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25 | 25 | the philosophy of science, since it can be used to study the |
26 | 26 | primitives by which a theory can or cannot be axiomatized. It also |
27 | 27 | implies Beth's definability theorem, which states that if a theory can |
28 | | -implicitly define a concept, if can define it explicitly. |
| 28 | +implicitly define a concept, it can define it explicitly. |
29 | 29 |
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30 | 30 | Like many results in mathematical logic, the interpolation theorem can |
31 | 31 | be proved both using the methods of model theory and those of proof |
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42 | 42 | \begin{thm}[Craig's Interpolation Theorem]\ollabel{thm:interpolation} |
43 | 43 | Let the !!{language} of~$!A$ be $\Lang L(!A)$ and that of $!B$ be |
44 | 44 | $\Lang L(!B)$. If $\Log{G3c} \Proves !A \Sequent !B$ then there is |
45 | | -!!a{formula}~$!C$ in the !!{language}~$\Lang L(!A) \cap \Lang L(!B)$ |
46 | | -such that $\Log{G3c} \Proves !A \Sequent !C$ and $\Log{G3c} \Proves !C \Sequent !B$. |
| 45 | +!!a{formula}~$!C$ in the !!{language}~$\Lang L(!C) \subseteq \Lang |
| 46 | +L(!A) \cap \Lang L(!B)$ such that $\Log{G3c} \Proves !A \Sequent !C$ |
| 47 | +and $\Log{G3c} \Proves !C \Sequent !B$. |
47 | 48 | \end{thm} |
48 | 49 |
|
49 | 50 | We prove a generalization of the interpolation theorem to make use of |
50 | 51 | the structure of !!{proof}s in~\Log{G3c}. To this end, imagine the |
51 | 52 | antecendent and succedents of sequents to be separated into two parts |
52 | | -each. We write $\maeh{\Gamma_1}{\Gamma_2} \Sequent \maeh{\Delta_1}{\Delta_2}$ to |
53 | | -indicate that $\Gamma_1$ and $\Delta_1$ belong to the first part and |
54 | | -$\Gamma_2$ and $\Delta_2$ to the second part. The interpolation |
55 | | -theorem then follows immediately from the following lemma. |
| 53 | +each. We write $\maeh{\Gamma_1}{\Gamma_2} \Sequent |
| 54 | +\maeh{\Delta_1}{\Delta_2}$ to indicate that $\Gamma_1$ and $\Delta_1$ |
| 55 | +belong to the first part and $\Gamma_2$ and $\Delta_2$ to the second |
| 56 | +part. The interpolation theorem then follows immediately from the |
| 57 | +following lemma. |
56 | 58 |
|
57 | 59 | \begin{lem}[Maehara's Lemma]\ollabel{lem:maehara} If $\Log{G3c} |
58 | 60 | \Proves \maeh{\Gamma_1}{\Gamma_2} \Sequent \maeh{\Delta_1}{\Delta_2}$ |
59 | | -then there is !!a{formula}~$!C$ in the !!{language}~$\Lang L(\Gamma_1, |
| 61 | +then there is !!a{formula}~$!C$ in the !!{language}~$\Lang L(!C) |
| 62 | +\subseteq \Lang L(\Gamma_1, |
60 | 63 | \Delta_1) \cap \Lang L(\Gamma_2, \Delta_2)$ such that $\Log{G3c} |
61 | 64 | \Proves \Gamma_1 \Sequent \Delta_1, !C$ and $\Log{G3c} \Proves !C, |
62 | 65 | \Gamma_2 \Sequent \Delta_2$. |
|
150 | 153 | already told us that the sequent on the right is !!{provable}, and |
151 | 154 | the sequent on the left follows from the sequent we got from the |
152 | 155 | inductive hypothesis by~$\RightR{\lnot}$. The inductive hypothesis |
153 | | - also tells us that $\Lang L(!C_1) \in \Lang L(!A, \Gamma_1, |
| 156 | + also tells us that $\Lang L(!C_1) \subseteq \Lang L(!A, \Gamma_1, |
154 | 157 | \Delta_1) \cap \Lang L(\Gamma_2, \Delta_2)$. Since $\Lang L(\lnot |
155 | 158 | !A) = \Lang L(!A)$ and $\Lang L(!C) = \Lang L(!C_1)$ we have that |
156 | | - $\Lang L(!C_1) \in \Lang L(\Gamma_1, \Delta_1, \lnot !A) \cap \Lang |
| 159 | + $\Lang L(!C_1) \subseteq \Lang L(\Gamma_1, \Delta_1, \lnot !A) \cap \Lang |
157 | 160 | L(\Gamma_2, \Delta_2)$. |
158 | 161 |
|
159 | 162 | In the second scenario the situation is slightly different |
|
169 | 172 | \end{align*} |
170 | 173 | These are again exactly the sequents we want to be !!{provable} if |
171 | 174 | $!C_1$ is an interpolant; we again let $!C \ident !C_1$. $\Lang |
172 | | - L(!C_1) \in \Lang L(\Gamma_1, \Delta_1) \cap \Lang L(\Gamma_2, |
| 175 | + L(!C_1) \subseteq \Lang L(\Gamma_1, \Delta_1) \cap \Lang L(\Gamma_2, |
173 | 176 | \Delta_2, \lnot !A)$ as before. |
174 | 177 | \item $\pi$ ends in $\RightR{\lif}$ with principal !!{formula}~$!A |
175 | 178 | \lif !B$. We again have two cases, depending on whether $!A \lif |
