof idea
+[The overall logical structure of the proof, key lemmas, and core techniques]
 
-## 证明
-[完整的逻辑推演，包含所有中间步骤和引用]
+## prove
+[Complete logical deduction, including all intermediate steps and references]
 
-## 反例检查与边界情况
-[检查命题的各种边界情形，验证是否有反例]
+## Counterexample checking and edge cases
+[Check various boundary cases of the proposition and verify whether there are counterexamples]
 
-## 结论
-[命题证明完成的总结]
+## in conclusion
+[Summary of completion of proposition proof]
 ```
 
-### 完整示例
+### Complete example
 
-```markdown
-## 命题
-对于任意正整数 n，证明：n³ - n 能被 6 整除。
+ ```markdown
+## Proposition
+For any positive integer n, prove that: n³ - n is divisible by 6.
 
-## 证明思路
-核心是对 n³ - n = n(n-1)(n+1) 进行因式分解，注意到这是三个连续整数的乘积。
-三个连续整数中必有一个是 2 的倍数，必有一个是 3 的倍数，
-因此乘积必有因子 2 × 3 = 6。
+## Proof idea
+The core is to factor n³ - n = n(n-1)(n+1), noting that this is the product of three consecutive integers.
+One of the three consecutive integers must be a multiple of 2, and one must be a multiple of 3.
+Therefore the product must have factors 2 × 3 = 6.
 
-## 证明
-**步骤 1**：因式分解
+## prove
+**Step 1**: Factorization
 n³ - n = n(n² - 1) = n(n-1)(n+1)
 
-**步骤 2**：三个连续整数的性质
-n-1, n, n+1 是三个连续整数，因此：
-- 其中至少有一个偶数（被 2 整除）
-- 其中恰好有一个被 3 整除
+**Step 2**: Properties of three consecutive integers
+n-1, n, n+1 are three consecutive integers, therefore:
+- There is at least one even number (divisible by 2)
+- Exactly one of them is divisible by 3
 
-**步骤 3**：组合因子
-因为乘积同时含有因子 2 和因子 3，且 2 和 3 互质，
-所以 n³ - n 含有因子 6 = 2 × 3。
+**Step 3**: Combining Factors
+Since the product contains both factors 2 and 3, and 2 and 3 are relatively prime,
+So n³ - n contains the factor 6 = 2 × 3.
 
-即：6 | (n³ - n)，对所有 n ∈ ℕ 成立。
+That is: 6 | (n³ - n), true for all n ∈ ℕ.
 
-## 反例检查与边界情况
-- n = 1：1³ - 1 = 0，0 可被 6 整除 ✓
-- n = 2：8 - 2 = 6，可被 6 整除 ✓
-- n = 3：27 - 3 = 24 = 6×4，可被 6 整除 ✓
-- n = 0：如果 n=0 被认为在正整数范围内（此处不包括），0-0=0 可被 6 整除
+## Counterexample checking and edge cases
+- n = 1: 1³ - 1 = 0, 0 is divisible by 6 ✓
+- n = 2: 8 - 2 = 6, divisible by 6 ✓
+- n = 3: 27 - 3 = 24 = 6×4, divisible by 6 ✓
+- n = 0: 0-0=0 is divisible by 6 if n=0 is considered to be in the range of positive integers (not included here)
 
-无边界反例。
+Unbounded counterexamples.
 
-## 结论
-命题得证：对任意正整数 n，6 | (n³ - n)。∎
+## in conclusion
+Proof of the proposition: For any positive integer n, 6 | (n³ - n). ∎
 ```
 
-### 特殊注意事项
-- 命题陈述必须精确，量化词完整
-- "反例检查与边界"部分必须存在，即使没有反例
-- 使用适当的结束标记（如 ∎ 或 Q.E.D.）
+### Special Notes
+- The proposition statement must be precise and the quantification words must be complete
+- The "Counterexample Checking and Bounds" section must be present even if there is no counterexample
+- Use an appropriate closing tag (such as ∎ or Q.E.D.)
 
 ---
 
-## 模板 D：解答检查 (Solution Checking)
+## Template D: Solution Checking
 
-### 何时使用
-- 用户提供了自己的解答，要求检查是否正确
-- "我这样算对吗？"
-- "看看错在哪里"
-- 需要指出并修正错误
+### When to use
+- Users provided their own answers and asked to check if they are correct
+- "Am I right?"
+- "See what's wrong"
+- Errors need to be pointed out and corrected
 
-### 标准结构
+### Standard structure
 
-```markdown
-## 检查结论
-[总体判断：正确 / 有误 / 部分正确]
+ ```markdown
+## Check conclusion
+[Overall judgment: correct / incorrect / partially correct]
 
-## 问题定位
-[如果有错误，精确指出错误步骤和错误类型]
-[如果正确，确认每一步的合理性]
+## Problem location
+[If there is an error, identify the exact step and type of error]
+[If correct, confirm the rationality of each step]
 
-## 正确做法
-[展示正确推导（如果是部分步骤修正，只需改正错误部分）]
+## Correct approach
+[Show correct derivation (if it is a partial step correction, only correct the wrong part)]
 
-## 验算
-[对正确解答进行独立验证]
+## Check calculation
+[Independent verification of correct answer]
 
-## 最终答案
-[如果是错的，给出正确答案；如果是正确的，确认用户答案]
+## Final answer
+[If it is wrong, give the correct answer; if it is correct, confirm the user's answer]
 ```
 
-### 完整示例
+### Complete example
 
-```markdown
-## 检查结论
-您的解答有误。前两步正确，但在第 3 步出现了关键错误。
+ ```markdown
+## Check conclusion
+Your answer is wrong. The first two steps are correct, but in step 3 a critical error occurs.
 
-## 问题定位
-**第 3 步**：您写了 √(x²) = x
-这是错误的。正确的关系是 √(x²) = |x|。
+## Problem location
+**Step 3**: You write √(x²) = x
+This is wrong. The correct relationship is √(x²) = |x|.
 
-因为平方根函数返回的是非负值，而 x 可能为负。
+Because the square root function returns a nonnegative value, x may be negative.
 
-在这个问题中，积分区间是 [-2, 1]，x 在此区间内可取负值，
-所以 √(x²) 不能简单等于 x。
+In this problem, the integration interval is [-2, 1], and x can take negative values ​​within this interval,
+So √(x²) cannot simply be equal to x.
 
-## 正确做法
-**修正第 3 步起**：
-将积分区间拆分为 [-2, 0] 和 [0, 1]：
+## Correct approach
+**Correction from step 3**:
+Split the integration interval into [-2, 0] and [0, 1]:
 
-在 [-2, 0] 上：√(x²) = |x| = -x（因为 x ≤ 0）
-在 [0, 1] 上：√(x²) = |x| = x（因为 x ≥ 0）
+On [-2, 0]: √(x²) = |x| = -x (since x ≤ 0)
+On [0, 1]: √(x²) = |x| = x (since x ≥ 0)
 
 ∫_{-2}^{1} |x| dx = ∫_{-2}^{0} (-x) dx + ∫_{0}^{1} x dx
 = [-x²/2]_{-2}^{0} + [x²/2]_{0}^{1}
@@ -243,365 +243,365 @@ n-1, n, n+1 是三个连续整数，因此：
 = 2 + 0.5
 = 2.5
 
-## 验算
-几何验证：|x| 在 [-2,0] 下的面积为三角形面积 = 2×2/2 = 2
-|x| 在 [0,1] 下的面积为三角形面积 = 1×1/2 = 0.5
-总面积 = 2 + 0.5 = 2.5 ✓
+## Check calculation
+Geometric verification: The area of ​​|x| under [-2,0] is the area of ​​the triangle = 2×2/2 = 2
+The area of ​​|x| under [0,1] is the area of ​​the triangle = 1×1/2 = 0.5
+Total area = 2 + 0.5 = 2.5 ✓
 
-## 最终答案
-正确答案是 2.5，而不是您得到的 1.5。
+## Final answer
+The correct answer is 2.5, not the 1.5 you got.
 
-您的错误主要在于忽略了平方根函数的非负性约定。
+Your mistake mainly lies in ignoring the non-negativity convention of the square root function.
 ```
 
-### 特殊注意事项
-- 错误定位要具体到行号或步骤编号
-- 指出错误时保持建设性，不说"你太笨了"之类的话
-- 如果解答完全正确，只需确认和给出验算即可
+### Special Notes
+- The error location should be specific to the line number or step number
+- Be constructive when pointing out mistakes and don't say things like "You're so stupid"
+- If the answer is completely correct, just confirm and give the verification
 
 ---
 
-## 模板 E：高等数学解答 (Higher Math)
+## Template E: Higher Math Solutions (Higher Math)
 
-### 何时使用
-- 问题需要使用大学及以上水平的数学工具
-- 涉及实分析、抽象代数、拓扑等高级课程内容
-- 用户明确要求用高等数学方法
+### When to use
+- Questions require the use of college level or above mathematical tools
+- Involving advanced course content such as real analysis, abstract algebra, topology, etc.
+- Users explicitly request advanced mathematical methods
 
-### 标准结构
+### Standard structure
 
-```markdown
-## 问题类型
-[领域分类：实分析 / 抽象代数 / 拓扑 / 高等概率...]
+ ```markdown
+## Question type
+[Field classification: real analysis / abstract algebra / topology / advanced probability...]
 
-## 条件与定义域
-[问题中所有条件的形式化表述，定义域的详细界定]
+## Conditions and domains
+[Formal expression of all conditions in the problem, detailed definition of the domain]
 
-## 使用的定理及适用条件
-[列出的定理名称、陈述、以及在本题中条件满足的验证]
+## Theorems used and applicable conditions
+[Listed theorem names, statements, and verification that the conditions in this question are met]
 
-## 推导
-[基于所选定理的严格推导过程]
+## Derivation
+[Strict derivation process based on selected theorem]
 
-## 验算与一致性检查
-[多种方式交叉验证结果，检查与其他已知结论的一致性]
+## Calculation and consistency check
+[Cross-validate results in multiple ways to check consistency with other known conclusions]
 
-## 最终结论
-[结论的形式化陈述]
+## Final conclusion
+[Formal statement of conclusion]
 ```
 
-### 完整示例
+### Complete example
 
-```markdown
-## 问题类型
-实分析——一致收敛与逐点收敛的关系
+ ```markdown
+## Question type
+Real analysis - the relationship between uniform convergence and point-wise convergence
 
-## 条件与定义域
-设有函数序列 f_n(x) = x^n，定义域 x ∈ [0, 1]。
-讨论 f_n 在 [0, 1] 上是否一致收敛。
+## Conditions and domains
+Let the function sequence f_n(x) = x^n and the domain x ∈ [0, 1].
+Discuss whether f_n converges uniformly on [0, 1].
 
-## 使用的定理及适用条件
-**采用的判定方法**：一致收敛的充要条件是 sup|f_n(x) - f(x)| → 0
-其中 f(x) 是逐点极限。
+## Theorems used and applicable conditions
+**Judgment method adopted**: The necessary and sufficient condition for consistent convergence is sup|f_n(x) - f(x)| → 0
+where f(x) is the pointwise limit.
 
-**逐点极限**：
-对任意 x ∈ [0, 1)：
+**Pointwise Limit**:
+For any x ∈ [0, 1):
   lim_{n→∞} x^n = 0
-对 x = 1：
+For x = 1:
   lim_{n→∞} 1^n = 1
 
-所以 f(x) = { 0, x ∈ [0, 1); 1, x = 1 }
+So f(x) = { 0, x ∈ [0, 1); 1, x = 1 }
 
-## 推导
-计算上确界：
-对任意 n，sup_{x∈[0,1]} |f_n(x) - f(x)|
+## Derivation
+Compute the exact bound:
+For any n, sup_{x∈[0,1]} |f_n(x) - f(x)|
 
-当 x ∈ [0, 1)：|f_n(x) - 0| = x^n
-当 x = 1：|f_n(1) - 1| = |1 - 1| = 0
+When x ∈ [0, 1): |f_n(x) - 0| = x^n
+When x = 1: |f_n(1) - 1| = |1 - 1| = 0
 
-在 [0, 1) 上，x^n 的上确界 = 1（当 x → 1⁻ 时），
-因此 sup |f_n(x) - f(x)| = 1
+On [0, 1), the supremum of x^n = 1 (when x → 1⁻),
+Therefore sup |f_n(x) - f(x)| = 1
 
-由于 lim_{n→∞} 1 = 1 ≠ 0，
-根据充要条件，f_n 在 [0,1] 上**不一致收敛**。
+Since lim_{n→∞} 1 = 1 ≠ 0,
+According to necessary and sufficient conditions, f_n **inconsistently converges** on [0,1].
 
-## 验算与一致性检查
-1. **连续性检验**：每个 f_n(x) = x^n 在 [0,1] 上连续。
-   若一致收敛，极限函数也应连续。
-   但 f(x) 在 x=1 处不连续（极限为 0，函数值为 1）。
-   这反向证实了不一致收敛。——结论自洽 ✓
+## Calculation and consistency check
+1. **Continuity Test**: Every f_n(x) = x^n is continuous on [0,1].
+If there is uniform convergence, the limit function should also be continuous.
+But f(x) is discontinuous at x=1 (the limit is 0 and the function value is 1).
+This inversely confirms inconsistent convergence. ——The conclusion is self-consistent ✓
 
-2. **紧集上的一致收敛**：如果缩小定义域到 [0, c]（c < 1），
-   则 sup x^n = c^n → 0（n → ∞），此时一致收敛。
-   这与理论完全一致 ✓
+2. **Uniform convergence on compact sets**: If the domain is reduced to [0, c] (c < 1),
+Then sup x^n = c^n → 0 (n → ∞), which converges uniformly.
+This is completely consistent with the theory ✓
 
-## 最终结论
-函数序列 f_n(x) = x^n 在 [0, 1] 上逐点收敛但不一致收敛。
-极限函数在 x = 1 处有跳跃间断点，破坏了连续性。
+## Final conclusion
+The sequence of functions f_n(x) = x^n converges pointwise but inconsistently on [0, 1].
+The limit function has a jump discontinuity at x = 1, breaking continuity.
 ```
 
-### 特殊注意事项
-- 定理的条件验证必须逐条完整列出，不可遗漏
-- 需要区分概念（如点态与一致收敛、紧致与列紧）
-- 一致性检查是高等数学解答中不可或缺的环节
+### Special Notes
+- The conditional verification of the theorem must be listed in full one by one and cannot be omitted.
+- Concepts need to be distinguished (such as point state and consistent convergence, compactness and column compaction)
+- Consistency checking is an indispensable part of advanced mathematics solutions
 
 ---
 
-## 模板 F：研究/开放问题 (Research/Open Problem)
+## Template F: Research/Open Problem (Research/Open Problem)
 
-### 何时使用
-- 问题被确认为已知开放问题
-- 问题属于研究级别，没有已知的标准解答
-- 用户的数学水平可能不足以判断问题的难度时，需要诚实说明
+### When to use
+- The issue is confirmed as a known open issue
+- Question is research level and has no known standard answer
+- When the user’s mathematical level may not be sufficient to judge the difficulty of the problem, an honest explanation is required
 
-### 标准结构
+### Standard structure
 
-```markdown
-## 初步判断
-[该问题在当前数学知识体系中的地位]
+ ```markdown
+## Preliminary judgment
+[The status of this problem in the current mathematical knowledge system]
 
-## 已知信息与可能相关理论
-[与该问题相关的已知结论、定理和理论框架]
+## Known information and possible related theories
+[Known conclusions, theorems and theoretical frameworks related to this problem]
 
-## 搜索或查证结果
-[搜索该问题的结果摘要，是否找到类似讨论、相关文献]
+## Search or check results
+[Search the summary of results for this issue and see if similar discussions and related literature are found]
 
-## 从第一原则出发的分析
-[从基本定义出发的分析，展示可做出的严格推导]
+## Analysis from first principles
+[Analysis starting from basic definitions, showing rigorous derivations that can be made]
 
-## 已验证的部分结论
-[在特定条件下确实可证的结论]
+## Some verified conclusions
+[A conclusion that is truly verifiable under specific conditions]
 
-## 反例搜索
-[数值搜索的结果，发现/未发现反例]
+## Counterexample search
+[Results of numerical search, found/not found counterexamples]
 
-## 当前障碍
-[阻碍完整证明的关键步骤，为何现有工具不足]
+## Current obstacle
+[Critical steps that hinder complete proofs, why existing tools are insufficient]
 
-## 可继续推进的方向
-[建议的下一步研究方向]
+## Directions that can be continued
+[Suggested next research directions]
 
