typose reported by Julia

Author: rzach

first-order-logic/syntax-and-semantics/extensionality.tex

@@ -27,8 +27,8 @@
 \begin{prop}[Extensionality]
 Let $!A$ be a sentence, and $\Struct M$ and $\Struct M'$ be !!{structure}s. 
 If $\Assign{c}{M} = \Assign{c}{M'}$, $\Assign{R}{M}=\Assign{R}{M'}$, and 
-$\Assign{f}{M} = \Assign{f}{M'}$ for every !!{constant} $c$, relation 
-symbol $R$, and !!{function} $f$ occurring in $!A$, then $\Sat{M}{!A}$ 
+$\Assign{f}{M} = \Assign{f}{M'}$ for every !!{constant}~$c$, relation 
+symbol~$R$, and !!{function} $f$ occurring in~$!A$, then $\Sat{M}{!A}$ 
 iff $\Sat{M'}{!A}$.
 \end{prop}
 
@@ -37,8 +37,8 @@
 
 \begin{prop}\ollabel{prop:ext-terms}
 Let $\Struct M$ be a !!{structure}, $t$ and $t'$ terms, and $s$ a
-variable assignment. Let $s' \sim_x s$ be the $x$-variant of $s$ given
-by~$s'(x) = \Value{t}{M}[s]$. Then $\Value{\Subst{t}{t'}{x}}{M}[s] =
+variable assignment. Let $s' \sim_x s$ be the $x$-variant of~$s$ given
+by~$s'(x) = \Value{t'}{M}[s]$. Then $\Value{\Subst{t}{t'}{x}}{M}[s] =
 \Value{t}{M}[s']$.
 \end{prop}
 
@@ -50,7 +50,7 @@
 
 \item If $t$ is a variable other than~$x$, say, $t \ident y$, then
   $\Subst{t}{t'}{x} = y$, and $\Value{y}{M}[s] = \Value{y}{M}[s']$
-  since $s' \sim_x s'$.
+  since $s' \sim_x s$.
 
 \item If $t \ident x$, then $\Subst{t}{t'}{x} = t'$.  But
   $\Value{x}{M}[s'] = \Value{t'}{M}[s]$ by definition of~$s'$.

first-order-logic/syntax-and-semantics/semantic-notions.tex

@@ -64,18 +64,19 @@
 \end{prop}
 
 \begin{proof}
-For the forward direction, let $\Gamma$ entail $!A$ and suppose to the
+For the forward direction, suppose $\Gamma \Entails !A$ and suppose to the
 contrary that there is a !!{structure} $\Struct M$ so that $\Sat{M}{\Gamma
   \cup \{ \lnot !A \}}$. Since $\Sat{M}{\Gamma}$ and $\Gamma \Entails
 !A$, $\Sat{M}{!A}$. Also, since $\Sat{M}{\Gamma\cup \{ \lnot !A \}}$,
 $\Sat{M}{\lnot !A}$, so we have both $\Sat{M}{!A}$ and $\Sat/{M}{!A}$,
 a contradiction. Hence, there can be no such !!{structure} $\Struct M$, so
 $\Gamma \cup \{ !A \}$ is unsatisfiable.
 
-For the reverse direction, let $\Gamma \cup \{ \lnot !A \}$. So for
-every !!{structure} $\Struct M$, either $\Sat/{M}{\Gamma}$ or
-$\Sat{M}{!A}$. Hence, for every !!{structure} $\Struct M$ with
-$\Sat{M}{\Gamma}$, $\Sat{M}{!A}$, so $\Gamma \Entails !A$.
+For the reverse direction, suppose $\Gamma \cup \{ \lnot !A \}$ is
+unsatisfiable. So for every !!{structure} $\Struct M$, either
+$\Sat/{M}{\Gamma}$ or $\Sat{M}{!A}$. Hence, for every !!{structure}
+$\Struct M$ with $\Sat{M}{\Gamma}$, $\Sat{M}{!A}$, so $\Gamma \Entails
+!A$.
 \end{proof}
 
 \begin{prob}
@@ -98,7 +99,7 @@
 Suppose that $\Gamma \subseteq \Gamma'$ and $\Gamma \Entails !A$. Let
 $\Struct M$ be such that $\Sat{M}{\Gamma'}$; then $\Sat{M}{\Gamma}$,
 and since $\Gamma \Entails !A$, we get that $\Sat{M}{!A}$. Hence,
-whenever $\Sat{M}{\Gamma}$, $\Sat{M}{!A}$, so $\Gamma' \Entails !A$.
+whenever $\Sat{M}{\Gamma'}$, $\Sat{M}{!A}$, so $\Gamma' \Entails !A$.
 \end{proof}