|
306 | 309 | !!{formula} built from $!C_1$ and $!C_2$, but this time we need it |
307 | 310 | in the antecedent (since we're now dealing with the second part). |
308 | 311 | Applying $\LeftR{\land}$ does the trick; we should take $!C \ident |
309 | | - (!C_1 \land !C_2)$ as the interpolant. |
310 | | - Fortuitously, we can use the remaining sequents (1) and (3) to |
311 | | - !!{prove} the corresponding sequent for the other part: |
| 312 | + (!C_1 \land !C_2)$ as the interpolant. Fortuitously, we can use |
| 313 | + the remaining sequents (1) and (3) to !!{prove} the corresponding |
| 314 | + sequent for the other part: |
312 | 315 | \[ |
313 | 316 | \AxiomC{} |
314 | 317 | \Deduce$\Gamma_1 \fCenter \Delta_1, !C_1$ |
|
321 | 324 | As before, we have $\Lang L(!C_1 \land !C_2) \subseteq \Lang |
322 | 325 | L(\Gamma_1, \Delta_1) \cap \Lang L(\Gamma_2, \Delta_2, !A \lif |
323 | 326 | !B)$. |
324 | | - \item $\pi$ ends in $\RightR{\lforall}$: |
325 | | - \[ |
| 327 | + |
| 328 | + \item $\pi$ ends in $\RightR{\lforall}$ with principal !!{formula}~$\lforall[x][!A(x)]$: |
| 329 | + \[ |
326 | 330 | \AxiomC{} |
327 | 331 | \RightLabel{$\pi_1$} |
328 | 332 | \Deduce$\maeh{\Gamma_1}{\Gamma_2} \fCenter |
|
362 | 366 | condition. |
363 | 367 |
|
364 | 368 | The other scenario is similar. |
365 | | - \item $\pi$ ends in $\RightR{\lexists}$. We again have two |
366 | | - scenarios. We consider the scenario where $\lexists[x][!A(x)]$ |
367 | | - belongs to the first part, and leave the other as an exercise. |
| 369 | + |
| 370 | + \item $\pi$ ends in $\RightR{\lexists}$ with principal |
| 371 | + !!{formula}~$\lexists[x][!A(x)]$. We again have two scenarios. We |
| 372 | + consider the scenario where $\lexists[x][!A(x)]$ belongs to the |
| 373 | + first part, and leave the other as an exercise. |
368 | 374 |
|
369 | 375 | In the first scenario, $\pi$ ends in |
370 | 376 | \[ |
|
454 | 460 | !!{proof} is satisfied because $b \notin \Lang L(\Gamma_2, |
455 | 461 | \Delta_2)$, and $!C_1'$ was defined so it does not contain~$b$. |
456 | 462 | \end{enumerate} |
| 463 | + The other cases are similary; we leave them as exercises. |
457 | 464 | \end{proof} |
458 | 465 |
|
459 | 466 | \begin{prob} |
|
527 | 534 | Replace $R'$ everywhere by~$R$ in the !!{proof} of the second sequent |
528 | 535 | to get !!a{proof} of} |
529 | 536 | !A(c_1, \dots, c_n), \Gamma & \Sequent R(c_1, \dots, c_n) |
530 | | - \intertext{Apply \RightR{\lif} and \RightR{\lforall} to get !!a{proof} of} |
531 | | - \Gamma & \Sequent \lforall[x_1][\dots\lforall[x_n][(R(x_1,\dots,x_n) \liff !A(x_1, \dots, x_n))]]. |
532 | 537 | \end{align*} |
| 538 | + Apply \RightR{\lif} and \RightR{\lforall} to get !!a{proof} of |
| 539 | + \[ |
| 540 | + \Gamma \Sequent \lforall[x_1][\dots\lforall[x_n][(R(x_1,\dots,x_n) \liff !A(x_1, \dots, x_n))]]. |
| 541 | + \] |
533 | 542 | This shows that $!A(x_1, \dots, x_n)$ explicitly defines $!R$ |
534 | 543 | in~$\Gamma$. |
535 | 544 | \end{proof} |
536 | 545 |
|
537 | 546 | A third result which is equivalent to interpolation (and, like it, can |
538 | 547 | be proved using Maehara's Lemma) is the following theorem: If two |
539 | 548 | consistent theories~$\Gamma_1$, $\Gamma_2$ contain a complete theory |
540 | | -$\Gamma$ in the language~$\Lang L(\Gamma_1) \cap \Lang L(\Gamma_2)$, |
541 | | -then $\Gamma_1 \cup \Gamma_2$ is consistent. |
| 549 | +$\Gamma$ in the language~$\Lang L(\Gamma) \subseteq \Lang L(\Gamma_1) |
| 550 | +\cap \Lang L(\Gamma_2)$, then $\Gamma_1 \cup \Gamma_2$ is consistent. |
542 | 551 |
|
543 | 552 | Recall that a complete theory is one such that for every |
544 | 553 | !!{sentence}~$!A$ in its language either $\Gamma \Proves !A$ or |
545 | | -$\Gamma \Proves \lnot !A$. Since $\Gamma_1$, $\Gamma_2$ are consistent |
| 554 | +$\Gamma \Proves \lnot !A$. Since $\Gamma_1$ and~$\Gamma_2$ are consistent |
546 | 555 | extensions of~$\Gamma$, $\Gamma$ must be consistent. It follows that |
547 | 556 | if $!A$ is !!a{sentence} in the language $\Lang L(\Gamma_1) \cap \Lang |
548 | 557 | L(\Gamma_2)$, we can't have $\Gamma_1 \Entails !A$ and $\Gamma_2 |
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