-## 结论状态
-[明确该问题的当前状态：开放/部分解决/特定条件下已解决]
+## Conclusion status
+[Specify the current status of this issue: Open/Partially Resolved/Resolved under certain conditions]
 ```
 
-### 完整示例
-
-```markdown
-## 初步判断
-此问题与 **Collatz 猜想（3n+1 问题）** 本质上等价。
-该问题自 1937 年提出以来，至今未被证明或证否。
-目前已有数值验证到约 2⁶⁸，未发现反例。
-
-## 已知信息与可能相关理论
-- Jeffery Lagarias (2010) 的综述《The Ultimate Challenge: The 3x+1 Problem》
-- Terence Tao (2019) 证明了对数密度趋于 0 的结果
-- 已有结论：几乎所有数经过 Collatz 迭代最终小于原数
-
-## 搜索或查证结果
-在 MathOverflow 上找到相关讨论（高赞回答 by Tao 等）：
-确认该问题中提出的特例未超出 Collatz 猜想的范围，
-目前无已知的完整证明。
-
-## 从第一原则出发的分析
-对您提出的特定形式，可以建立与标准 Collatz 迭代的等价映射：
-设 T(n) = n/2 (n 偶), 3n+1 (n 奇)，则……
-[具体的等价推导]
-
-这表明您的问题和 Collatz 猜想同时为真或同时为假。
-
-## 已验证的部分结论
-对于 n ≤ 10⁶ 范围内的数：
-- 经过有限步迭代后进入 4→2→1 循环
-- 未发现循环长度 > 10⁴ 的其他循环
-使用 Python 验证（需约为 5 秒）可复现此结果。
-
-## 反例搜索
-在 n ∈ [1, 10⁶] 范围内未发现反例。
-此结果与已知的大规模数值验证结果一致。
-
-## 当前障碍
-要完成一般性证明，需要处理 Collatz 迭代中的任意长轨道问题，
-目前没有任何已知的数学工具可以完全解决这一问题。
-
-## 可继续推进的方向
-1. 研究该问题的统计行为（如 Tao 所做的工作）
-2. 探索与 p-进数分析或其他数论工具的联系
-3. 研究特定子集的枚举性质
-4. 如果您对此问题有新的视角，建议查阅 Lagarias 的综述以了解已知进展
-
-## 结论状态
-**该问题目前不存在被数学界公认的完整证明。**
-以上分析为已有知识的整理和特例的数值实验，
-不代表对原问题的证明。
+### Complete example
+
+ ```markdown
+## Preliminary judgment
+This problem is essentially equivalent to **Collatz Conjecture (3n+1 problem)**.
+This question has not been proven or disproven since it was raised in 1937.
+At present, it has been numerically verified to about 2⁶⁸, and no counterexamples have been found.
+
+## Known information and possible related theories
+- Review of "The Ultimate Challenge: The 3x+1 Problem" by Jeffery Lagarias (2010)
+- Terence Tao (2019) proved that the logarithmic density tends to 0
+- It has been concluded that almost all numbers are ultimately smaller than the original number after Collatz iteration
+
+## Search or check results
+Find related discussions on MathOverflow (highly praised answers by Tao, etc.):
+Confirm that the special case raised in this question is within the scope of Collatz's conjecture,
+No complete proof is known.
+
+## Analysis from first principles
+For the specific form you propose, an equivalent mapping to standard Collatz iterations can be established:
+Assume T(n) = n/2 (n is even), 3n+1 (n is odd), then...
+[Specific equivalent derivation]
+
+This shows that both your question and Collatz's conjecture are either true or false at the same time.
+
+## Some verified conclusions
+For numbers in the range n ≤ 10⁶:
+- After finite step iteration, enter the 4→2→1 loop
+- No other loops found with loop length > 10⁴
+This result can be reproduced using Python verification (takes ~5 seconds).
+
+## Counterexample search
+No counterexamples were found in the range n ∈ [1, 10⁶].
+This result is consistent with known large-scale numerical verification results.
+
+## Current obstacle
+To complete the general proof, one needs to deal with the problem of arbitrarily long orbits in Collatz iterations,
+There are currently no known mathematical tools that can completely solve this problem.
+
+## Directions that can be continued
+1. Study the statistical behavior of the problem (as done by Tao)
+2. Explore connections to p-radical number analysis or other number theory tools
+3. Study the enumeration properties of specific subsets
+4. If you have a new perspective on this issue, it is recommended to check out Lagarias’ review to understand the known progress
+
+## Conclusion status
+**There is currently no complete proof of this problem recognized by the mathematical community. **
+The above analysis is a compilation of existing knowledge and numerical experiments of special cases.
+It does not represent a proof of the original question.
 ```
 
-### 特殊注意事项
-- 必须明确标注"目前不存在完整证明"
-- 区分"已验证事实"和"推测"
-- 提供给用户的后续建议应具体可操作
-- 引用文献时必须有具体可查的来源（不要虚构）
+### Special Notes
+- Must be clearly marked "A complete certificate does not currently exist"
+- Distinguish between "verified facts" and "speculations"
+- Follow-up suggestions provided to users should be specific and actionable
+- When citing documents, there must be specific and verifiable sources (no fictitious ones)
 
 ---
 
-## 模板 G：研究级问题深度分析 (Research-Level Deep Analysis)
+## Template G: Research-Level Deep Analysis
 
-### 何时使用
-- 用户明确提出研究级别的数学问题
-- 问题涉及多个数学分支的交叉
-- 需要展示深入的理论分析和多种可能的解决路径
-- 用户具备研究生或以上数学背景
+### When to use
+- Users explicitly ask research-level mathematical questions
+- Questions involving the intersection of multiple branches of mathematics
+- Need to demonstrate in-depth theoretical analysis and multiple possible solution paths
+- Users have a graduate degree or above in mathematics background
 
-### 标准结构
+### Standard structure
 
-```markdown
-## 问题分类与定位
-[问题所属的数学领域，与已知问题的关联]
+ ```markdown
+## Problem classification and positioning
+[The mathematical field to which the problem belongs, and its relationship to known problems]
 
-## 核心难点分析
-[问题的本质困难在哪里，为何标准方法失效]
+## Analysis of core difficulties
+[What is the essential difficulty of the problem and why the standard method fails]
 
-## 相关理论框架
-[可能需要用到的深层理论和工具]
+## Related theoretical framework
+[In-depth theories and tools that may be needed]
 
-## 已知结果综述
-[与该问题最接近的已知定理和结论]
+## Summary of known results
+[The closest known theorem and conclusion to the problem]
 
-## 尝试路径一：[方法名称]
-[第一种解决思路的详细推导]
-### 进展
-[该方法能走多远]
-### 障碍
-[遇到的具体困难]
+## Try path one: [method name]
+[Detailed derivation of the first solution]
+### Progress
+[How far can this approach go]
+### Obstacles
+[Specific difficulties encountered]
 
-## 尝试路径二：[方法名称]
-[第二种解决思路的详细推导]
-### 进展
-### 障碍
+## Try path two: [method name]
+[Detailed derivation of the second solution idea]
+### Progress
+### Obstacles
 
-## 尝试路径三：[方法名称]
-[第三种解决思路的详细推导]
-### 进展
-### 障碍
+## Try path three: [method name]
+[Detailed derivation of the third solution idea]
+### Progress
+### Obstacles
 
-## 部分结果与引理
-[可以严格证明的中间结果]
+## Some results and lemmas
+[Intermediate results that can be rigorously proven]
 
-## 数值/符号计算验证
-[计算机辅助验证的结果]
+## Numerical/symbolic calculation verification
+[Results of computer-aided verification]
 
-## 可能的突破方向
-[基于上述分析，最有希望的研究方向]
+## Possible breakthrough direction
+[Based on the above analysis, the most promising research direction]
 
-## 开放性结论
-[诚实说明当前分析的局限性]
+## Open conclusion
+[Be honest about the limitations of the current analysis]
 ```
 
-### 完整示例
+### Complete example
 
-```markdown
-## 问题分类与定位
-**领域**: 解析数论 × 代数几何  
-**问题类型**: L-函数零点分布与算术对象的关联  
-**关联猜想**:广义 Riemann 假设，Birch and Swinnerton-Dyer 猜想
+ ```markdown
+## Problem classification and positioning
+**Field**: Analytic Number Theory × Algebraic Geometry
+**Problem type**: Association of zero point distribution of L-functions with arithmetic objects
+**Correlation Conjecture**: Generalized Riemann hypothesis, Birch and Swinnerton-Dyer conjecture
 
-## 核心难点分析
-1. **解析延拓的存在性**: 该 L-函数的解析延拓尚未被证明
-2. **函数方程未知**: 缺乏函数方程使得零点分布分析极为困难
-3. **算术信息编码**: L-函数系数与椭圆曲线秩的关系不明确
+## Analysis of core difficulties
+1. **Existence of analytic continuation**: The analytic continuation of this L-function has not been proven yet
+2. **Unknown functional equation**: The lack of functional equation makes zero-point distribution analysis extremely difficult
+3. **Arithmetic Information Encoding**: The relationship between L-function coefficients and elliptic curve rank is unclear
 
-## 相关理论框架
-- **模形式理论**: 若该 L-函数来自模形式，可利用 modularity theorem
-- **Iwasawa 理论**: 研究 p-进 L-函数的性质
-- **Trace formula**: Arthur-Selberg trace formula 可能提供谱信息
+## Related theoretical framework
+- **Modular form theory**: If the L-function comes from modular form, modularity theorem can be used
+- **Iwasawa Theory**: Study of the properties of p-radical L-functions
+- **Trace formula**: Arthur-Selberg trace formula may provide spectral information
 
-## 已知结果综述
-- **最接近的定理**: Wiles (1995) 证明了半稳定椭圆曲线的模性
-- **部分结果**: Kolyvagin (1988) 对秩为 0 或 1 的情况证明了 BSD 猜想
-- **数值证据**: Cremona 数据库验证了 conductor < 500000 的情况
+## Summary of known results
+- **Nearest Theorem**: Wiles (1995) proved the modularity of semi-stable elliptic curves
+- **Partial results**: Kolyvagin (1988) proved the BSD conjecture for the case of rank 0 or 1
+- **Numerical evidence**: Cremona database verified for conductor < 500000
 
-## 尝试路径一：模性提升
-通过已知的模性提升定理，尝试将该椭圆曲线与模形式关联。
+## Try path one: improve modularity
+An attempt is made to relate this elliptic curve to a modular form via the known modularity promotion theorem.
 
-### 进展
-对于 good reduction 的素数 p，可以计算局部因子...
+### Progress
+For a prime p of good reduction, the local factors can be calculated...
 
-### 障碍
-在 p = 2, 3 处的 bad reduction 类型不明，无法应用现有的提升定理。
+### Obstacles
+The bad reduction at p = 2, 3 is of unknown type and the existing lifting theorem cannot be applied.
 
-## 尝试路径二：p-进方法
-构造 p-进 L-函数并研究其零点。
+## Try path two: p-advanced method
+Construct p-adic L-functions and study their zeros.
 
-### 进展
-利用 Klingen-Stackelberg 构造可以得到 p-进插值公式...
+### Progress
+The p-adic interpolation formula can be obtained using the Klingen-Stackelberg construction...
 
-### 障碍
-p-进 L-函数的非平凡零点与原 L-函数的对应关系未明。
+### Obstacles
+The corresponding relationship between the non-trivial zeros of the p-radical L-function and the original L-function is not clear.
 
-## 尝试路径三：平均结果
-考虑一族类似的 L-函数，研究统计性质。
+## Try path three: average results
+Consider a family of similar L-functions and study the statistical properties.
 
-### 进展
-可以证明平均意义下的零点密度估计...
+### Progress
+It can be shown that the zero-point density estimate in the mean sense...
 
-### 障碍
-平均结果无法推出单个 L-函数的性质。
+### Obstacles
+The properties of individual L-functions cannot be deduced from the average results.
 
-## 部分结果与引理
-**引理 1**: 在 Re(s) > 3/2 区域，该 L-函数绝对收敛。
-**引理 2**: 若存在函数方程，则临界线为 Re(s) = 1。
-**命题**: 对于 90% 的素数 p，局部因子满足预期界限。
+## Some results and lemmas
+**Lemma 1**: In the region Re(s) > 3/2, the L-function absolutely converges.
+**Lemma 2**: If there is a functional equation, then the critical line is Re(s) = 1.
+**Proposition**: For 90% of primes p, the local factors satisfy the expected bounds.
 
-## 数值/符号计算验证
-使用 SageMath 计算前 1000 个 Dirichlet 系数：
-- 系数增长符合 Ramanujan-Petersson 猜想
-- 前 100 个非平凡零点均位于临界线上（精度 10⁻¹⁰）
+## Numerical/symbolic calculation verification
+Use SageMath to calculate the first 1000 Dirichlet coefficients:
+- The coefficient growth is consistent with the Ramanujan-Petersson conjecture
+- The first 100 non-trivial zeros are all on the critical line (accuracy 10⁻¹⁰)
 
-## 可能的突破方向
-1. **确定 bad reduction 类型**: 计算 discriminant 和 conductor
-2. **寻找函数方程**: 通过数值拟合猜测函数方程形式
-3. **应用 converse theorem**: 验证足够的 analytic properties 以推出模性
+## Possible breakthrough direction
+1. **Determine bad reduction type**: Calculate discriminant and conductor
+2. **Find function equation**: Guess the form of function equation through numerical fitting
+3. **Application of converse theorem**: Verify sufficient analytic properties to derive modularity
 
-## 开放性结论
-**当前状态**: 该问题的完整解决超出已知数学工具的范围。
-上述分析展示了多种可能的研究路径，但每条路径都存在实质性障碍。
-建议将此问题分解为若干可处理的子问题逐个击破。
+## Open conclusion
+**Current Status**: A complete solution of this problem is beyond the scope of known mathematical tools.
+The above analysis illustrates a variety of possible research paths, each of which presents substantial obstacles.
+It is recommended to decompose this problem into several tractable sub-problems and solve them one by one.
 ```
 
-### 特殊注意事项
-- 必须明确区分"已证明"、"推测"、"数值证据"三种断言强度
-- 引用的定理必须准确陈述条件，不可模糊处理
-- 对于开放问题，不提供虚假的"解答"
-- 建议的研究方向应具体且可行
-- 如涉及计算，应说明所用工具和精度
+### Special Notes
+- It is necessary to clearly distinguish between the three assertion strengths of "proven", "speculation" and "numerical evidence"
+- The cited theorem must state the conditions accurately and cannot be blurred
+- Do not provide false "answers" to open questions
+- Proposed research directions should be specific and feasible
+- If calculations are involved, the tools and precision used should be stated
 
 ---
 
-## 模板选择决策树
+## Template selection decision tree
 
-```
-收到数学问题
+ ```
+received a math question
 │
-├─ 用户指定模板 → 使用指定模板
+├─ User specified template → Use specified template
 │
-├─ 未指定模板 ↓
+├─ No template specified ↓
 │  │
-│  ├─ "只要答案" → 模板 B：仅答案
-│  ├─ "帮我看看对不对" → 模板 D：解答检查
-│  ├─ "证明……" → 模板 C：证明
-│  ├─ 已知开放问题 / 一般研究级 → 模板 F：研究/开放
-│  ├─ 深度研究级问题（多分支交叉） → 模板 G：研究级深度分析
-│  ├─ 高等数学领域 → 模板 E：高等数学
-│  └─ 其他情况 → 模板 A：标准解答（默认）
+│ ├─ "Just the answer" → Template B: Only the answer
+│ ├─ "Help me see if I'm right" → Template D: Answer Check
+│ ├─ "Proof..." → Template C: Proof
+│ ├─ Known Open Issues/General Research Level → Template F: Research/Open
+│ ├─ In-depth research-level problem (multi-branch crossover) → Template G: Research-level in-depth analysis
+│ ├─ Advanced Mathematics Field → Template E: Advanced Mathematics
+│ └─ Other situations → Template A: Standard answer (default)
 ```
 
 ---
 
-## 跨模板通用规范
+## Common specifications across templates
 
-1. **数学符号**：使用标准 LaTeX 语法（$...$ 行内，$$...$$ 独立行）
-2. **格式一致性**：同一模板内部的编号、缩进、标记方式必须一致
-3. **语言**：使用简体中文，专业术语使用中文或英文取决于读者可能更熟悉的表述
-4. **验算**：所有模板都必须包含验算环节（模板 B 以"简单验算"形式存在）
-5. **答案标注**：最终答案必须醒目标注，方便定位
-6. **引用规范**：引用定理时必须注明定理名称，引用外部来源时注明出处
+1. **Mathematical notation**: Use standard LaTeX syntax ($...$ inline, $$...$$ independent line)
+2. **Format Consistency**: The numbering, indentation, and marking methods within the same template must be consistent
+3. **Language**: Simplified Chinese is used, and professional terms are used in Chinese or English depending on the expression that the reader may be more familiar with.
+4. **Check Calculation**: All templates must include a verification link (Template B exists in the form of "Simple Calculation")
+5. **Answer Marking**: The final answer must be clearly marked for easy location.
+6. **Citation Standards**: When citing a theorem, the name of the theorem must be indicated, and when citing external sources, the source must be indicated.
diff --git a/modules/search_policy.md b/modules/search_policy.md
index 0f6509d..39bf3f2 100644
--- a/modules/search_policy.md
+++ b/modules/search_policy.md
@@ -1,221 +1,221 @@
-# Math.skill 搜索策略 (Search Policy)
-
-## 1. 何时搜索 (When to Search)
-
-遇到以下任一情况时，必须执行搜索：
-
-### 触发条件 1：数学事实不确定
-当对以下内容存在不确定性时：
-- 定理的精确陈述、前提条件或结论
-- 公式的形式、适用范围或推导路径
-- 定义的标准表述或等价形式
-- 数学常数的精确值或近似值
-- 命名定理的作者、年份或标准名称
-
-**示例**："斯托尔茨定理的精确条件是什么？需要数列严格单调吗？"——应搜索确认。
-
-### 触发条件 2：高度专业化的数学对象
-当问题涉及以下内容时：
-- 高度专业化的数学对象（如 p-进数、模形式、同调代数、层论）
-- 近 5 年内发表的论文或研究成果
-- 现代数学研究的前沿课题
-- 非标准记号或领域特定的约定
-
-**示例**：用户问到关于 Perverse Sheaves 的性质——应搜索确认定义和已知结论。
-
-### 触发条件 3：用户明确要求
-当用户明确表达以下意图时：
-- "帮我查一下……"
-- "有没有类似的题目？"
-- "这个结果有出处吗？"
-- "能帮我找一下……的原始论文吗？"
-- "网上有没有标准答案？"
-- "帮我验证一下这个定理"
-
-### 触发条件 4：怀疑题目来源已知
-当有理由怀疑问题时：
-- 可能来自已知竞赛（IMO、Putnam、中国数学奥林匹克等）
-- 可能来自知名教材的习题（如 Rudin、Artin、同济高数等）
-- 可能来自某篇论文的结论
-- 可能在 Math StackExchange 或 AOPS 上有讨论
-
-### 触发条件 5：需要确认是否为已知未解决问题
-当问题看起来过于困难或与已知猜想有关时：
-- 涉及素数分布、哥德巴赫、黎曼猜想等
-- 涉及 NP vs P、Collatz 猜想等开放问题
-- 结构简单但已知无初等解的问题
-
-### 触发条件 6：需要查找标准定义或权威表述
-当需要引用公认的数学表述时：
-- 查标准定义（如"紧致性的等价定义"）
-- 查命名定理的权威表述（如"隐函数定理的完整陈述"）
-- 比较不同教材或文献中同一概念的表述差异
-
-### 触发条件 7：需要比较解题方法或查找已知技术
-当目标是比较不同解法或发现已知技术时：
-- 希望比较 2 种以上的解题路径
-- 需要了解某一类问题的标准解法
-- 需要找到某一定理的标准应用模式
+# Math.skill Search Policy
+
+## 1. When to Search (When to Search)
+
+A search must be performed when any of the following conditions are encountered:
+
+### Trigger condition 1: Uncertain mathematical facts
+When there is uncertainty about:
+- precise statement of a theorem, prerequisite or conclusion
+- The form, scope of application or derivation path of the formula
+- A standard expression or equivalent form of a definition
+- Exact or approximate values ​​of mathematical constants
+- author, year or standard name of the naming theorem
+
+**Example**: "What are the precise conditions for Stoltz's theorem? Does the sequence need to be strictly monotonic?" - should be searched for confirmation.
+
+### Trigger 2: Highly specialized mathematical objects
+When the question involves:
+- Highly specialized mathematical objects (e.g. p-radical numbers, modular forms, homologous algebra, sheath theory)
+- Papers or research results published in the past 5 years
+- Cutting edge topics in modern mathematics research
+- Non-standard notation or domain-specific conventions
+
+**Example**: User asked about the properties of Perverse Sheaves - should search for confirmed definitions and known conclusions.
+
+### Trigger condition 3: User’s explicit request
+When a user explicitly expresses the following intent:
+- "Check it for me..."
+- "Are there any similar questions?"
+- "Is there a source for this result?"
+- "Can you help me find the original paper of...?"
+- "Are there any standard answers online?"
+- "Help me verify this theorem"
+
+### Trigger condition 4: The source of the suspected question is known
+When there is reason to suspect a problem:
+- May be from known competitions (IMO, Putnam, China Mathematical Olympiad, etc.)
+- Exercises that may come from well-known textbooks (such as Rudin, Artin, Tongji Mathematics, etc.)
+- Conclusions that may come from a paper
+- May be discussed on Math StackExchange or AOPS
+
+### Trigger condition 5: Need to confirm whether it is a known unresolved issue
+When a problem seems too difficult or involves known conjecture:
+- Involving prime number distribution, Goldbach, Riemann conjecture, etc.
+- Involving NP vs P, Collatz conjecture and other open problems
+- Problems with simple structures but no known elementary solutions
+
+### Trigger condition 6: Need to find standard definitions or authoritative expressions
+When it is necessary to cite a generally accepted mathematical expression:
+- Look up standard definitions (such as "equivalent definition of compactness")
+- Look up the authoritative statement of the naming theorem (such as "the complete statement of the implicit function theorem")
+- Compare the differences in expressions of the same concept in different textbooks or documents
+
+### Trigger condition 7: Need to compare problem-solving methods or find known technologies
+When the goal is to compare different solutions or discover known techniques:
+- Want to compare more than 2 problem-solving paths
+- Need to know the standard solution to a certain type of problem
+- Need to find a standard application pattern of a certain theorem
 
 ---
 
-## 2. 搜索权威层级 (Search Authority Hierarchy)
+## 2. Search Authority Hierarchy
 
-搜索结果按可信度分为四个层级：
+Search results are divided into four levels based on credibility:
 
-### 优先级 1：最高可信度
-可直接信任并引用的来源：
-- **经典教材**：如 Rudin《数学分析原理》、Artin《代数》、同济《高等数学》等
-- **大学官方讲义**：知名大学发布的课程笔记或讲义（如 MIT OCW、Stanford 课程笔记）
-- **数学百科类权威网站**：如 ProofWiki（严谨证明）、nLab（范畴论等）、Springer Encyclopedia of Mathematics
+### Priority 1: Highest credibility
+Sources you can trust and cite directly:
+- **Classic textbooks**: such as Rudin's "Principles of Mathematical Analysis", Artin's "Algebra", Tongji's "Advanced Mathematics", etc.
+- **Official university handouts**: course notes or handouts issued by well-known universities (such as MIT OCW, Stanford course notes)
+- **Mathematical encyclopedia authoritative websites**: such as ProofWiki (rigorous proof), nLab (category theory, etc.), Springer Encyclopedia of Mathematics
 
-**使用方式**：直接引用定理编号、页码、标准表述。
+**Usage**: Directly quote the theorem number, page number, and standard expression.
 
-### 优先级 2：高可信度
-可信任但需交叉验证的来源：
-- **同行评审的数学论文**：发表在公认期刊上的论文
-- **MathWorld (Wolfram)**：公认的数学参考资源
-- **OEIS (oeis.org)**：数列百科全书，数论/组合问题的权威参考
-- **arXiv**：预印本，需注意是否为正式发表版本
-- **竞赛官方解答**：IMO、Putnam 等官方发布的解答
-- **Terence Tao 博客、Timothy Gowers 博客**等公认数学家维护的博客
+### Priority 2: High Confidence
+Trusted but cross-verified sources:
+- **Peer-reviewed mathematics papers**: papers published in recognized journals
+- **MathWorld (Wolfram)**: The recognized mathematical reference resource
+- **OEIS (oeis.org)**: Encyclopedia of Sequences, the authoritative reference for number theory/combinatorial problems
+- **arXiv**: Preprint, please note whether it is an officially published version
+- **Official answers to the competition**: Answers officially released by IMO, Putnam, etc.
+- Blogs maintained by recognized mathematicians such as **Terence Tao Blog and Timothy Gowers Blog**
 
-**使用方式**：以这些来源为主要参考，但遇到矛盾时以自身推导为准。
+**How ​​to use**: Use these sources as the main reference, but in the event of conflict, your own derivation shall prevail.
 
-### 优先级 3：中等可信度（需谨慎）
-可作为参考但不能盲从的来源：
-- **Math StackExchange**：答案质量参差，需看投票数和评论
-- **MathOverflow**：研究级数学问答，质量通常高于 MSE，但仍需验证
-- **Art of Problem Solving (AoPS)**：竞赛社区解答
-- **Wikipedia（数学条目）**：作为起点可读，但需交叉验证关键陈述
+### Priority 3: Medium confidence (caution required)
+Sources that can be used as a reference but not to be followed blindly:
+- **Math StackExchange**: The quality of answers varies, depending on the number of votes and comments
+- **MathOverflow**: Research-grade math Q&A, usually higher quality than MSE, but still needs to be verified
+- **Art of Problem Solving (AoPS)**: Contest community answers
+- **Wikipedia (Mathematical Entry)**: Readable as a starting point, but requires cross-validation of key statements
 
-**使用方式**：作为启发参考，关键步骤必须独立重新推导验证。
+**How ​​to use**: As a heuristic reference, key steps must be independently re-derived and verified.
 
-### 优先级 4：不可使用
-绝对不能使用以下来源：
-- **随机个人博客**：无明显学术资质的个人博客
-- **未经验证的论坛**：如 Zhihu（知乎）、Baidu Zhidao（百度知道）等非专业数学论坛的未经证实的答案
-- **AI 生成内容**：其他 AI 或 LLM 生成的数学内容，ChatGPT 输出等
-- **社交媒体帖子**：微博、Twitter、Reddit 等非结构化讨论
-- **内容农场**：以 SEO 为目的拼凑的数学内容
+### Priority 4: Not available
+The following sources must not be used:
+- **Random Personal Blog**: Personal blog without obvious academic qualifications
+- **Unverified Forum**: Unverified answers from non-professional mathematics forums such as Zhihu (Zhihu) and Baidu Zhidao (Baidu Zhidao)
+- **AI generated content**: mathematical content generated by other AI or LLM, ChatGPT output, etc.
+- **Social media posts**: Unstructured discussions on Weibo, Twitter, Reddit, etc.
+- **Content Farm**: Math content cobbled together for SEO purposes
 
 ---
 
-## 3. 搜索伦理与规则 (Search Ethics and Rules)
-
-### 规则 1：严禁直接复制
-**绝对禁止**直接复制网上找到的解答。搜索结果只能作为参考，最终呈现的解答必须是从原理出发独立重新推导的。
-
-### 规则 2：搜索结果为参考
-搜索结果仅作为理解问题结构和方向的手段，最终解答必须用自己的语言重新表述，每一步都经过独立验证。
-
-### 规则 3：处理冲突
-如果网上找到的答案与自身推导不一致：
-1. 标注冲突存在的每一步
-2. 列出自身推导和网上答案的分歧点
-3. 重新检查两边的每一个条件
-4. 尝试用反例区分两种方法
-5. 如果无法确定哪一方正确，如实说明
-
-### 规则 4：找不到类似问题时
-如果搜索后找不到任何类似问题或方法：
-1. 从定义出发推导
-2. 使用已知标准定理
-3. 运用基本数学方法（归纳、反证、构造等）
-4. 明确标注"此问题在公开资源中未找到类似题目"
-5. 不因此降低推导的严谨性
-
-### 规则 5：已知开放问题
-如果发现问题是已知的开放问题：
-1. 明确写明"此为已知未解决问题，目前不存在被数学界公认的完整证明。"
-2. 提供该问题的已知标准名称
-3. 概述已知的部分结果（如果有）
-4. 指出已知的证明尝试及其失败原因（如果有）
-5. 绝不对开放问题声称"已解决"
-
-### 规则 6：禁止虚构引用
-**绝对禁止**虚构以下内容：
-- 不存在的论文标题、作者或期刊
-- 不存在的定理编号或名称
-- 不存在的教材引用
-- 不存在的 DOI 或 URL
-
-如果确实需要引用但无法找到确切来源，使用"根据已知的数学结论……"的表述方式。
+## 3. Search Ethics and Rules
+
+### Rule 1: Direct copying is strictly prohibited
+**Absolutely prohibited** Directly copying solutions found online. The search results can only be used as a reference, and the final solution must be independently re-derived from the principles.
+
+### Rule 2: Search results are for reference
+Search results only serve as a means of understanding the structure and direction of the question; the final solution must be restated in your own words, with each step independently verified.
+
+### Rule 3: Handling Conflicts
+If the answer found online is inconsistent with your own derivation:
+1. Mark each step where a conflict exists
+2. List the differences between your own derivation and online answers
+3. Recheck each condition on both sides
+4. Try to use counterexamples to distinguish between the two methods
+5. If you are not sure which side is correct, explain it truthfully.
+
+### Rule 4: When no similar questions can be found
+If you can't find any similar questions or methods after searching:
+1. Derivation from the definition
+2. Use known standard theorems
+3. Use basic mathematical methods (induction, disproof, construction, etc.)
+4. Clearly mark "This question has no similar questions found in public resources"
+5. Do not reduce the rigor of the derivation.
+
+### Rule 5: Known Open Issues
+If the issue is found to be a known open issue:
+1. Clearly state that "This is a known unsolved problem, and there is currently no complete proof recognized by the mathematical community."
+2. Provide a known standard name for the problem
+3. Summarize some of the known results (if any)
+4. Indicate known attempts at proof and why they failed (if any)
+5. Never claim that an open issue is "solved"
+
+### Rule 6: No fictitious references allowed
+**Absolutely prohibited** The following fictional content:
+- Non-existent paper title, author or journal
+- non-existent theorem number or name
+- References to non-existent textbooks
+- Non-existent DOI or URL
+
+If you really need to cite but the exact source cannot be found, use the expression "based on known mathematical results..."
 
 ---
 
-## 4. 无网络时的应对策略
+## 4. Countermeasures when there is no network
 
-当无法访问互联网时，遵循以下策略：
+When Internet access is not available, follow these strategies:
 
-### 可用的内部资源
-1. **数学推理本身是第一资源**——从第一性原理出发的推导不需要联网
-2. **标准定理的知识储备**——利用训练数据中内化的数学知识
-3. **基本方法的系统性应用**——归纳、反证、构造、变换等方法论不依赖联网
+### Available internal resources
+1. **Mathematical reasoning itself is the first resource** - derivation from first principles does not require an Internet connection
+2. **Knowledge reserve of standard theorems** - using the mathematical knowledge internalized in the training data
+3. **Systematic application of basic methods** - Methodologies such as induction, disproof, construction, and transformation do not rely on the Internet
 
-### 应对框架
-1. **从定义和公理出发**：写下所有相关定义
-2. **引用已知标准定理**：虽无法查证精确编号，但可使用定理名称和标准表述
-3. **明确陈述假设**：如果某个定理的条件记忆不确定，明确说明"假设……定理的条件被满足"
-4. **标注需要网络验证的部分**：在解答末尾添加`[待联网验证]`标记
+### Response Framework
+1. **Start with definitions and axioms**: Write down all relevant definitions
+2. **Quoting known standard theorems**: Although the exact number cannot be verified, the theorem name and standard expression can be used
+3. **Explicitly state assumptions**: If the condition memory of a certain theorem is uncertain, clearly state "Assume...the conditions of the theorem are satisfied"
+4. **Mark the part that requires network verification**: Add the `[to be network verified]` mark at the end of the answer
 
-### 标注规范
-```
-## 联网验证建议
-- [ ] 验证斯托尔茨定理对分母单调性的精确要求
-- [ ] 确认该积分的解析解是否存在
-- [ ] 查证该组合恒等式的标准名称
+### Labeling specifications
+ ```
+## Internet verification suggestions
+- [ ] Verify the precise requirement of Stoltz's theorem on the monotonicity of the denominator
+- [ ] Confirm whether the analytical solution of the integral exists
+- [ ] Check the standard name of the combined identity
 ```
 
 ---
 
-## 5. 搜索查询构造指南 (Search Query Formulation)
+## 5. Search Query Formulation
 
-### 基本原则
-1. **使用英文关键词**：数学文献以英文为主，英文查询的命中率显著高于中文
-2. **包含 LaTeX 语法**：直接在查询中使用`$a_n$`、`$\sum$`等 LaTeX 片段
-3. **特定于目标来源**：如果希望搜索 StackExchange，加上`site:math.stackexchange.com`
-4. **包含定理名称**：如有标准名称，直接使用标准名称查询
+### Basic principles
+1. **Use English keywords**: Mathematics literature is mainly in English, and the hit rate of English queries is significantly higher than that of Chinese
+2. **Contains LaTeX syntax**: Use `$a_n$`, `$\sum$` and other LaTeX fragments directly in the query
+3. **Target source specific**: If you want to search StackExchange, add `site:math.stackexchange.com`
+4. **Contains theorem name**: If there is a standard name, directly use the standard name to query
 
-### 查询构造模板
+### Query construction template
 
-#### 查定义/定理表述
-```
+#### Check definition/theorem expression
+ ```
 "theorem_name" statement conditions
 ```
-示例：`"Stolz Cesaro theorem" statement conditions`
+Example: `"Stolz Cesaro theorem" statement conditions`
 
-#### 查类似题目
+#### Check similar questions
+ ```
+"[Simplified mathematical expression]" similar problem solution
 ```
-"[简化的数学表达式]" similar problem solution
-```
-示例：`sum_(k=1)^n k*k! similar problem`
+Example: `sum_(k=1)^n k*k! similar problem`
 
-#### 查是否为开放问题
-```
-"[问题描述]" open problem unsolved conjecture
+#### Check if it is an open issue
+ ```
+"[Problem description]" open problem unsolved conjecture
 ```
-示例：`"is every even number sum of two primes" open problem`
+Example: `"is every even number sum of two primes" open problem`
 
-#### 查特定竞赛题目
-```
-"[关键词]" IMO OR Putnam OR "Chinese Mathematical Olympiad" problem solution
+#### Check specific competition questions
+ ```
+"[Keywords]" IMO OR Putnam OR "Chinese Mathematical Olympiad" problem solution
 ```
 
-#### 查标准方法
-```
-"[问题类型]" standard technique OR method OR approach
+#### Check the standard method
+ ```
+"[Problem type]" standard technique OR method OR approach
 ```
-示例：`"evaluate improper integral" standard technique contour`
+Example: `"evaluate improper integral" standard technique contour`
 
-#### 使用 OEIS
-```
-"[数列前几项]" OEIS
+#### Using OEIS
+ ```
+"[First few items of the sequence]" OEIS
 ```
-示例：`1, 1, 2, 3, 5, 8, 13 OEIS`
+Example: `1, 1, 2, 3, 5, 8, 13 OEIS`
 
-### 搜索后处理
-1. 阅读搜索结果摘要而非逐条点击
-2. 优先阅读 StackExchange 的高赞答案（>10 upvotes）和 MathOverflow 答案
-3. 验证关键公式是否与自身理解一致
-4. 记录搜索到的关键结论和来源
+### Post-search processing
+1. Read summaries of search results instead of clicking through them one by one
+2. Prioritize reading StackExchange’s highly praised answers (>10 upvotes) and MathOverflow answers
+3. Verify whether the key formulas are consistent with your own understanding
+4. Record the key conclusions and sources found
diff --git a/modules/verification_engine.md b/modules/verification_engine.md
index 6e859b9..c2cbbbb 100644
--- a/modules/verification_engine.md
+++ b/modules/verification_engine.md
@@ -581,7 +581,12 @@ When any verification method signals an error:
 4. **Fix**: Correct the error and propagate the change forward
 5. **Re-verify**: Apply the same verification methods again, plus at least one additional method
 6. **If error persists** after two correction cycles: Switch to an independent solution method (Method H)
-7. **If still failing**: State uncertainty explicitly; do not output an unverified answer
+7. **If still failing**: You MUST explicitly admit failure and decline to provide a final answer. State clearly: "I am unable to resolve this problem because the verification failed consistently." Do NOT output an unverified answer, do NOT fabricate steps, and do NOT invent "fake" verifications to pass.
+
+### Strict Anti-Hallucination Constraints in Verification
+- **Do NOT fake verifications**: Never claim a verification method passed if it did not or if you did not actually perform the calculation.
+- **Do NOT force a pass**: If a calculation results in a mismatch, do not invent algebraic rules or rounding reasons to justify it. A failure is a failure.
+- **No BS rule**: If you do not have high confidence in an intermediate step, do not invent "plausible" math (Plausible BS) to bridge the gap.
 
 ### Diagnostic Heuristics
 
diff --git a/tests/adversarial_cases.md b/tests/adversarial_cases.md
index 37ce39b..1410230 100644
--- a/tests/adversarial_cases.md
+++ b/tests/adversarial_cases.md
@@ -1,458 +1,458 @@
-# Math.skill 对抗性测试用例
+# Math.skill adversarial test cases
 
-> 本文档包含 15 个专门设计用于"欺骗"系统的对抗性测试用例。
-> 每个用例都包含容易被忽视的陷阱，用于测试 Math.skill 的防御能力和严谨性。
+> This document contains 15 adversarial test cases specifically designed to "fool" the system.
+> Each use case contains easily overlooked pitfalls that test the defensibility and rigor of Math.skill.
 
 ---
 
-## 1. 平方引入增根
+## 1. Square introduces increasing roots
 
 **Test ID:** ADV-001  
-**Input:** "解方程 \(\sqrt{x+3} = x+1\)"  
+**Input:** "Solve the equation \(\sqrt{x+3} = x+1\)"
 **Expected Classification:** algebra, radical_equation, extraneous_root_trap  
 **Expected Behavior:**
-- 平方：x+3 = (x+1)² = x² + 2x + 1
-- x² + x - 2 = 0 → x = 1 或 x = -2
-- 回代 x=1：√4 = 2, 1+1 = 2 ✓
-- 回代 x=-2：√1 = 1, -2+1 = -1 ✗
-- x = -2 是增根 → 最终解 x = 1
+- Square: x+3 = (x+1)² = x² + 2x + 1
+- x² + x - 2 = 0 → x = 1 or x = -2
+- Substitute x=1: √4 = 2, 1+1 = 2 ✓
+- Substitute x=-2: √1 = 1, -2+1 = -1 ✗
+- x = -2 is increasing root → final solution x = 1
 
 **Required Verification:**
-- 必须回代到原方程
-- 必须显式说明 x=-2 是增根以及原因
+- must be substituted back into the original equation
+- Must explicitly state that x=-2 is an increasing root and why
 
 **Key Checks:**
-- 平方两边是不是等价变换？
-- 每个解是否都回代了？
+- Are both sides of the square equivalent transformations?
+- Is every solution resubstituted?
 
 **Failure Modes:**
-- 直接给两个解，未检查
-- 解释增根原因不够清晰
+- Give two solutions directly without checking
+- The explanation of the reasons for root increase is not clear enough
 
 ---
 
-## 2. 除以可能为零的表达式
+## 2. Divide by an expression that may be zero
 
 **Test ID:** ADV-002  
-**Input:** "解方程 \(\frac{x^2-1}{x-1} = x+1\)，注意 x≠1"  
+**Input:** "Solve the equation \(\frac{x^2-1}{x-1} = x+1\), note that x≠1"
 **Expected Classification:** algebra, rational_equation, removable_singularity_trap  
 **Expected Behavior:**
-- 注意：即使提示了 x≠1，仍需严谨
-- 当 x≠1 时，分子因式分解 (x-1)(x+1)
-- 左 = x+1（x≠1 时）
-- 方程变为 x+1 = x+1（对所有 x≠1 成立）
-- 解为：x ∈ ℝ \ {1}（所有不等于 1 的实数）
-- 注意：x=1 时分母为零，原式无定义
+- Note: Even if x≠1 is prompted, you still need to be rigorous
+- When x≠1, the numerator factors (x-1)(x+1)
+- Left = x+1 (when x≠1)
+- The equation becomes x+1 = x+1 (true for all x≠1)
+- Solution is: x ∈ ℝ \ {1} (all real numbers not equal to 1)
+- Note: When x=1, the denominator is zero and the original formula is undefined.
 
 **Required Verification:**
-- 不能给出 x=1
-- 需要理解这不是"无解"，而是"无穷多解（除 1 外）"
-- 可能要用极限观点：lim_{x→1} 左 = 2 = 右，但 x=1 处函数无定义
+- cannot give x=1
+- Need to understand that this is not "no solution", but "infinitely many solutions (except 1)"
+- You may need to use the limit view: lim_{x→1} left = 2 = right, but the function at x=1 is undefined
 
 **Key Checks:**
-- 是否给出 x 可以是任何实数（除 1 外）？
-- 是否说明 x=1 被排除的原因？
+- Is it given that x can be any real number (except 1)?
+- Explain why x=1 is excluded?
 
 **Failure Modes:**
-- 说"恒等式，x ∈ ℝ"而遗漏 x≠1（严重）
-- 错误地称"无解"
-- 直接代入 x=1 认为成立
+- Say "Identity, x ∈ ℝ" and omit x≠1 (serious)
+- Incorrectly called "no solution"
+- Directly substitute x=1 and it is considered established.
 
 ---
 
-## 3. 有理不等式中的符号表陷阱
+## 3. Symbol table trap in rational inequalities
 
 **Test ID:** ADV-003  
-**Input:** "解不等式 \(\frac{x-1}{x^2-4} \leq 0\)"  
+**Input:** "Solve the inequality \(\frac{x-1}{x^2-4} \leq 0\)"
 **Expected Classification:** algebra, rational_inequality, sign_chart_trap  
 **Expected Behavior:**
-- 分子零点：x = 1
-- 分母零点（必须排除！）：x = -2, x = 2
-- 符号表检验区间：(-∞,-2), (-2,1), (1,2), (2,∞)
-- 注意 ≤0：分子为 0 时（x=1）不等式成立（0 ≤ 0 ✓）
-- 分母为 0 时必须排除（函数无定义）
-- 解集：(-2, 1] ∪ (2, ∞)
-  - 验证：x=-3 负数/正数=负<0 → (-∞,-2): 否！
-  - 重新算：x=-3: (-4)/(5)=-0.8<0 ✓
+- Numerator zero point: x = 1
+- Denominator zeros (must be excluded!): x = -2, x = 2
+- Symbol table test interval: (-∞,-2), (-2,1), (1,2), (2,∞)
+- Note ≤0: When the numerator is 0 (x=1) the inequality holds (0 ≤ 0 ✓)
+- Must be excluded when the denominator is 0 (function is undefined)
+- Solution set: (-2, 1] ∪ (2, ∞)
+- Verification: x=-3 negative number/positive number = negative <0 → (-∞,-2): No!
+- Recalculate: x=-3: (-4)/(5)=-0.8<0 ✓
   - x=0: (-1)/(-4)=0.25>0 ✗
   - x=1.5: 0.5/(-1.75)<0 ✓
   - x=3: 2/5>0 ✗
-- 正确解集：(-∞,-2) ∪ (1, 2) ？需要仔细算
+- Correct solution set: (-∞,-2) ∪ (1, 2) ? Need to calculate carefully
   - x=-3: (-4)/(5) = -0.8 ✓ → (-∞,-2) ✓
   - x=0: (-1)/(-4) = 0.25 ✗ → (-2,1) ✗
   - x=1: 0/(-3) = 0 ✓ → x=1 ✓
   - x=1.5: 0.5/(-1.75) = -0.29 ✓ → (1,2) ✓
   - x=3: 2/5 = 0.4 ✗ → (2,∞) ✗
-- 解集：(-∞, -2) ∪ [1, 2)
+- Solution set: (-∞, -2) ∪ [1, 2)
 
 **Required Verification:**
-- 符号表的每个区间是否正确
-- 含等于号时分子零点的处理
-- 分母零点的正确处理
+- Whether each interval of the symbol table is correct
+- Treatment of the zero point of the numerator when it contains the equal sign
+- Correct handling of denominator zeros
 
 **Key Checks:**
-- 符号表制作是否正确
-- x=1 处的不等式方向（含等号）
-- x=-2 和 x=2 处的定义域排除
+- Is the symbol table created correctly?
+- Direction of inequality at x=1 (including equal sign)
+- domain exclusion at x=-2 and x=2
 
 **Failure Modes:**
-- 符号表错误
-- x=1 处的开闭区间搞混
-- 分母零点未正确排除
+- Symbol table error
+- The opening and closing intervals at x=1 are confused
+- The denominator zero is not excluded correctly
 
 ---
 
-## 4. 隐藏定义域限制的复合函数
+## 4. Composite functions that hide domain restrictions
 
 **Test ID:** ADV-004  
-**Input:** "求函数 \(f(x) = \frac{\ln(x^2-1)}{\sqrt{x-2}}\) 的定义域。"  
+**Input:** "Find the domain of the function \(f(x) = \frac{\ln(x^2-1)}{\sqrt{x-2}}\)."
 **Expected Classification:** algebra, domain_analysis, composite_function_trap  
 **Expected Behavior:**
-- 分母 √(x-2) 条件：x-2 > 0 → x > 2（平方根内的必须 > 0 因为还在分母中）
-- 等一下：√(x-2) 作为分母 → x-2 > 0 → x > 2
-- 对数条件：ln(x²-1) 要求 x²-1 > 0 → x < -1 或 x > 1
-- 取交集：x > 2 与 (x < -1 或 x > 1) = x > 2
-- 定义域：(2, ∞)
+- Denominator √(x-2) Condition: x-2 > 0 → x > 2 (the square root must be > 0 because it is still in the denominator)
+- Wait a minute: √(x-2) as denominator → x-2 > 0 → x > 2
+- Logarithmic condition: ln(x²-1) requires x²-1 > 0 → x < -1 or x > 1
+- Take the intersection: x > 2 and (x < -1 or x > 1) = x > 2
+- Domain: (2, ∞)
 
 **Required Verification:**
-- 每个组成部分的条件
-- 交集计算是否考虑了两部分
-- 端点检验：x=2 时分母为零
+- Conditions for each component
+- Whether the intersection calculation considers both parts
+- Endpoint test: denominator is zero when x=2
 
 **Key Checks:**
-- 平方根内必须是 ≥0 还是 >0（取决于是否在分母）
-- 对数的条件
-- 正确取交集
+- The square root must be ≥0 or >0 (depending on whether it is in the denominator)
+- Conditions for logarithms
+- Get the intersection correctly
 
 **Failure Modes:**
-- 忘记分母中平方根的条件
-- 交集计算错误
+- Forgot the condition for the square root in the denominator
+- Intersection calculation error
 
 ---
 
-## 5. L'Hôpital 法则看起来适用但不行
+## 5. L'Hôpital's Law seems to apply but doesn't
 
 **Test ID:** ADV-005  
-**Input:** "求极限 \(\lim_{x \to 0} \frac{x^2 \sin(1/x)}{x}\)"  
+**Input:** "Find the limit \(\lim_{x \to 0} \frac{x^2 \sin(1/x)}{x}\)"
 **Expected Classification:** calculus, limit, LHopital_not_applicable  
 **Expected Behavior:**
-- 尝试 L'Hôpital：分子导数 = 2x sin(1/x) + x² cos(1/x) · (-1/x²) = 2x sin(1/x) - cos(1/x)
-- 分母导数 = 1
-- 但 lim_{x→0} cos(1/x) 不存在！所以 L'Hôpital 不适用
-- 正确方法：约分 x：lim_{x→0} x sin(1/x) = 0（夹逼定理：-x ≤ x sin(1/x) ≤ x）
-- **L'Hôpital 的逆命题不成立**：导数之比的极限存在是充分非必要条件
+- Try L'Hôpital: Nolecular derivative = 2x sin(1/x) + x² cos(1/x) · (-1/x²) = 2x sin(1/x) - cos(1/x)
+- Denominator derivative = 1
+- But lim_{x→0} cos(1/x) does not exist! So L'Hôpital does not apply
+- Correct method: reduce x: lim_{x→0} x sin(1/x) = 0 (pinch theorem: -x ≤ x sin(1/x) ≤ x)
+- **The converse of L'Hôpital is not true**: the existence of the limit of the ratio of derivatives is a sufficient but not necessary condition
 
 **Required Verification:**
-- 必须指出为什么 L'Hôpital 不适用
-- 用夹逼定理验证
-- 说明 L'Hôpital 法则的条件
+- It must be stated why L'Hôpital does not apply
+- Verify using the pinch theorem
+- Explain the conditions of L'Hôpital's law
 
 **Key Checks:**
-- 是否滥用了 L'Hôpital？
-- 是否理解了 L'Hôpital 的使用条件？
+- Has L'Hôpital been abused?
+- Do you understand L'Hôpital's conditions of use?
 
 **Failure Modes:**
-- 误用 L'Hôpital 并得出不存在/错误的极限
-- 声称极限不存在（实际上存在且为 0）
+- Misusing L'Hôpital and arriving at non-existent/wrong limits
+- claims that the limit does not exist (actually it does and is 0)
 
 ---
 
-## 6. 误导性几何图：不可能三角形
+## 6. Misleading Geometry: Impossible Triangle
 
 **Test ID:** ADV-006  
-**Input:** "三角形 ABC 中，AB=5, BC=7, CA=13。文字描述，求三角形面积。"  
+**Input:** "In triangle ABC, AB=5, BC=7, CA=13. Text description, find the area of ​​the triangle."
 **Expected Classification:** geometry, triangle_inequality_check, impossible_figure  
 **Expected Behavior:**
-- 检查三角不等式：5+7 = 12 < 13
-- 这不满足三角不等式！
-- 这样的三角形不存在
-- 不应给出面积，而应指出三角形不可能存在
-- 如果强行用海伦公式，s = (5+7+13)/2 = 12.5
-- √(12.5·7.5·5.5·(-0.5)) 会有负的被开方数
+- Check the triangle inequality: 5+7 = 12 < 13
+- This does not satisfy the triangle inequality!
+- Such a triangle does not exist
+- Instead of giving the area, point out that the triangle cannot exist
+- If you use Heron's formula, s = (5+7+13)/2 = 12.5
+- √(12.5·7.5·5.5·(-0.5)) will have a negative radicand number
 
 **Required Verification:**
-- 三角不等式检查
-- 明确声明三角形不存在
-- 解释为何海伦公式会给出虚数
+- Triangle inequality check
+- Explicitly declare that the triangle does not exist
+- Explain why Heron's formula gives imaginary numbers
 
 **Key Checks:**
-- 关键陷阱：是否会盲目用海伦公式？
-- 是否会忽略最基本的几何公理？
+- Key trap: Will Heron's formula be used blindly?
+- Are the most basic geometric axioms ignored?
 
 **Failure Modes:**
-- 不对边长作合理性检查
-- 强行用海伦公式
-- 给出一个面积值
+- No plausibility checks on side lengths
+- Forced use of Helen's formula
+- gives an area value
 
 ---
 
-## 7. 微妙的独立性假设
+## 7. Subtle independence assumption
 
 **Test ID:** ADV-007  
-**Input:** "事件 A 和 B 满足 P(A)=0.5, P(B)=0.4, P(A∪B)=0.7。问 A 和 B 是否独立？"  
+**Input:** "Events A and B satisfy P(A)=0.5, P(B)=0.4, P(A∪B)=0.7. Are A and B independent?"
 **Expected Classification:** probability, independence_test, common_trap  
 **Expected Behavior:**
-- 由加法公式：P(A∪B) = P(A) + P(B) - P(A∩B)
+- From the addition formula: P(A∪B) = P(A) + P(B) - P(A∩B)
 - 0.7 = 0.5 + 0.4 - P(A∩B) → P(A∩B) = 0.2
-- 检查独立性：P(A)·P(B) = 0.5 × 0.4 = 0.2
+- Check independence: P(A)·P(B) = 0.5 × 0.4 = 0.2
 - P(A∩B) = 0.2 = P(A)·P(B)
-- 所以 A 和 B 是独立的！
+- So A and B are independent!
 
 **Required Verification:**
-- 正确应用加法公式
-- 正确的独立性检验
-- 数值演算
+- Correct application of addition formulas
+- Correct independence test
+- Numerical calculation
 
 **Key Checks:**
-- 陷阱：许多人会直接代入 P(A∪B)=P(A)+P(B)（互斥假设）
-- 必须用加法公式正确推导
+- Trap: Many people will directly substitute P(A∪B)=P(A)+P(B) (mutually exclusive assumption)
+- Must use addition formula to derive correctly
 
 **Failure Modes:**
-- 误认为不独立（直觉有时误导）
-- 计算 P(A∩B) 错误
+- Mistaken for not being independent (intuition is sometimes misleading)
+- Error in calculating P(A∩B)
 
 ---
 
-## 8. 换元积分中的定义域陷阱
+## 8. Domain trap in substitution integrals
 
 **Test ID:** ADV-008  
-**Input:** "计算定积分 \(\int_{-1}^1 \frac{dx}{1+x^2}\)"  
+**Input:** "Calculate definite integral \(\int_{-1}^1 \frac{dx}{1+x^2}\)"
 **Expected Classification:** calculus, definite_integral, substitution_domain  
 **Expected Behavior:**
-- 反导数：arctan(x)
+- Antiderivative: arctan(x)
 - ∫_{-1}^{1} dx/(1+x²) = arctan(1) - arctan(-1) = π/4 - (-π/4) = π/2
-- 如果换元 x = tan θ，需要注意 θ 的范围
+- If you change the element x = tan θ, you need to pay attention to the range of θ
 - x ∈ [-1,1] → θ ∈ [-π/4, π/4]
 - ∫ dx/(1+x²) = θ|_{θ=-π/4}^{θ=π/4} = π/4 - (-π/4) = π/2
 
 **Required Verification:**
-- 如果换元，定义域变换是否正确
-- 答案 π/2
-- 直接方法和换元方法一致
+- If the element is changed, is the domain transformation correct?
+- Answer π/2
+- The direct method is the same as the substitution method
 
 **Key Checks:**
-- 换元时区间变换是否正确
-- 积分值是否正确
+- Is the conversion time interval conversion correct?
+- Are the points values ​​correct?
 
 **Failure Modes:**
-- 换元时区间映射错误
-- 忘记改变积分限
+- Swap time interval mapping error
+- Forgot to change the points limit
 
 ---
 
-## 9. 比率检验法失效的级数
+## 9. The series where the ratio test method fails
 
 **Test ID:** ADV-009  
-**Input:** "判断级数 \(\sum_{n=1}^\infty \frac{n!}{n^n}\) 的敛散性。"  
+**Input:** "Judge the convergence and divergence of the series \(\sum_{n=1}^\infty \frac{n!}{n^n}\)."
 **Expected Classification:** calculus, series_convergence, ratio_test_edge  
 **Expected Behavior:**
-- 比率检验：a_n/a_{n+1} = ... 
-- 实际上：lim_{n→∞} a_{n+1}/a_n = lim_{n→∞} (n+1)!/(n+1)^(n+1) · n^n/n!
+- Ratio test: a_n/a_{n+1} = ...
+- Actually: lim_{n→∞} a_{n+1}/a_n = lim_{n→∞} (n+1)!/(n+1)^(n+1) · n^n/n!
   = lim n^n/(n+1)^n = lim (n/(n+1))^n = lim (1 - 1/(n+1))^n = 1/e
-- 1/e < 1，由比率检验知级数收敛
-- 这和直觉不同（一些人会觉得 n! 增长很快所以发散）
+- 1/e < 1, the series convergence is known from the ratio test
+- This is different from intuition (some people will think that n! grows quickly and diverges)
 
 **Required Verification:**
-- 比率检验的极限是否正确
-- 结果 < 1 的含义
-- 其他检验方法（如斯特林公式）
+- Are the limits of the ratio test correct?
+- Meaning of result < 1
+- Other testing methods (such as Stirling's formula)
 
 **Key Checks:**
-- 陷阱：直觉判断可能出错
-- 比率检验极限计算的代数是否正确
+- Pitfall: Intuitive judgment may be wrong
+- Whether the algebra of the ratio test limit calculation is correct
 
 **Failure Modes:**
-- 凭直觉判断发散
-- 比率检验极限计算错误
+- Judgment of divergences based on intuition
+- Ratio test limit calculation error
 
 ---
 
-## 10. 矩阵维度不匹配
+## 10. Matrix dimensions do not match
 
 **Test ID:** ADV-010  
-**Input:** "设 A 是 2×3 矩阵，B 是 3×2 矩阵。求证 AB 和 BA 都有意义但不经历相同。计算 tr(AB) 和 tr(BA) 的关系。"  
+**Input:** "Let A be a 2×3 matrix and B be a 3×2 matrix. Verify that AB and BA are both meaningful but not experienced the same way. Compute the relationship between tr(AB) and tr(BA)."
 **Expected Classification:** linear_algebra, matrix_multiplication, trace_properties  
 **Expected Behavior:**
-- AB 是 2×2，BA 是 3×3
-- tr(AB) = tr(BA) = Σ_i Σ_j a_{ij} b_{ji}（迹的循环性质，即使维度不同也成立）
-- 但 AB ≠ BA，因为它们维度不同
+- AB is 2×2, BA is 3×3
+- tr(AB) = tr(BA) = Σ_i Σ_j a_{ij} b_{ji} (the cyclic property of the trace is true even if the dimensions are different)
+- But AB ≠ BA because they have different dimensions
 
 **Required Verification:**
-- 维度推理
-- 迹的循环性质的应用
-- 不能错误地说 AB = BA
+- Dimensional reasoning
+- Application of cyclic properties of traces
+- You cannot incorrectly say AB = BA
 
 **Key Checks:**
-- 陷阱：可能错误地认为 AB = BA
-- 迹的性质
-- 维度理解
+- Pitfall: Possibly incorrectly assuming AB = BA
+- nature of trace
+- Dimensional understanding
 
 **Failure Modes:**
-- 说 AB = BA
-- 不理解迹的循环性质
-- 维度混乱
+- Say AB = BA
+- Not understanding the cyclic nature of traces
+- Dimensional chaos
 
 ---
 
-## 11. 隐藏除以零的"证明"
+## 11. Hide the "proof" of division by zero
 
 **Test ID:** ADV-011  
-**Input:** "以下证明有什么问题？设 a=b，则 a²=ab, a²-b²=ab-b², (a-b)(a+b)=b(a-b), a+b=b, 2b=b, 2=1。"  
+**Input:** "What's wrong with the following proof? Suppose a=b, then a²=ab, a²-b²=ab-b², (a-b)(a+b)=b(a-b), a+b=b, 2b=b, 2=1."
 **Expected Classification:** algebra, error_detection, division_by_zero  
 **Expected Behavior:**
-- 从 (a-b)(a+b) = b(a-b) 到 a+b = b 这一步除以了 (a-b)
-- 但已知 a=b，所以 a-b = 0
-- 除以零的步骤是非法的
-- 这解释了为什么得到了矛盾的结果
+- The step from (a-b)(a+b) = b(a-b) to a+b = b is divided by (a-b)
+- But we know a=b, so a-b = 0
+- The step of dividing by zero is illegal
+- This explains why the contradictory results were obtained
 
 **Required Verification:**
-- 精确定位除以零的步骤
-- 说明为什么非法
-- 展示正确的推导（不能消去零因子）
+- Pinpoint the steps to divide by zero
+- Explain why it is illegal
+- Show correct derivation (cannot eliminate zero factors)
 
 **Key Checks:**
-- 是否能找到错误的精确位置？
-- 是否解释了错误的原理？
+- Can the precise location of the error be found?
+- Was the rationale for the error explained?
 
 **Failure Modes:**
-- 找不到隐藏的除以零
-- 认为"证明显然有问题"但不精确指出
+- Hidden divide by zero not found
+- Thinks "there is something obviously wrong with the proof" but does not point out precisely
 
 ---
 
-## 12. 贝叶斯定理的误解
+## 12. Misunderstandings of Bayes’ Theorem
 
 **Test ID:** ADV-012  
-**Input:** "某种疾病在人群中的患病率是 0.1%。检测方法的灵敏度是 99%（有病被检出阳性的概率），特异度也是 99%（没病被检出阴性的概率）。某人检测结果为阳性，问他实际患病的概率是多少？"  
+**Input:** "The prevalence of a certain disease in the population is 0.1%. The sensitivity of the detection method is 99% (the probability of being detected as positive if the disease is present), and the specificity is also 99% (the probability of being detected as negative without the disease). When someone tests positive, what is the actual probability of having the disease?"
 **Expected Classification:** probability, Bayes_theorem, base_rate_fallacy  
 **Expected Behavior:**
-- 这不是 99%！
-- P(患病|阳性) = P(阳性|患病)P(患病) / P(阳性)
+- It's not 99%!
+- P(disease|positive) = P(positive|disease)P(disease) / P(positive)
   = 0.99 × 0.001 / (0.99×0.001 + 0.01×0.999)
   = 0.00099 / (0.00099 + 0.00999)
   = 0.00099 / 0.01098
   ≈ 0.0902 ≈ 9%
-- 尽管检测很准，但因为基础患病率极低，阳性结果中只有约 9% 真正患病
+- Although the test is very accurate, only about 9% of positive results are actually sick because the underlying prevalence is so low
 
 **Required Verification:**
-- 贝叶斯公式应用
-- 全概率公式计算 P(阳性)
-- 数值计算
+- Application of Bayesian formula
+- Total probability formula calculation P (positive)
+- Numerical calculations
 
 **Key Checks:**
-- 陷阱：很多人直接说 99%
-- 基础概率谬误
-- 贝叶斯推理
+- Trap: Many people simply say 99%
+- Basic probability fallacy
+- Bayesian inference
 
 **Failure Modes:**
-- 直接回答 99%（最常见的贝叶斯错误）
-- 公式使用错误
+- Direct answer 99% (most common Bayesian error)
+- Wrong use of formulas
 
 ---
 
-## 13. 看似不活跃的约束条件
+## 13. Constraints that appear to be inactive
 
 **Test ID:** ADV-013  
-**Input:** "在约束 x² + y² ≤ 1 下，求 f(x,y) = x + y 的最大值和最小值。"  
+**Input:** "Under the constraint x² + y² ≤ 1, find the maximum and minimum values ​​of f(x,y) = x + y."
 **Expected Classification:** optimization, constrained_optimization, boundary_check  
 **Expected Behavior:**
-- 无约束临界点：∇f = (1,1) ≠ (0,0) → 无内部临界点
-- 约束在边界 x²+y²=1 上达到极值
-- Lagrange 乘子法或直接解：x=cos θ, y=sin θ → f(θ)=cos θ + sin θ = √2 sin(θ+π/4)
-- 最大值 √2（θ=π/4），最小值 -√2（θ=5π/4）
+- Unconstrained critical point: ∇f = (1,1) ≠ (0,0) → No internal critical point
+- The constraint reaches its extreme value on the boundary x²+y²=1
+- Lagrange multiplier method or direct solution: x=cos θ, y=sin θ → f(θ)=cos θ + sin θ = √2 sin(θ+π/4)
+- Maximum value √2 (θ=π/4), minimum value -√2 (θ=5π/4)
 
 **Required Verification:**
-- 检查内部是否有临界点
-- 边界上求极值
-- 约束是否活跃
+- Check whether there are critical points inside
+- Find the extreme value on the boundary
+- Whether the constraint is active
 
 **Key Checks:**
-- 陷阱：可能误以为约束不活跃
-- 边界分析
+- Pitfall: Possibility of mistakenly thinking that a constraint is inactive
+- Boundary analysis
 
 **Failure Modes:**
-- 忽略约束条件
-- 不用 Lagrange 乘子法或参数化
+- Ignore constraints
+- No Lagrange multiplier method or parameterization required
 
 ---
 
-## 14. 参数方程中不完整的分类讨论
+## 14. Discussion of incomplete classification in parametric equations
 
 **Test ID:** ADV-014  
-**Input:** "解关于 x 的方程 \(ax^2 + (a-1)x - 1 = 0\)（a 为参数）。"  
+**Input:** "Solve the equation \(ax^2 + (a-1)x - 1 = 0\) for x (a is a parameter)."
 **Expected Classification:** algebra, parametric_equation, case_analysis  
 **Expected Behavior:**
-- 情况 1：a = 0 → 方程变为 -x - 1 = 0 → x = -1
-- 情况 2：a ≠ 0 → 二次方程
-  - 判别式：Δ = (a-1)² + 4a = a² - 2a + 1 + 4a = a² + 2a + 1 = (a+1)² ≥ 0
+- Case 1: a = 0 → equation becomes -x - 1 = 0 → x = -1
+- Case 2: a ≠ 0 → quadratic equation
+- Discriminant: Δ = (a-1)² + 4a = a² - 2a + 1 + 4a = a² + 2a + 1 = (a+1)² ≥ 0
   - x = [-(a-1) ± |a+1|] / (2a)
-  - 子情况 2a：a+1 ≥ 0（a ≥ -1 且 a≠0）
+- Subcase 2a: a+1 ≥ 0 (a ≥ -1 and a≠0)
     - x₁ = -(a-1+a+1)/(2a) = -1
     - x₂ = -(a-1-a-1)/(2a) = 2/(2a) = 1/a
-  - 子情况 2b：a+1 < 0（a < -1）
+- Subcase 2b: a+1 < 0 (a < -1)
     - x₁ = 1/a
     - x₂ = -1
 
 **Required Verification:**
-- a=0 的处理
-- a≠0 的情况覆盖
-- 判别式的含参分析
-- 讨论的完整性
+- Processing of a=0
+- The case of a≠0 is covered
+- Parametric analysis of discriminant
+- Completeness of discussion
 
 **Key Checks:**
-- 陷阱：忘记 a=0 时方程退化
-- 判别式中的 a 的处理
-- 分类的完整性
+- Pitfall: forgetting that the equation degenerates when a=0
+- Processing of a in the discriminant
+- Completeness of classification
 
 **Failure Modes:**
-- 忘记 a=0 的情况
-- 判别式开方时忘记绝对值
-- 分类讨论不完整
+- Forget the case a=0
+- Forget the absolute value when taking the square root of the discriminant
+- Classification discussion is incomplete
 
 ---
 
-## 15. 答案正确但理由错误
+## 15. Correct answer but wrong reason
 
 **Test ID:** ADV-015  
-**Input:** "学生回答：因为 \(\sqrt{16} = \pm 4\)。这个说法对吗？"  
+**Input:** "Student answer: Because \(\sqrt{16} = \pm 4\). Is this statement correct?"
 **Expected Classification:** arithmetic, notation, conceptual_error  
 **Expected Behavior:**
-- 不对！
-- 符号 √(16) 在实数范围内**定义为**非负平方根，即 4
-- 方程 x² = 16 的解是 x = ±4，但这和 √16 = ±4 不同
-- √ 符号是函数，定义为主平方根（非负的那个）
-- 如果学生说 √16 = ±4，这是符号的误用
+- No!
+- The symbol √(16) is defined in the real range as a non-negative square root, which is 4
+- The solution to the equation x² = 16 is x = ±4, but this is different from √16 = ±4
+- The √ symbol is a function, defined as the principal square root (the non-negative one)
+- If a student says √16 = ±4, this is a misuse of symbols
 
 **Required Verification:**
-- 区分 √ 符号和"平方根"概念的区别
-- 指出学生的概念混淆
+- Distinguish the difference between the √ symbol and the concept of "square root"
+- Point out student confusion
 
 **Key Checks:**
-- 是否理解 √ 符号的准确定义？
-- 是否能区分开方运算和方程求解？
+- Do you understand the precise definition of the √ symbol?
+- Can you differentiate between squaring operations and equation solving?
 
 **Failure Modes:**
-- 同意学生说法
-- 未能清晰区分两个概念
+- Agree with the student's statement
+- Failure to clearly differentiate between the two concepts
 
 ---
 
-## 对抗性测试用例汇总
+## Summary of adversarial test cases
 
-| Test ID | 陷阱类型 | 风险等级 | 核心防御 |
+| Test ID | Trap Type | Risk Level | Core Defense |
 |---------|----------|----------|----------|
-| ADV-001 | 平方引入增根 | 高 | 回代检验 |
-| ADV-002 | 除以零 | 高 | 定义域检查 |
-| ADV-003 | 符号表陷阱 | 高 | 逐区间检验 |
-| ADV-004 | 隐藏定义域 | 中 | 逐条件分析 |
-| ADV-005 | L'Hôpital 滥用 | 高 | 条件验证 |
-| ADV-006 | 不可能图形 | 高 | 三角不等式 |
-| ADV-007 | 独立性直觉陷阱 | 中 | 公式计算 |
-| ADV-008 | 换元定义域 | 中 | 区间变换 |
-| ADV-009 | 直觉与公式矛盾 | 中 | 严格计算 |
-| ADV-010 | 矩阵维度陷阱 | 中 | 维度检查 |
-| ADV-011 | 隐藏除以零 | 高 | 代数审查 |
-| ADV-012 | 贝叶斯谬误 | 高 | 基础概率 |
-| ADV-013 | 约束活跃性 | 中 | 边界分析 |
-| ADV-014 | 参变量分类不完 | 高 | 完整讨论 |
-| ADV-015 | 概念混淆 | 中 | 定义准确性 |
+| ADV-001 | Square introduction of increasing roots | High | Back-substitution test |
+| ADV-002 | Divide by Zero | High | Domain Check |
+| ADV-003 | Symbol table trap | High | Interval-by-interval testing |
+| ADV-004 | Hidden domain | Medium | Condition-by-condition analysis |
+| ADV-005 | L'Hôpital Abuse | High | Condition Verification |
+| ADV-006 | Impossible Figure | Height | Triangle Inequality |
+| ADV-007 | Independence Intuition Trap | Medium | Formula Calculation |
+| ADV-008 | Change domain | Medium | Interval transformation |
+| ADV-009 | Conflict between intuition and formula | Medium | Strict calculation |
+| ADV-010 | Matrix Dimension Trap | Medium | Dimension Check |
+| ADV-011 | Hidden division by zero | High | Algebra Review |
+| ADV-012 | Bayesian Fallacy | High | Base Probability |
+| ADV-013 | Constraint Liveness | Medium | Boundary Analysis |
+| ADV-014 | Incomplete parameter classification | High | Full discussion |
+| ADV-015 | Concept Confusion | Medium | Definition Accuracy |
diff --git a/tests/test_cases.md b/tests/test_cases.md
index b28f528..a433e51 100644
--- a/tests/test_cases.md
+++ b/tests/test_cases.md
@@ -1,892 +1,892 @@
-# Math.skill 综合测试用例
+# Math.skill comprehensive test case
 
-> 本文档包含 30 个覆盖所有数学类别和技能维度的综合测试用例。
-> 每个测试用例用于评估 Math.skill 的正确性、推理透明度、严谨性、验证质量等。
+> This document contains 30 comprehensive test cases covering all mathematics categories and skill dimensions.
+> Each test case is used to evaluate Math.skill's correctness, reasoning transparency, rigor, verification quality, etc.
 
 ---
 
-## 1. 简单算术：小数乘除混合运算
+## 1. Simple arithmetic: decimal multiplication and division mixed operations
 
 **Test ID:** TC-ARITH-001  
-**Input:** "计算 3.14 × 2.5 ÷ 0.5"  
+**Input:** "Calculate 3.14 × 2.5 ÷ 0.5"
 **Expected Classification:** calculation
-**Expected Behavior:** 逐步计算乘法与除法，给出精确结果 15.7  
+**Expected Behavior:** Calculate multiplication and division step by step, giving accurate results 15.7
 **Required Verification:**
-- 验证 15.7 × 0.5 = 7.85，7.85 ÷ 2.5 = 3.14（逆运算验证）
-- 检查中间步骤是否保持足够精度
+- Verify that 15.7 × 0.5 = 7.85, 7.85 ÷ 2.5 = 3.14 (inverse operation verification)
+- Check whether intermediate steps maintain sufficient accuracy
 
 **Key Checks:**
-- 中间步骤不应舍入过早
-- 最终结果应精确
+- Intermediate steps should not be rounded too early
+- The final result should be accurate
 
 **Failure Modes:**
-- 中间舍入导致误差累积
-- 计算器模式，不展示步骤
+- Intermediate rounding leads to error accumulation
+- Calculator mode, no steps shown
 
 ---
 
-## 2. 代数化简与定义域限制
+## 2. Algebraic simplification and domain restrictions
 
 **Test ID:** TC-ALG-001  
-**Input:** "化简表达式 \(\frac{x^2 - 4}{x^2 + x - 6}\) 并说明定义域限制"  
+**Input:** "Simplify the expression \(\frac{x^2 - 4}{x^2 + x - 6}\) and state the domain restrictions"
 **Expected Classification:** algebra_simplification
 **Expected Behavior:**
-- 因式分解分子 \( (x-2)(x+2) \)
-- 因式分解分母 \( (x+3)(x-2) \)
-- 约去 \( (x-2) \)，得 \( \frac{x+2}{x+3} \)
-- 声明 \( x \neq 2, x \neq -3 \)（即使约去后 x=2 处的原函数无定义）
+- Factor the numerator \( (x-2)(x+2) \)
+- Factoring denominator \( (x+3)(x-2) \)
+- About \( (x-2) \), we get \( \frac{x+2}{x+3} \)
+- Statement \( x \neq 2, x \neq -3 \) (even if the original function at x=2 is undefined after deducting it)
 
 **Required Verification:**
-- 代入 x=2 检验原函数无定义
-- 代入 x=1 检验化简前后值相等
-- 代入 x=0 检验化简前后值相等
+- Substitute x=2 to check that the original function is undefined
+- Substitute x=1 to test that the values ​​before and after simplification are equal
+- Substitute x=0 to test that the values ​​before and after simplification are equal
 
 **Key Checks:**
-- **关键陷阱**：约去 (x-2) 后是否仍然声明 x≠2
-- 分母因式分解是否正确
+- **Key Trap**: Whether to still declare x≠2 after removing (x-2)
+- Is the denominator factorization correct?
 
 **Failure Modes:**
-- 忘记声明 x≠2 的定义域限制（最常见的错误）
-- 分母因式分解错误
+- Forgot to declare the domain restriction of x≠2 (most common mistake)
+- Denominator factoring error
 
 ---
 
-## 3. 二次方程：判别式为零
+## 3. Quadratic equation: discriminant is zero
 
 **Test ID:** TC-ALG-002  
-**Input:** "解方程 \(x^2 - 6x + 9 = 0\)"  
+**Input:** "Solve the equation \(x^2 - 6x + 9 = 0\)"
 **Expected Classification:** equation_solving
 **Expected Behavior:**
-- 计算判别式 \( \Delta = 36 - 36 = 0 \)
-- 指出判别式为零意味着一个重根
-- 得到 \( x = 3 \)（二重根）
-- 使用求根公式或完全平方：\( (x-3)^2 = 0 \)
+- Calculate the discriminant \( \Delta = 36 - 36 = 0 \)
+- Point out that a discriminant of zero means a multiple root
+- get \( x = 3 \) (double root)
+- Use the root formula or the perfect square: \( (x-3)^2 = 0 \)
 
 **Required Verification:**
-- 回代验证：\( 3^2 - 6 \times 3 + 9 = 9 - 18 + 9 = 0 \)
-- 显式说明判别式为零 = 一个实数根（而非无根）
+- Back-substitution verification: \( 3^2 - 6 \times 3 + 9 = 9 - 18 + 9 = 0 \)
+- Explicitly state that the discriminant is zero = a real root (rather than no root)
 
 **Key Checks:**
-- 不应说"无解"
-- 应说明是二重根
+- Don't say "no solution"
+- It should be stated that it is a double root
 
 **Failure Modes:**
-- 将 Δ=0 误判为无实数解
-- 只说 x=3 而不提二重根的性质
+- Misjudgment of Δ=0 as no real solution
+- Just say x=3 without mentioning the properties of the double root
 
 ---
 
-## 4. 分式方程与增根
+## 4. Fractional equations and increasing roots
 
 **Test ID:** TC-ALG-003  
-**Input:** "解方程 \(\frac{x}{x-2} + \frac{1}{x} = \frac{4}{x(x-2)}\)"  
+**Input:** "Solve the equation \(\frac{x}{x-2} + \frac{1}{x} = \frac{4}{x(x-2)}\)"
 **Expected Classification:** equation_solving
 **Expected Behavior:**
-- 乘以公分母 \( x(x-2) \)：\( x^2 + (x-2) = 4 \)
-- 得 \( x^2 + x - 6 = 0 \)
-- 解得 \( x = 2 \) 或 \( x = -3 \)
-- **验证**：x=2 使分母为零，是增根
-- 最终解：\( x = -3 \)
+- Multiply by the common denominator \( x(x-2) \): \( x^2 + (x-2) = 4 \)
+- Get \( x^2 + x - 6 = 0 \)
+- Solve for \( x = 2 \) or \( x = -3 \)
+- **Verification**: x=2 makes the denominator zero, which is increasing the root
+- Final solution: \( x = -3 \)
 
 **Required Verification:**
-- 显式回代 x=2 到原方程，分母 \( x-2 = 0 \) 无定义
-- 回代 x=-3：左 = \( \frac{-3}{-5} + \frac{1}{-3} = 0.6 - 0.333... = 0.2666... \) = 右
+- Explicitly back-substitute x=2 into the original equation, the denominator \( x-2 = 0 \) is undefined
+- Back substitution x=-3: left = \( \frac{-3}{-5} + \frac{1}{-3} = 0.6 - 0.333... = 0.2666... ​​\) = right
 
 **Key Checks:**
-- **必须**识别增根
-- 必须在解题后显式验根
-- 不应跳过定义域分析
+- **MUST** identify root augmentation
+- Must explicitly verify the root after solving the problem
+- Domain analysis should not be skipped
 
 **Failure Modes:**
-- 直接给出两个解，未检查增根（最严重错误）
-- 未说明为什么 x=2 不是解
+- Give two solutions directly without checking for increasing roots (the most serious error)
+- Doesn't explain why x=2 is not a solution
 
 ---
 
-## 5. 线性方程组：唯一解
+## 5. System of linear equations: unique solution
 
 **Test ID:** TC-LIN-001  
-**Input:** "解方程组 \begin{cases} 2x + 3y = 8 \\ 4x - y = 2 \end{cases}"  
+**Input:** "Solve the system of equations \begin{cases} 2x + 3y = 8 \\ 4x - y = 2 \end{cases}"
 **Expected Classification:** equation_solving, system_of_equations
 **Expected Behavior:**
-- 使用消元法或代入法
-- 解出 \( x = 1, y = 2 \)
-- 指出方程组的秩为 2，有唯一解
+- Use elimination method or substitution method
+- Solve \( x = 1, y = 2 \)
+- Show that the system of equations has rank 2 and has a unique solution
 
 **Required Verification:**
-- 回代第一条方程：\( 2(1) + 3(2) = 2 + 6 = 8 \) ✓
-- 回代第二条方程：\( 4(1) - 2 = 2 \) ✓
-- 用矩阵方法交叉验证（可选）
+- Back-substitute the first equation: \( 2(1) + 3(2) = 2 + 6 = 8 \) ✓
+- Back-substitute the second equation: \( 4(1) - 2 = 2 \) ✓
+- Cross-validation with matrix method (optional)
 
 **Key Checks:**
-- 两个解都应满足两个方程
-- 如果有唯一解，显式说明
+- Both solutions should satisfy both equations
+- If there is a unique solution, state it explicitly
 
 **Failure Modes:**
-- 消元过程中算术错误
-- 忘记验证
+- Arithmetic errors during elimination
+- Forgot to verify
 
 ---
 
-## 6. 不等式与符号表
+## 6. Inequalities and symbol tables
 
 **Test ID:** TC-INEQ-001  
-**Input:** "解不等式 \(x^3 - 3x^2 - 4x + 12 > 0\)"  
+**Input:** "Solve the inequality \(x^3 - 3x^2 - 4x + 12 > 0\)"
 **Expected Classification:** equation_solving, inequality_solving
 **Expected Behavior:**
-- 因式分解：尝试 x=2 是根，得 \( (x-2)(x^2-x-6) = (x-2)(x-3)(x+2) \)
-- 关键点：x = -2, 2, 3
-- 制作符号表，检验四个区间
-- 给出解集 \( (-2, 2) \cup (3, +\infty) \)
+- Factoring: try x=2 as a root, we get \( (x-2)(x^2-x-6) = (x-2)(x-3)(x+2) \)
+- Key points: x = -2, 2, 3
+- Make a symbol table and check four intervals
+- gives the solution set \( (-2, 2) \cup (3, +\infty) \)
 
 **Required Verification:**
-- 在每个区间取代表值检验（如 x=-3, 0, 2.5, 4）
-- 端点检验：x=-2, 2, 3 处函数值为 0
-- 符号表逻辑是否完整
+- replace the representative test in each interval (e.g. x=-3, 0, 2.5, 4)
+- Endpoint test: the function value at x=-2, 2, 3 is 0
+- Whether the symbol table logic is complete
 
 **Key Checks:**
-- 符号表是否正确
-- 区间端点是否正确处理（开区间/闭区间）
-- 因式分解是否正确
+- Is the symbol table correct?
+- Whether the interval endpoints are handled correctly (open interval/closed interval)
+- Is the factorization correct?
 
 **Failure Modes:**
-- 因式分解错误导致符号表错误
-- 漏掉某个区间
-- 端点的开闭搞混
+- Factorization error leads to symbol table error
+- Missing a certain interval
+- The opening and closing of endpoints is confused
 
 ---
 
-## 7. 函数定义域与值域
+## 7. Function domain and value range
 
 **Test ID:** TC-FUNC-001  
-**Input:** "求函数 \(f(x) = \sqrt{4 - x^2} + \ln(x-1)\) 的定义域"  
+**Input:** "Find the domain of function \(f(x) = \sqrt{4 - x^2} + \ln(x-1)\)"
 **Expected Classification:** equation_solving, function_analysis
 **Expected Behavior:**
-- 对 \( \sqrt{4-x^2} \)：\( 4-x^2 \geq 0 \) → \( x \in [-2, 2] \)
-- 对 \( \ln(x-1) \)：\( x-1 > 0 \) → \( x > 1 \)
-- 取交集：\( x \in (1, 2] \)
+- For \( \sqrt{4-x^2} \): \( 4-x^2 \geq 0 \) → \( x \in [-2, 2] \)
+- For \( \ln(x-1) \): \( x-1 > 0 \) → \( x > 1 \)
+- Take the intersection: \( x \in (1, 2] \)
 
 **Required Verification:**
-- 端点 x=1：验证 ln(0) 无定义
-- 端点 x=2：验证 √(4-4)=0 和 ln(1)=0，f(2)=0
-- 区间内取 x=1.5 检验
+- Endpoint x=1: Verify that ln(0) is undefined
+- Endpoint x=2: Verify √(4-4)=0 and ln(1)=0, f(2)=0
+- Take x=1.5 test within the interval
 
 **Key Checks:**
-- 平方根条件是 ≥0 而不是 >0
-- 对数条件是 >0 而不是 ≥0
-- 交集是否正确
+- The square root condition is ≥0 instead of >0
+- The logarithmic condition is >0 instead of ≥0
+- Is the intersection correct?
 
 **Failure Modes:**
-- 平方根条件写成 >0，遗漏 x=2
-- 对数条件写成 ≥0
-- 不懂取交集
+- The square root condition is written as >0, leaving out x=2
+- The logarithmic condition is written as ≥0
+- Don’t know how to intersect
 
 ---
 
-## 8. 几何证明：三角形全等
+## 8. Geometric proof: triangles are congruent
 
 **Test ID:** TC-GEOM-001  
-**Input:** "如图，在 △ABC 中，AB=AC，D 是 BC 中点。求证 AD⊥BC。文字描述，无图。"  
+**Input:** "As shown in the figure, in △ABC, AB=AC, D is the midpoint of BC. Prove AD⊥BC. Text description, no picture."
 **Expected Classification:** geometry, proof
 **Expected Behavior:**
-- 利用等腰三角形性质
-- 已知 AB=AC，D 是 BC 中点
-- 在 △ABD 和 △ACD 中：AB=AC, BD=CD, AD=AD
-- 由 SSS 得 △ABD ≅ △ACD
-- 故 ∠ADB = ∠ADC
-- 又 ∠ADB + ∠ADC = 180°
-- 所以 ∠ADB = ∠ADC = 90°，即 AD⊥BC
+- Utilize the properties of isosceles triangles
+- It is known that AB=AC, D is the midpoint of BC
+- In △ABD and △ACD: AB=AC, BD=CD, AD=AD
+- Obtain △ABD ≅ △ACD from SSS
+- Therefore ∠ADB = ∠ADC
+- Also ∠ADB + ∠ADC = 180°
+- So ∠ADB = ∠ADC = 90°, which is AD⊥BC
 
 **Required Verification:**
-- 全等的 SSS 条件是否完整（三条边对应相等）
-- 逻辑链是否完整：全等 → 对应角相等 → 邻补角 → 直角
-- 是否有跳步
+- Is the congruent SSS condition complete (three sides are equal)
+- Whether the logical chain is complete: congruent → corresponding angles are equal → adjacent supplementary angles → right angles
+- Whether there is a skip step
 
 **Key Checks:**
-- SSS 全等条件正确应用
-- 等腰三角形性质的推理是否正确
-- 最终结论的逻辑完备性
+- SSS congruence conditions are applied correctly
+- Is the reasoning about the properties of an isosceles triangle correct?
+- The logical completeness of the final conclusion
 
 **Failure Modes:**
-- 直接声称"等腰三角形底边中线即高"而无证明（循环论证）
-- SSS 条件陈述不全
+- Directly claiming that "the midline of the base of an isosceles triangle is the height" without proof (circular argument)
+- SSS condition statement is incomplete
 
 ---
 
-## 9. 三角恒等式
+## 9. Trigonometric Identities
 
 **Test ID:** TC-TRIG-001  
-**Input:** "证明 \(\frac{\sin 2x}{1 + \cos 2x} = \tan x\)"  
+**Input:** "Prove \(\frac{\sin 2x}{1 + \cos 2x} = \tan x\)"
 **Expected Classification:** trigonometry, proof
 **Expected Behavior:**
-- 使用倍角公式：\( \sin 2x = 2\sin x \cos x \)
+- Use the double angle formula: \( \sin 2x = 2\sin x \cos x \)
 - \( \cos 2x = 2\cos^2 x - 1 \)
-- 左 = \( \frac{2\sin x \cos x}{1 + (2\cos^2 x - 1)} = \frac{2\sin x \cos x}{2\cos^2 x} \)
-- 约分：\( \frac{\sin x}{\cos x} = \tan x \) = 右
-- 定义域说明：\( \cos 2x \neq -1 \) 即 \( x \neq \frac{\pi}{2} + k\pi \) 且 \( \cos x \neq 0 \)
+- Left = \( \frac{2\sin x \cos x}{1 + (2\cos^2 x - 1)} = \frac{2\sin x \cos x}{2\cos^2 x} \)
+- Reduction: \( \frac{\sin x}{\cos x} = \tan x \) = right
+- Domain description: \( \cos 2x \neq -1 \) that is \( x \neq \frac{\pi}{2} + k\pi \) and \( \cos x \neq 0 \)
 
 **Required Verification:**
-- 从右到左倒推验证
-- 代入特殊值检验：x=π/4 时左=1，右=1
-- 定义域说明
+- Verify backwards from right to left
+- Substitute special values ​​to test: when x=π/4, left=1, right=1
+- Domain description
 
 **Key Checks:**
-- 倍角公式是否正确
-- 化简过程代数错误概率
-- 是否说明定义域
+- Is the angle multiplier formula correct?
+- Algebraic error probability of simplification process
+- Whether to specify the domain
 
 **Failure Modes:**
-- 倍角公式符号错误（sin2x 或 cos2x）
-- 忘记说明定义域
-- 代数化简错误
+- Wrong sign of multiple angle formula (sin2x or cos2x)
+- Forgot to specify the domain
+- algebraic simplification errors
 
 ---
 
-## 10. 等差数列：求通项与求和
+## 10. Arithmetic sequence: finding general terms and summation
 
 **Test ID:** TC-SEQ-001  
-**Input:** "等差数列的前 5 项和为 30，第 10 项为 25。求通项公式和前 n 项和公式。"  
+**Input:** "The sum of the first 5 terms of the arithmetic sequence is 30, and the 10th term is 25. Find the general formula and the sum formula of the first n terms."
 **Expected Classification:** equation_solving, sequence, system_of_equations
 **Expected Behavior:**
-- 设首项 a，公差 d
+- Let the leading term a be the tolerance d
 - S₅ = \( \frac{5}{2}(2a + 4d) = 30 \) → \( 2a + 4d = 12 \)
 - a₁₀ = a + 9d = 25
-- 解方程组：得 a = -4, d = 3.222...? 
-  - 等一下，验证：2(-4) + 4d = 12 → -8 + 4d = 12 → d = 5
+- Solving the system of equations: we get a = -4, d = 3.222...?
+- Wait, verify: 2(-4) + 4d = 12 → -8 + 4d = 12 → d = 5
   - a + 9d = -4 + 45 = 41 ≠ 25
-  - 看来不一致。重新：S₅ = 5a + 10d = 30 → a + 2d = 6
+- Seems inconsistent. Re: S₅ = 5a + 10d = 30 → a + 2d = 6
   - a + 9d = 25
-  - 相减：7d = 19 → d = 19/7 ≈ 2.714...
+- Subtraction: 7d = 19 → d = 19/7 ≈ 2.714...
   - a = 6 - 2(19/7) = 6 - 38/7 = 4/7
-  - 验证：S₅ = 5(4/7) + 10(19/7) = 20/7 + 190/7 = 210/7 = 30 ✓
+- Verification: S₅ = 5(4/7) + 10(19/7) = 20/7 + 190/7 = 210/7 = 30 ✓
 
 **Required Verification:**
-- 验证 a₁₀：4/7 + 9(19/7) = 4/7 + 171/7 = 175/7 = 25 ✓
-- 通项：\( a_n = \frac{4}{7} + (n-1)\frac{19}{7} \)
-- 前 n 项和：\( S_n = \frac{n}{2}(2 \cdot \frac{4}{7} + (n-1)\frac{19}{7}) \)
+- Verify a₁₀: 4/7 + 9(19/7) = 4/7 + 171/7 = 175/7 = 25 ✓
+- General term: \( a_n = \frac{4}{7} + (n-1)\frac{19}{7} \)
+- Sum of the first n terms: \( S_n = \frac{n}{2}(2 \cdot \frac{4}{7} + (n-1)\frac{19}{7}) \)
 
 **Key Checks:**
-- 方程组列写是否正确
-- 解出的 a 和 d 是否正确
-- 验证步骤完整性
+- Is the system of equations written correctly?
+- Are the solutions for a and d correct?
+- Verify step completeness
 
 **Failure Modes:**
-- 等差数列公式错误
-- 方程组求解错误
+- Arithmetic sequence formula error
+- Error in solving system of equations
 
 ---
 
-## 11. 组合数学：排列限制
+## 11. Combinatorial Mathematics: Permutation Restrictions
 
 **Test ID:** TC-COMB-001  
-**Input:** "从 5 个男生和 4 个女生中选 4 人组成委员会，要求至少包含 1 名女生。有多少种不同的选法？"  
+**Input:** "Select 4 people from 5 boys and 4 girls to form a committee, including at least 1 girl. How many different selection methods are there?"
 **Expected Classification:** combinatorics
 **Expected Behavior:**
-- 方法一（补集法）：总数 C(9,4) - 全男生 C(5,4)
+- Method 1 (complement method): Total number C(9,4) - All boys C(5,4)
   - C(9,4) = 126, C(5,4) = 5 → 121
-- 方法二（分情况）：1女3男 + 2女2男 + 3女1男 + 4女0男
+- Method 2 (different situations): 1 woman, 3 men + 2 women, 2 men + 3 women, 1 man + 4 women, 0 men
   - C(4,1)C(5,3) + C(4,2)C(5,2) + C(4,3)C(5,1) + C(4,4)C(5,0)
   - 4×10 + 6×10 + 4×5 + 1×1 = 40+60+20+1 = 121
 
 **Required Verification:**
-- 两种方法结果一致（121）
-- 逐个验证各分情况的计算
-- 确保无遗漏或重复
+- The results of both methods are consistent (121)
+- Verify the calculations of each situation one by one
+- Make sure there are no omissions or duplications
 
 **Key Checks:**
-- 补集法的应用是否正确
-- C(n,k) 计算是否正确
-- 是否考虑了所有情况
+- Is the complement method applied correctly?
+- Is the calculation of C(n,k) correct?
+- Have all circumstances been considered?
 
 **Failure Modes:**
-- 直接加而不考虑组合
-- 补集法中总数减去的方式有误
-- 分情况讨论遗漏
+- Add directly without considering the combination
+- The way of subtracting the total in the complement method is wrong
+- Discuss omissions on a case-by-case basis
 
 ---
 
-## 12. 概率：条件概率
+## 12. Probability: Conditional probability
 
 **Test ID:** TC-PROB-001  
-**Input:** "袋中有 3 个红球和 2 个蓝球。不放回地摸两次。已知第一次摸到红球，求第二次也摸到红球的概率。"  
+**Input:** "There are 3 red balls and 2 blue balls in the bag. Touch it twice without replacing it. Knowing that the red ball was touched the first time, find the probability that the red ball is also touched the second time."
 **Expected Classification:** probability_statistics
 **Expected Behavior:**
-- 第一次摸到红球后，剩余 2 红 2 蓝共 4 球
-- P(第二次红 | 第一次红) = 2/4 = 1/2
-- 也可用定义：P(A∩B)/P(A) = (3/5 × 2/4) / (3/5) = (6/20)/(3/5) = 6/20 × 5/3 = 1/2
+- After touching the red ball for the first time, there are 4 balls left, 2 red and 2 blue.
+- P(second red | first red) = 2/4 = 1/2
+- Also available definition: P(A∩B)/P(A) = (3/5 × 2/4) / (3/5) = (6/20)/(3/5) = 6/20 × 5/3 = 1/2
 
 **Required Verification:**
-- 两种方法结果一致
-- 说明条件概率的直观意义
-- 不放回抽样的影响
+- The results of both methods are consistent
+- Explain the intuitive meaning of conditional probability
+- Effect of sampling without replacement
 
 **Key Checks:**
-- 条件概率定义的正确应用
-- 不放回的影响是否正确体现
-- 结果 1/2 是否合理（直观检验）
+- Correct application of the definition of conditional probability
+- Whether the impact of not replacing is reflected correctly
+- Is the result 1/2 reasonable (visual inspection)
 
 **Failure Modes:**
-- 误用放回抽样公式
-- 条件概率定义混淆 P(A|B) 和 P(B|A)
+- Misuse of replacement sampling formula
+- Conditional probability definition confuses P(A|B) and P(B|A)
 
 ---
 
-## 13. 微积分：L'Hôpital 法则求极限
+## 13. Calculus: L'Hôpital's rule for finding limits
 
 **Test ID:** TC-CALC-001  
-**Input:** "求极限 \(\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}\)"  
+**Input:** "Find the limit \(\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}\)"
 **Expected Classification:** limit
 **Expected Behavior:**
-- 代入 x→0：分子 1-1-0=0，分母 0，0/0 型
-- 第一次 L'Hôpital：\( \lim \frac{e^x - 1}{2x} \)，仍是 0/0
-- 第二次 L'Hôpital：\( \lim \frac{e^x}{2} = \frac{1}{2} \)
-- 或者展开 \( e^x = 1 + x + x^2/2 + x^3/6 + ... \)
+- Substitute x→0: numerator 1-1-0=0, denominator 0, 0/0 type
+- First L'Hôpital: \( \lim \frac{e^x - 1}{2x} \), still 0/0
+- Second time L'Hôpital: \( \lim \frac{e^x}{2} = \frac{1}{2} \)
+- Or expand \( e^x = 1 + x + x^2/2 + x^3/6 + ... \)
 
 **Required Verification:**
-- 验证每次 L'Hôpital 前确实是 0/0 或 ∞/∞ 不定式
-- 用泰勒展开交叉验证：\( e^x - 1 - x = x^2/2 + x^3/6 + ... \)，除以 x^2 → 1/2
-- 用数值验证：x=0.001 时近似值为 0.500000167...
+- Verify that each time L'Hôpital is preceded by 0/0 or ∞/∞ infinitive
+- Cross-validation with Taylor expansion: \( e^x - 1 - x = x^2/2 + x^3/6 + ... \), divided by x^2 → 1/2
+- Verify numerically: when x=0.001 the approximate value is 0.500000167...
 
 **Key Checks:**
-- 两次 L'Hôpital 的条件是否满足
-- 最终答案 1/2
-- 交叉验证方法
+- Whether the conditions of L'Hôpital are met twice
+- Final Answer 1/2
+- Cross-validation method
 
 **Failure Modes:**
-- 不检查 L'Hôpital 条件就应用
-- 只做一次 L'Hôpital 就停止
+- Apply without checking L'Hôpital conditions
+- Do L'Hôpital just once and stop
 
 ---
 
-## 14. 导数应用：最优化
+## 14. Derivative Application: Optimization
 
 **Test ID:** TC-CALC-002  
-**Input:** "用一块边长为 12cm 的正方形铁皮，在四个角各剪去一个相同的小正方形，然后折成一个无盖长方体盒子。问剪去的小正方形边长为多少时，盒子的容积最大？"  
+**Input:** "Use a square piece of iron sheet with a side length of 12cm, cut off an identical small square at each of the four corners, and then fold it into a rectangular box without a lid. What is the maximum volume of the box when the side length of the small square is cut off?"
 **Expected Classification:** limit, optimization, differentiation
 **Expected Behavior:**
-- 设剪去小正方形边长为 x (0 < x < 6)
-- 盒子底面边长 12-2x，高 x
-- 体积 V(x) = x(12-2x)² = 4x(6-x)²
+- Let the side length of the small square be cut off as x (0 < x < 6)
+- The bottom side of the box is 12-2x, and the height is x
+- Volume V(x) = x(12-2x)² = 4x(6-x)²
 - V'(x) = 4(6-x)² - 8x(6-x) = 4(6-x)[(6-x) - 2x] = 4(6-x)(6-3x)
-- 令 V'(x) = 0：x=6 或 x=2
-- x=6 对应体积 0（最小值），x=2 对应最大值
-- 验证：V''(2) < 0
-- 最大体积 V(2) = 2 × 8² = 128 cm³
+- Let V'(x) = 0: x=6 or x=2
+- x=6 corresponds to volume 0 (minimum), x=2 corresponds to maximum
+- Verification: V''(2) < 0
+- Maximum volume V(2) = 2 × 8² = 128 cm³
 
 **Required Verification:**
-- 给出 V'(x) 的详细推导
-- 临界点的二阶导数检验
-- 端点检验：x→0 时 V→0，x→6 时 V→0
-- 答案的合理性检验
+- Give a detailed derivation of V'(x)
+- Second derivative test for critical points
+- Endpoint test: when x→0, V→0, when x→6, V→0
+- Reasonability test of answers
 
 **Key Checks:**
-- 导数计算是否正确
-- 极值判断是正确的最大值而非最小值
-- 实际意义：边长不能超过 6
+- Is the derivative calculation correct?
+- The extreme value judgment is the correct maximum value rather than the minimum value
+- Practical meaning: Side length cannot exceed 6
 
 **Failure Modes:**
-- 忘记检查二阶导数
-- 乘积求导错误
-- 不懂最优化的基本步骤
+- Forgot to check the second derivative
+- product derivation error
+- Not understanding the basic steps of optimization
 
 ---
 
-## 15. 换元积分法
+## 15. Substitution integral method
 
 **Test ID:** TC-CALC-003  
-**Input:** "计算不定积分 \(\int x\sqrt{2x+1}\,dx\)"  
+**Input:** "Calculate indefinite integral \(\int x\sqrt{2x+1}\,dx\)"
 **Expected Classification:** limit, integration
 **Expected Behavior:**
-- 换元：令 u = 2x+1，则 x = (u-1)/2, dx = du/2
+- Substitution: Let u = 2x+1, then x = (u-1)/2, dx = du/2
 - \( \int \frac{u-1}{2} \cdot \sqrt{u} \cdot \frac{du}{2} = \frac{1}{4} \int (u-1)u^{1/2} du \)
 - = \( \frac{1}{4} \int (u^{3/2} - u^{1/2}) du \)
 - = \( \frac{1}{4} \left( \frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2} \right) + C \)
 - = \( \frac{1}{10}(2x+1)^{5/2} - \frac{1}{6}(2x+1)^{3/2} + C \)
-- 可进一步化简为 \( \frac{(2x+1)^{3/2}(3x-1)}{15} + C \)
+- can be further simplified to \( \frac{(2x+1)^{3/2}(3x-1)}{15} + C \)
 
 **Required Verification:**
-- 求导验证：对结果求导，检验是否等于原被积函数
-- 代入特殊值数值验算
+- Derivative verification: Derive the result and check whether it is equal to the original integrand
+- Substitute special values ​​for numerical verification
 
 **Key Checks:**
-- 换元步骤中 dx 的转换
-- 积分后的代数化简
-- 不要忘记 +C
+- Conversion of dx in the substitution step
+- Algebraic simplification after integration
+- Don't forget +C
 
 **Failure Modes:**
-- 换元时 dx 转换错误
-- 积分公式应用错误
-- 代数化简中的错误
+- dx conversion error when changing yuan
+- Incorrect integration formula application
+- Errors in algebraic simplification
 
 ---
 
-## 16. 反常积分收敛性
+## 16. Abnormal integral convergence
 
 **Test ID:** TC-CALC-004  
-**Input:** "判断反常积分 \(\int_1^\infty \frac{\ln x}{x^2}\,dx\) 的敛散性，若收敛求其值。"  
+**Input:** "Judge the convergence and divergence of the anomalous integral \(\int_1^\infty \frac{\ln x}{x^2}\,dx\), and find its value if it converges."
 **Expected Classification:** limit, integration
 **Expected Behavior:**
-- 先判断收敛性：\( \frac{\ln x}{x^2} \) 的增长阶
-  - 对任意 ε>0，ln x = o(x^ε)
-  - \( \frac{\ln x}{x^2} = \frac{\ln x}{x^{1/2}} \cdot \frac{1}{x^{3/2}} \leq \frac{C}{x^{3/2}} \)（对充分大的 x）
-  - 由 p-积分，\( \int_1^\infty \frac{1}{x^{3/2}} dx \) 收敛（p=3/2 > 1）
-  - 由比较判别法，原积分收敛
-- 计算积分值：分部积分法
-  - 令 u = ln x, dv = dx/x²
+- First determine the convergence: the growth order of \( \frac{\ln x}{x^2} \)
+- For any ε>0, ln x = o(x^ε)
+- \( \frac{\ln x}{x^2} = \frac{\ln x}{x^{1/2}} \cdot \frac{1}{x^{3/2}} \leq \frac{C}{x^{3/2}} \) (for sufficiently large x)
+- By p-integration, \( \int_1^\infty \frac{1}{x^{3/2}} dx \) converges (p=3/2 > 1)
+- According to the comparative discrimination method, the original integral converges
+- Calculate integral value: Integration by parts method
+- Let u = ln x, dv = dx/x²
   - du = dx/x, v = -1/x
   - \( \int \frac{\ln x}{x^2} dx = -\frac{\ln x}{x} + \int \frac{1}{x^2} dx = -\frac{\ln x}{x} - \frac{1}{x} + C \)
   - \( \lim_{t \to \infty} [-\frac{\ln x}{x} - \frac{1}{x}]_1^t = 0 - 0 - (-0 - (-1)) = 1 \)
 
 **Required Verification:**
-- 比较判别法的基准 p-积分选择是否合理
-- 分部积分步骤是否正确
-- 极限计算：lim(t→∞) ln(t)/t = 0
-- 数值验证：部分和近似
+- Is the selection of the benchmark p-point of the comparative discrimination method reasonable?
+- Are the partial integration steps correct?
+- Limit calculation: lim(t→∞) ln(t)/t = 0
+- Numerical verification: partial sum approximations
 
 **Key Checks:**
-- 先证明收敛再计算（对反常积分的关键步骤）
-- 比较判别法的应用
-- 分部积分的正确性
+- Prove convergence first and then calculate (key step for anomalous integration)
+- Application of comparative discrimination method
+- Correctness of integrals by parts
 
 **Failure Modes:**
-- 不验证收敛性直接计算（最常见的错误）
-- 发散但给出有限值
+- Direct calculation without verifying convergence (most common mistake)
+- diverges but gives finite values
 
 ---
 
-## 17. 多元函数：临界点与分类
+## 17. Multivariate functions: critical points and classification
 
 **Test ID:** TC-MULTI-001  
-**Input:** "求函数 \(f(x,y) = x^3 + 3xy^2 - 15x - 12y\) 的所有临界点，并分类（极大值、极小值或鞍点）。"  
+**Input:** "Find all critical points of the function \(f(x,y) = x^3 + 3xy^2 - 15x - 12y\) and classify them (maximum, minimum or saddle point)."
 **Expected Classification:** multivariable_calculus
 **Expected Behavior:**
-- 求偏导数：
+- Find partial derivatives:
   - \( f_x = 3x^2 + 3y^2 - 15 = 3(x^2 + y^2 - 5) \)
   - \( f_y = 6xy - 12 = 6(xy - 2) \)
-- 解 f_x = 0: \( x^2 + y^2 = 5 \)
-- 解 f_y = 0: \( xy = 2 \)
-- 解方程组：x² + y² = 5, xy = 2
+- Solution f_x = 0: \( x^2 + y^2 = 5 \)
+- Solve f_y = 0: \( xy = 2 \)
+- Solve the system of equations: x² + y² = 5, xy = 2
   - (x+y)² = x² + y² + 2xy = 5 + 4 = 9 → x+y = ±3
   - (x-y)² = x² + y² - 2xy = 5 - 4 = 1 → x-y = ±1
-  - 四组合：4 个临界点
+- Four combinations: 4 critical points
   - (2,1), (1,2), (-1,-2), (-2,-1)
-- Hessian 矩阵：H = [[6x, 6y], [6y, 6x]], det(H) = 36(x² - y²)
-  - (2,1): det(H)=108>0, f_{xx}=12>0 → 局部极小值
-  - (1,2): det(H)=-108<0 → 鞍点
-  - (-1,-2): det(H)=-108<0 → 鞍点
-  - (-2,-1): det(H)=108>0, f_{xx}=-12<0 → 局部极大值
+- Hessian matrix: H = [[6x, 6y], ​​[6y, 6x]], det(H) = 36(x² - y²)
+- (2,1): det(H)=108>0, f_{xx}=12>0 → local minimum
+- (1,2): det(H)=-108<0 → saddle point
+- (-1,-2): det(H)=-108<0 → saddle point
+- (-2,-1): det(H)=108>0, f_{xx}=-12<0 → local maximum
 
 **Required Verification:**
-- 方程组求解是否完整
-- Hessian 行列式分类是否正确
-- 每个临界点的 f_{xx} 符号检查
+- Whether the solution to the system of equations is complete
+- Is the Hessian determinant classification correct?
+- f_{xx} sign check for each critical point
 
 **Key Checks:**
-- 不能漏解
-- Hessian 判别法的应用
-- 区分鞍点与极值点
+- Can't miss a solution
+- Application of Hessian discriminant method
+- Distinguish between saddle points and extreme points
 
 **Failure Modes:**
-- 方程组求解漏解
-- Hessian 计算错误
-- 分类弄反（极大/极小）
+- Missing solution in solving system of equations
+- Hessian calculation error
+- The classification is reversed (very large/very small)
 
 ---
 
-## 18. 矩阵特征值问题
+## 18. Matrix eigenvalue problem
 
 **Test ID:** TC-LINALG-001  
-**Input:** "求矩阵 \(A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}\) 的特征值和特征向量。"  
+**Input:** "Find the eigenvalues ​​and eigenvectors of the matrix \(A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}\)."
 **Expected Classification:** linear_algebra
 **Expected Behavior:**
-- 特征方程：det(A - λI) = 0
+- Characteristic equation: det(A - λI) = 0
   - \( \begin{vmatrix} 2-λ & 1 & 1 \\ 1 & 2-λ & 1 \\ 1 & 1 & 2-λ \end{vmatrix} = 0 \)
-  - 计算行列式：\( (2-λ)^3 + 1 + 1 - (2-λ) - (2-λ) - (2-λ) \)
+- Calculate the determinant: \( (2-λ)^3 + 1 + 1 - (2-λ) - (2-λ) - (2-λ) \)
   - = \( (2-λ)^3 - 3(2-λ) + 2 \)
-  - 令 t = 2-λ：t³ - 3t + 2 = (t-1)²(t+2) = 0
-  - t = 1 (二重) 或 t = -2
-  - λ = 1 (二重) 或 λ = 4
-- 验证迹：tr(A) = 6, 特征值和 = 1+1+4 = 6 ✓
-- λ=1 的特征向量（解 (A-I)x=0）
-- λ=4 的特征向量（解 (A-4I)x=0）
+- Let t = 2-λ: t³ - 3t + 2 = (t-1)²(t+2) = 0
+- t = 1 (double) or t = -2
+- λ = 1 (double) or λ = 4
+- Verification trace: tr(A) = 6, sum of eigenvalues ​​= 1+1+4 = 6 ✓
+- Eigenvector of λ=1 (solution (A-I)x=0)
+- Eigenvector of λ=4 (solution (A-4I)x=0)
 
 **Required Verification:**
-- 迹等于特征值之和
-- 行列式等于特征值之积
-- 代入检验 Av=λv
+- The trace is equal to the sum of eigenvalues
+- The determinant is equal to the product of eigenvalues
+- Substitution test Av=λv
 
 **Key Checks:**
-- 行列式计算正确
-- 特征多项式因式分解
-- 特征向量的基础解系
+- Determinant calculation is correct
+- Characteristic polynomial factorization
+- Basic solution system of eigenvectors
 
 **Failure Modes:**
-- 行列式计算错误
-- 特征值漏解（三重根 vs 二重根+单根）
-- 忘记特征向量
+- Determinant calculation error
+- Missing solution for eigenvalues ​​(triple root vs double root + single root)
+- Forget about eigenvectors
 
 ---
 
-## 19. 一阶线性常微分方程
+## 19. First-order linear ordinary differential equations
 
 **Test ID:** TC-ODE-001  
-**Input:** "解微分方程 \(y' + 2xy = x\)，满足 y(0) = 1。"  
+**Input:** "Solve the differential equation \(y' + 2xy = x\) such that y(0) = 1."
 **Expected Classification:** ordinary_differential_equation
 **Expected Behavior:**
-- 识别为一阶线性 ODE：\( y' + P(x)y = Q(x) \)，P(x) = 2x, Q(x) = x
-- 积分因子：\( \mu(x) = e^{\int 2x dx} = e^{x^2} \)
-- 两边同乘：\( e^{x^2}y' + 2xe^{x^2}y = xe^{x^2} \)
-- 左边 = \( \frac{d}{dx}(e^{x^2}y) \)
-- 积分：\( e^{x^2}y = \int xe^{x^2} dx = \frac{1}{2}e^{x^2} + C \)
+- Identified as a first-order linear ODE: \( y' + P(x)y = Q(x) \), P(x) = 2x, Q(x) = x
+- Integration factor: \( \mu(x) = e^{\int 2x dx} = e^{x^2} \)
+- Multiply both sides: \( e^{x^2}y' + 2xe^{x^2}y = xe^{x^2} \)
+- Left = \( \frac{d}{dx}(e^{x^2}y) \)
+- Integral: \( e^{x^2}y = \int xe^{x^2} dx = \frac{1}{2}e^{x^2} + C \)
 - \( y = \frac{1}{2} + Ce^{-x^2} \)
-- 用 y(0) = 1：1 = 1/2 + C → C = 1/2
-- 特解：\( y = \frac{1}{2}(1 + e^{-x^2}) \)
+- Using y(0) = 1: 1 = 1/2 + C → C = 1/2
+- Special solution: \( y = \frac{1}{2}(1 + e^{-x^2}) \)
 
 **Required Verification:**
-- 代入 ODE 验证：y' = -xe^{-x^2}, y' + 2xy = -xe^{-x^2} + x(1+e^{-x^2}) = x ✓
-- 检验初始条件
+- Substitute into ODE to verify: y' = -xe^{-x^2}, y' + 2xy = -xe^{-x^2} + x(1+e^{-x^2}) = x ✓
+- Check initial conditions
 
 **Key Checks:**
-- 积分因子计算正确
-- 积分步骤
-- 初始条件代入
+- The integration factor is calculated correctly
+- Integration steps
+- Substitution of initial conditions
 
 **Failure Modes:**
-- 积分因子公式记忆错误
-- 右边积分错误
-- 忘记用初始条件求 C
+- Memory error in the integral factor formula
+- Incorrect integration on the right
+- Forgot to use initial conditions to find C
 
 ---
 
-## 20. 实分析：证明一致连续性
+## 20. Real analysis: proving consistent continuity
 
 **Test ID:** TC-ANALYSIS-001  
-**Input:** "证明函数 \(f(x) = \sqrt{x}\) 在 \([0, \infty)\) 上一致连续吗？给出证明或反例。"  
+**Input:** "Prove that the function \(f(x) = \sqrt{x}\) is uniformly continuous on \([0, \infty)\)? Give a proof or a counterexample."
 **Expected Classification:** real_analysis
 **Expected Behavior:**
-- 结论：是，√x 在 [0,∞) 一致连续
-- 证明：对任意 ε>0，取 δ = ε²
-- 若 |x-y| < δ，则 |√x - √y| = |x-y|/(√x+√y) ≤ |x-y|/√|x-y|（利用 √x+√y ≥ √|x-y|）
-  - 且当 x,y≥0 时，|√x-√y|² = |x-y| - 2√xy(1 - sgn(xy)...) ≤ |x-y|
-  - 更简洁：|√x - √y| ≤ √|x-y|（√是 1/2-Hölder 连续的）
-- 所以当 |x-y| < δ = ε² 时，|√x - √y| ≤ √δ = ε
+- Conclusion: Yes, √x is uniformly continuous in [0,∞)
+- Proof: For any ε>0, take δ = ε²
+- If |x-y| < δ, then |√x - √y| = |x-y|/(√x+√y) ≤ |x-y|/√|x-y| (using √x+√y ≥ √|x-y|)
+- And when x,y≥0, |√x-√y|² = |x-y| - 2√xy(1 - sgn(xy)...) ≤ |x-y|
+- More concise: |√x - √y| ≤ √|x-y| (√ is 1/2-Hölder continuous)
+- So when |x-y| < δ = ε², |√x - √y| ≤ √δ = ε
 
 **Required Verification:**
-- δ 是否仅依赖于 ε 而不依赖于 x（一致连续的核心）
-- 不等式的推导是否正确
-- 另一种证明：在 [0,1] 用紧致性，在 [1,∞) 用导数有界
+- whether δ depends only on ε and not on x (uniformly continuous core)
+- Is the derivation of the inequality correct?
+- Another proof: use compactness in [0,1], bounded by derivatives in [1,∞)
 
 **Key Checks:**
-- 一致连续 vs 普通连续的理解
-- δ 的选择
-- 不等式的使用
+- Understanding of consistent continuity vs. ordinary continuity
+- Choice of δ
+- Use of inequalities
 
 **Failure Modes:**
-- 混淆一致连续和普通连续
-- 错误地判定为不一致连续
-- 证明中的不等式不严密
+- Confusing consistent continuity and ordinary continuity
+- Incorrectly determined to be inconsistently continuous
+- The inequality in the proof is not rigorous
 
 ---
 
-## 21. 抽象代数：群的性质
+## 21. Abstract Algebra: Properties of Groups
 
 **Test ID:** TC-ALGEBRA-001  
-**Input:** "设 G 是一个群，证明：若对任意 g∈G 有 g² = e，则 G 是交换群。"  
+**Input:** "Assume G is a group, prove: if g² = e for any g∈G, then G is a commutative group."
 **Expected Classification:** abstract_algebra
 **Expected Behavior:**
-- 对于任意 a,b∈G，需要证明 ab = ba
-- 已知：(ab)² = e → abab = e
-- 两边左乘 a，右乘 b：a(abab)b = aeb → a²bab² = ab
-- 因 a² = b² = e：ebab = ab → ab = ab（这不对，重新推导）
-- 正确推导：(ab)² = e → abab = e
-- 左乘 a：a(abab) = a → (a²)bab = a → bab = a
-- 右乘 b：bab·b = ab → ba·b² = ab → ba = ab
-- 得证 G 是交换群
+- For any a,b∈G, it is necessary to prove ab = ba
+- Given: (ab)² = e → abab = e
+- Multiply both sides by a on the left and b on the right: a(abab)b = aeb → a²bab² = ab
+- Since a² = b² = e: ebab = ab → ab = ab (this is wrong, re-derive)
+- Correct derivation: (ab)² = e → abab = e
+- Left multiplication of a: a(abab) = a → (a²)bab = a → bab = a
+- Right multiply b: bab·b = ab → ba·b² = ab → ba = ab
+- Prove that G is a commutative group
 
 **Required Verification:**
-- 每一步的逻辑依据
-- 检查群运算的结合律是否被正确使用
-- 结论：ba = ab
+- Logical basis for each step
+- Check whether the associativity of group operations is used correctly
+- Conclusion: ba = ab
 
 **Key Checks:**
-- 结合律的运用
-- 每一步的合理性
+- Application of associative law
+- Reasonableness of every step
 
 **Failure Modes:**
-- 跳过中间步骤，"显然"的推理
-- 运算顺序搞反
+- Skipping intermediate steps, "obviously" reasoning
+- Reverse the order of operations
 
 ---
 
-## 22. 拓扑学：开集与闭集
+## 22. Topology: open sets and closed sets
 
 **Test ID:** TC-TOPOL-001  
-**Input:** "在实数集 R（配备标准拓扑）中，集合 A = [0,1) 是开集吗？是闭集吗？请解释。"  
+**Input:** "In the set R of real numbers (with standard topology), is the set A = [0,1) open? Is it closed? Please explain."
 **Expected Classification:** topology
 **Expected Behavior:**
-- 不是开集：0 ∈ A 但 0 的任何邻域 (-ε,ε) 不完全包含于 A（包含负数不在 A 中）
-- 不是闭集：极限点 1 ∉ A（1 的每个邻域与 A 相交），但 A 不包含 1
-- A = [0,1) 既不开也不闭
-- 但若在子空间拓扑 [0,1] 中，则是开的……
+- is not an open set: 0 ∈ A but any neighborhood (-ε,ε) of 0 is not completely contained in A (including negative numbers is not in A)
+- is not a closed set: the limit point 1 ∉ A (every neighborhood of 1 intersects A), but A does not contain 1
+- A = [0,1) neither open nor closed
+- But if it is in the subspace topology [0,1], it is open...
 
 **Required Verification:**
-- 开集定义的逐点否定
-- 闭集定义：补集 (-∞,0)∪[1,∞) 不是开集（1 的邻域不完全在补集中）
+- Pointwise negation of open set definition
+- Definition of closed set: the complement (-∞,0)∪[1,∞) is not an open set (the neighborhood of 1 is not entirely in the complement set)
 
 **Key Checks:**
-- 区分"非开"和"闭"
-- 区分"非闭"和"开"
-- 准确应用定义
+- Distinguish between "non-open" and "closed"
+- Distinguish between "non-closed" and "open"
+- Accurately apply definitions
 
 **Failure Modes:**
-- 误判为开集
-- 误判为闭集
-- 混淆开闭的定义
+- Misjudged as an open set
+- Misjudgment as closed set
+- Confusing the definition of opening and closing
 
 ---
 
-## 23. 数论：模运算
+## 23. Number theory: modular arithmetic
 
 **Test ID:** TC-NUMTH-001  
-**Input:** "求 3^100 mod 7 的值。"  
+**Input:** "Find the value of 3^100 mod 7."
 **Expected Classification:** number_theory
 **Expected Behavior:**
-- 利用费马小定理：3⁶ ≡ 1 (mod 7)（因为 3 和 7 互素）
+- Use Fermat's little theorem: 3⁶ ≡ 1 (mod 7) (because 3 and 7 are relatively prime)
 - 100 = 6×16 + 4
 - 3¹⁰⁰ = 3⁶·¹⁶⁺⁴ ≡ (3⁶)¹⁶ · 3⁴ ≡ 1¹⁶ · 3⁴ ≡ 3⁴ (mod 7)
 - 3⁴ = 81 ≡ 81 - 7×11 = 81 - 77 = 4 (mod 7)
-- 答案：4
+- Answer: 4
 
 **Required Verification:**
-- 费马小定理的适用性：3 和 7 互素 ✓
-- 直接计算验证：3²=9≡2, 3³=6≡-1, 3⁴=18≡4 ✓
+- Applicability of Fermat's Little Theorem: 3 and 7 are relatively prime ✓
+- Direct calculation verification: 3²=9≡2, 3³=6≡-1, 3⁴=18≡4 ✓
 - 3¹⁰⁰ ≡ 4 (mod 7)
 
 **Key Checks:**
-- 费马小定理的引用
-- 指数分解
-- 模运算步骤
+- Reference to Fermat's Little Theorem
+- Exponential decomposition
+- Modulo operation steps
 
 **Failure Modes:**
-- 忘记验证互素条件
-- 指数计算错误
+- Forgot to verify the coprime condition
+- Index calculation error
 
 ---
 
-## 24. 条件不足问题
+## 24. Insufficient conditions problem
 
 **Test ID:** TC-INSUFF-001  
-**Input:** "三角形的一条边长为 5，求面积。"  
+**Input:** "The length of one side of the triangle is 5, find the area."
 **Expected Classification:** ambiguous_or_incomplete, geometry
 **Expected Behavior:**
-- **不应直接给出答案**
-- 应指出：仅知一条边长无法确定三角形面积
-- 说明需要什么额外信息（如高、角度、其他边长等）
-- 可能给出面积的取值范围（>0 且 ≤ 通过不等式的上界）
+- **No direct answers should be given**
+- It should be pointed out that the area of ​​a triangle cannot be determined by knowing only the length of one side.
+- Specify what additional information is needed (e.g. height, angle, other side lengths, etc.)
+- Possibly give a range of values ​​for area (>0 and ≤ upper bound by inequality)
 
 **Required Verification:**
-- 是否识别出条件不足
-- 是否请求更多信息
-- 是否说明原因
+- Whether insufficient conditions are identified
+- Whether to request more information
+- Whether to explain the reason?
 
 **Key Checks:**
-- 绝对不要猜测或编造条件
-- 诚实面对不确定性和不足条件
+- Never guess or make up conditions
+- Be honest about uncertainties and shortcomings
 
 **Failure Modes:**
-- 直接给出"答案"（最严重错误）
-- 未意识到条件不足
+- Give the "answer" directly (the worst mistake)
+- Unaware of insufficient conditions
 
 ---
 
-## 25. 超出范围问题
+## 25. Out of scope problem
 
 **Test ID:** TC-OUTOFSCOPE-001  
 **Input:** "What is the meaning of life?"  
 **Expected Classification:** out_of_scope
 **Expected Behavior:**
-- 识别出这不是数学问题
-- 礼貌地说明 Math.skill 的范围是数学问题
-- 不尝试回答（即使回答 42 也是不恰当的）
+- Recognize that this is not a math problem
+- Politely explain that the scope of Math.skill is math problems
+- No attempt to answer (even answer 42 is inappropriate)
 
 **Required Verification:**
-- 是否正确识别为非数学问题
-- 回复是否专业
+- Whether it is correctly identified as a non-mathematical question
+- Is the reply professional?
 
 **Key Checks:**
-- 不能当作数学问题处理
-- 不能强行找数学角度回答
+- Cannot be treated as a math problem
+- Don’t force the answer from a mathematical perspective
 
 **Failure Modes:**
-- 尝试用数学方式回答（"从概率角度看..."）
-- 没有边界意识
+- Try to answer mathematically ("Probabilistically...")
+- No sense of boundaries
 
 ---
 
-## 26. 用户要求"仅给答案"
+## 26. User request "only answer"
 
 **Test ID:** TC-META-001  
-**Input:** "解 x²-5x+6=0，只给答案就行。"  
+**Input:** "Solve x²-5x+6=0, just give the answer."
 **Expected Classification:** equation_solving
 **Expected Behavior:**
-- 理解用户要求"仅答案"
-- 但在给出答案 (x=2, 3) 后，可以简洁地建议验证步骤
-- 或者直接给出 x=2 或 x=3
+- Understand user requirements for "answers only"
+- But after giving the answer (x=2, 3), one can succinctly suggest the verification steps
+- Or just give x=2 or x=3
 
 **Required Verification:**
-- 答案是否正确
-- 是否尊重了用户的格式偏好
-- 是否仍提供最小必要信息
-- 是否可以回溯验证
+- Is the answer correct?
+- Whether the user's format preferences are respected
+- Whether the minimum necessary information is still provided
+- Is it possible to retroactively verify
 
 **Key Checks:**
-- 用户要求仅答案时，是否需要提供最小验证
-- 权衡简洁性与正确性
+- Whether minimum verification needs to be provided when the user asks for an answer only
+- Weigh simplicity and correctness
 
 **Failure Modes:**
-- 忽略用户要求，输出冗长的完整解答
-- 完全不给验证线索，用户无法确认
+- Ignore user requests and output lengthy complete answers
+- No verification clues are given at all, users cannot confirm
 
 ---
 
-## 27. 用户要求"用初中方法"
+## 27. User request "use junior high school method"
 
 **Test ID:** TC-META-002  
-**Input:** "求函数 f(x)=x²-4x+3 在 [0,3] 上的最大值和最小值。用初中方法，不要用微积分。"  
+**Input:** "Find the maximum and minimum values ​​of the function f(x)=x²-4x+3 on [0,3]. Use the junior high school method, do not use calculus."
 **Expected Classification:** equation_solving, function_analysis
 **Expected Behavior:**
-- 用配方法：f(x) = (x-2)² - 1
-- 对称轴 x=2 在区间 [0,3] 内
-- 最小值在顶点：f(2) = -1
-- 最大值在端点（离对称轴最远）：f(0) = 3, f(3) = 0 → 最大值为 3
-- **不使用导数**
+- Use the matching method: f(x) = (x-2)² - 1
+- The axis of symmetry x=2 is in the interval [0,3]
+- The minimum is at the vertex: f(2) = -1
+- The maximum value is at the endpoint (farthest from the axis of symmetry): f(0) = 3, f(3) = 0 → the maximum value is 3
+- **Do not use derivatives**
 
 **Required Verification:**
-- 是否使用了导数？（检查组）
-- 二次函数顶点法和端点比较是否正确
-- 端点比较是否完整
+- Are derivatives used? (inspection team)
+- Are the quadratic function vertex method and endpoint comparison correct?
+- Is the endpoint comparison complete?
 
 **Key Checks:**
-- 遵守了方法限制（不用微积分）
-- 用配方法/顶点法正确求解
+- Method constraints are observed (no calculus required)
+- Solve correctly using collocation method/vertex method
 
 **Failure Modes:**
-- 使用了导数（违反方法限制）
-- 忘记比较端点
+- Derivatives used (violation of method restrictions)
+- Forgot to compare endpoints
 
 ---
 
-## 28. 错误命题需要反例
+## 28. False propositions require counterexamples
 
 **Test ID:** TC-PROOF-001  
-**Input:** "判断真假：若数列 {a_n} 和 {b_n} 都发散，则 {a_n + b_n} 一定发散。证明或举反例。"  
+**Input:** "Judge true or false: If the sequence {a_n} and {b_n} both diverge, then {a_n + b_n} must diverge. Prove or give a counterexample."
 **Expected Classification:** real_analysis, counterexample, sequence
 **Expected Behavior:**
-- 判断：假
-- 反例：a_n = n, b_n = -n，则 a_n + b_n = 0（收敛）
-- 说明：发散数列的和可能收敛
+- Judgment: False
+- Counterexample: a_n = n, b_n = -n, then a_n + b_n = 0 (convergence)
+- Note: The sum of a diverging sequence may converge
 
 **Required Verification:**
-- 反例是否正确
-- 解释为什么是反例
+- Is the counterexample correct?
+-Explain why it is a counterexample
 
 **Key Checks:**
-- 必须给出具体反例
-- 不能说"显然为真"
+- Specific counterexamples must be given
+- Cannot say "Obviously true"
 
 **Failure Modes:**
-- 错误地判断为真并"证明"
-- 反例不合理
+- Falsely judge to be true and "prove"
+- The counterexample is unreasonable
 
 ---
 
-## 29. 学生解答中的细微错误
+## 29. Minor errors in student answers
 
 **Test ID:** TC-STUDENT-001  
-**Input:** "学生解方程 √(x+5) = x-1 得到 x=4 和 x=-1，认为两个都是解。请检查学生的解答。"  
+**Input:** "The student solves the equation √(x+5) = x-1 and gets x=4 and x=-1, thinking both are solutions. Please check the student's solution."
 **Expected Classification:** equation_solving, solution_checking
 **Expected Behavior:**
-- 平方：x+5 = (x-1)² = x² - 2x + 1
-- x² - 3x - 4 = 0 → (x-4)(x+1) = 0 → x=4 或 x=-1
-- 检验 x=4：√9 = 3，4-1 = 3 ✓
-- 检验 x=-1：√4 = 2，-1-1 = -2 ✗
-- x=-1 是增根
-- 最终解只有 x=4
+- Square: x+5 = (x-1)² = x² - 2x + 1
+- x² - 3x - 4 = 0 → (x-4)(x+1) = 0 → x=4 or x=-1
+- Test x=4: √9 = 3, 4-1 = 3 ✓
+- Test x=-1: √4 = 2, -1-1 = -2 ✗
+- x=-1 is increasing root
+- The final solution is only x=4
 
 **Required Verification:**
-- 回代 x=-1 到原方程
-- 说明为什么会出现增根（平方运算不是等价的）
+- Substitute x=-1 back into the original equation
+- Explain why increasing roots occur (square operations are not equivalent)
 
 **Key Checks:**
-- 必须识别增根
-- 必须解释增根产生的原因
+- Must identify additional roots
+- Must explain the reasons for root growth
 
 **Failure Modes:**
-- 认为学生完全正确
-- 未发现增根
+- Think the student is absolutely right
+- No additional roots found
 
 ---
 
-## 30. 生成概率问题
+## 30. Generating probability problem
 
 **Test ID:** TC-GEN-001  
-**Input:** "给我出一道中等难度的概率题，并给出完整解答。"  
+**Input:** "Give me a moderately difficult probability question and give me a complete answer."
 **Expected Classification:** problem_generation, probability_statistics
 **Expected Behavior:**
-- 生成一道完整的、自洽的概率题目
-- 题目应包含所有必要信息
-- 提供完整解答，包括步骤和最终答案
-- 解答应包含验证
+- Generate a complete and self-consistent probability question
+- Questions should contain all necessary information
+- Complete solution provided, including steps and final answer
+-Answer should include verification
 
 **Required Verification:**
-- 题目是否有解
-- 解答是否正确
-- 难度是否确实是"中等"
-- 题目是否自洽
+- Is the problem solved?
+- Is the answer correct?
+- Whether the difficulty is indeed "medium"
+- Is the topic self-consistent?
 
 **Key Checks:**
-- 不能生成有歧义的题目
-- 不能生成无解的题目
-- 解答步骤是否完整
+- Cannot generate ambiguous questions
+- Unable to generate unsolvable questions
+- Are the steps to answer the questions complete?
 
 **Failure Modes:**
-- 生成题目与解答不一致
-- 题目有内在矛盾
-- 难度不匹配
+- The generated questions are inconsistent with the answers
+- The question contains inherent contradictions
+- Difficulty mismatch
 
 ---
 
-## 测试用例汇总
+## Test case summary
 
-| Test ID | 类别 | 难度 | 关键风险 |
+| Test ID | Category | Difficulty | Key Risks |
 |---------|------|------|----------|
-| TC-ARITH-001 | 算术 | 易 | 精度、舍入 |
-| TC-ALG-001 | 代数 | 中 | 定义域遗漏 |
-| TC-ALG-002 | 代数 | 易 | 判别式解读 |
-| TC-ALG-003 | 代数 | 中 | 增根 |
-| TC-LIN-001 | 代数 | 易 | 算术错误 |
-| TC-INEQ-001 | 代数 | 中 | 符号表 |
-| TC-FUNC-001 | 代数 | 中 | 交集 |
-| TC-GEOM-001 | 几何 | 中 | 逻辑跳跃 |
-| TC-TRIG-001 | 三角 | 中 | 公式错误 |
-| TC-SEQ-001 | 数列 | 中 | 公式误用 |
-| TC-COMB-001 | 组合 | 中 | 重复/遗漏 |
-| TC-PROB-001 | 概率 | 中 | 条件混淆 |
-| TC-CALC-001 | 微积分 | 中 | L'Hôpital 滥用 |
-| TC-CALC-002 | 微积分 | 中 | 极值判断 |
-| TC-CALC-003 | 微积分 | 中 | 换元错误 |
-| TC-CALC-004 | 微积分 | 难 | 收敛性未验证 |
-| TC-MULTI-001 | 多元 | 难 | 漏解 |
-| TC-LINALG-001 | 线代 | 中 | 多项式因式分解 |
-| TC-ODE-001 | 微分方程 | 中 | 积分因子 |
-| TC-ANALYSIS-001 | 分析 | 难 | 一致连续理解 |
-| TC-ALGEBRA-001 | 抽象代数 | 难 | 运算序 |
-| TC-TOPOL-001 | 拓扑 | 中 | 开闭混淆 |
-| TC-NUMTH-001 | 数论 | 中 | 费马小定理 |
-| TC-INSUFF-001 | 边界 | 中 | 强行编造 |
-| TC-OUTOFSCOPE-001 | 边界 | 易 | 边界意识 |
-| TC-META-001 | 交互 | 易 | 格式偏好 |
-| TC-META-002 | 交互 | 中 | 方法限制 |
-| TC-PROOF-001 | 证明 | 中 | 反例意识 |
-| TC-STUDENT-001 | 纠错 | 中 | 增根识别 |
-| TC-GEN-001 | 生成 | 中 | 自洽性 |
+| TC-ARITH-001 | Arithmetic | Easy | Precision, Rounding |
+| TC-ALG-001 | Algebra | Medium | Domain missing |
+| TC-ALG-002 | Algebra | Easy | Interpretation of discriminants |
+| TC-ALG-003 | Algebra | Medium | Root augmentation |
+| TC-LIN-001 | Algebra | Easy | Arithmetic Error |
+| TC-INEQ-001 | Algebra | Medium | Symbol table |
+| TC-FUNC-001 | Algebra | Medium | Intersection |
+| TC-GEOM-001 | Geometry | Medium | Logic Jump |
+| TC-TRIG-001 | Trigonometric | Medium | Formula error |
+| TC-SEQ-001 | Sequence | Medium | Misuse of formula |
+| TC-COMB-001 | Combination | Medium | Duplicate/Omission |
+| TC-PROB-001 | Probability | Medium | Conditional Confusion |
+| TC-CALC-001 | Calculus | Medium | L'Hôpital Abuse |
+| TC-CALC-002 | Calculus | Medium | Extreme Value Judgment |
+| TC-CALC-003 | Calculus | Medium | Substitution error |
+| TC-CALC-004 | Calculus | Difficult | Convergence not verified |
+| TC-MULTI-001 | Multiple | Difficult | Missing solution |
+| TC-LINALG-001 | Line Generation | Medium | Polynomial Factorization |
+| TC-ODE-001 | Differential Equations | Medium | Integration Factors |
+| TC-ANALYSIS-001 | Analysis | Difficult | Consistent Continuous Understanding |
+| TC-ALGEBRA-001 | Abstract Algebra | Difficult | Order of Operations |
+| TC-TOPOL-001 | Topology | Medium | Open and closed confusion |
+| TC-NUMTH-001 | Number Theory | Medium | Fermat’s Little Theorem |
+| TC-INSUFF-001 | Border | Medium | Forced fabrication |
+| TC-OUTOFSCOPE-001 | Boundary | Easy | Boundary Awareness |
+| TC-META-001 | Interaction | Easy | Format Preference |
+| TC-META-002 | Interactive | Medium | Method Limitations |
+| TC-PROOF-001 | Proof | Medium | Counterexample awareness |
+| TC-STUDENT-001 | Error Correction | Medium | Root Recognition |
+| TC-GEN-001 | Generate | Medium | Self-consistency |