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Fix typos related to axiomatic derivations (#409)
* Fix typo in Section 12.1 of the Open Logic Text * Fix typo in Section 12.2 of the Open Logic Text * Fix typos in Section 12.3 of the Open Logic Text * Fix typo in Section 12.5 of the Open Logic Text * Fix typo in Section 22.11 of the Open Logic Text * Fix typos in proof of Theorem 22.36 * Add line break for screen reading In Section 27.4: The Definability Theorem
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content/first-order-logic/axiomatic-deduction/axioms-rules-propositional.tex

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{\olfileid{fol}{axd}{prp}}
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{\olfileid{pl}{axd}{prp}}
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\olsection{Axiom and Rules for the Propositional Connectives}
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\olsection{Axioms and Rules for the Propositional Connectives}
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\begin{defn}[Axioms]
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The set of $\PAx$ of \emph{axioms} for the propositional connectives comprises

content/first-order-logic/axiomatic-deduction/deduction-theorem.tex

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is an axiom, then $\Gamma \Proves !B$. We also have that $\Gamma
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\Proves !B \lif (!A \lif !B)$ by \olref[prp]{ax:lif1}, and
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\olref{prop:mp} gives $\Gamma \Proves !A \lif !B$. If $!B \in \{ !A\}$
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then $\Gamma \Proves !A \lif !B$ because then last !!{sentence}~$!A
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then $\Gamma \Proves !A \lif !B$ because the last !!{sentence}~$!A
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\lif !B$ is the same as $!A \lif !A$, and we have !!{derive}d that in
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\olref[pro]{ex:identity}.
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content/first-order-logic/axiomatic-deduction/provability-quantifiers.tex

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\begin{proof}
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By the deduction theorem, $\Gamma \Proves \ltrue \lif !A(c)$. Since
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$c$ does not occur in $\Gamma$ or~$\top$, we get $\Gamma \Proves
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\ltrue \lif !A(c)$. By the deduction theorem again, $\Gamma \Proves
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\ltrue \lif \lforall[x][!A(x)]$. By the deduction theorem again, $\Gamma \Proves
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\lforall[x][!A(x)]$.
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\end{proof}
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content/first-order-logic/axiomatic-deduction/proving-things.tex

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1. & $\lnot !D \lif (!D \lif !E)$ & \olref[prp]{ax:lnot2} \\
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2. & $(\lnot !D \lif (!D \lif !E)) \lif {}$\\
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&\qquad $((!E \lif (!D \lif !E)) \lif ((\lnot !D \lor !E) \lif (!D \lif !E)))$ & \olref[prp]{ax:lor3}\\
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3. & $((!E \lif (!D \lif !E)) \lif ((\lnot !D \lor !E) \lif (!D \lif !E))$ & 1, 2, \MP\\
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3. & $(!E \lif (!D \lif !E)) \lif ((\lnot !D \lor !E) \lif (!D \lif !E))$ & 1, 2, \MP\\
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4. & $!E \lif (!D \lif !E)$ & \olref[prp]{ax:lif1}\\
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5. & $(\lnot !D \lor !E) \lif (!D \lif !E)$ & 3, 4, \MP
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\end{derivation}
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lines 3--7 of the !!{derivation} in the preceding example. This is
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!!a{derivation} of $!A \lif !C$---which is the last line of the new
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!!{derivation}---from~$\Gamma$. Note that the justifications of
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lines 4 and~7 remain valid if the reference to line number~2 is
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lines 4 and~7 remain valid if the reference to line number~1 is
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replaced by reference to the last line of the !!{derivation} of~$!A
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\lif !B$, and reference to line number~1 by reference to the last
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line of the !!{derivation} of~$B \lif !C$.
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\lif !B$, and reference to line number~2 by reference to the last
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line of the !!{derivation} of~$!B \lif !C$.
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\end{proof}
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\begin{prob}

content/first-order-logic/axiomatic-deduction/rules-and-proofs.tex

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\item for some $j < i$, $!A_j$ is $!B \lif !A_i$, and for some $k < i$,
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$!A_k$ is~$!B$.
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\end{enumerate}
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The last clause says that $!A_i$ follows from~$!A_j$ ($!B$) and $!A_k$
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($!B \lif !A_i$) by modus ponens. If we can go from $1$ to~$n$, and
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The last clause says that $!A_i$ follows from~$!A_j$ ($!B \lif !A_i$) and $!A_k$
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($!B$) by modus ponens. If we can go from $1$ to~$n$, and
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each time we find !!a{formula}~$!A_i$ that is either in~$\Gamma$, an
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axiom, or which a rule of inference tells us that it is a correct
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inference step, then the entire sequence counts as a correct

content/first-order-logic/axiomatic-deduction/soundness.tex

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ponens, then there are !!{formula}s $!B$ and $!B \lif !A$ in the
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!!{derivation}, and the induction hypothesis applies to the part of
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the !!{derivation} ending in those !!{formula}s (since they contain at
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least one fewer steps justified by an inference). So, by induction
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least one fewer step justified by an inference). So, by induction
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hypothesis, $\Gamma \Entails !B$ and $\Gamma \Entails !B \lif
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!A$. Then $\Gamma \Entails !A$ by
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\iftag{FOL}
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\QR. Then that step has the form $!C \lif \lforall[x][B(x)]$ and
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there is a preceding step $!C \lif !B(c)$ with $c$ not in $\Gamma$,
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$!C$, or $\lforall[x][B(x)]$. By induction hypothesis, $\Gamma
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\Entails !C \lif \lforall[x][B(x)]$. By
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\Entails !C \lif !B(c)$. By
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\olref[syn][sem]{thm:sem-deduction}, $\Gamma \cup \{!C\} \Entails
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!B(c)$.
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not occur in~$\Gamma$ or~$!C$, $\Sat{M'}{\Gamma \cup \{!C\}}$ by
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\olref[syn][ext]{cor:extensionality-sent}. Since $\Gamma \cup \{!C\}
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\Entails !B(c)$, $\Sat{M'}{B(c)}$. Since $!B(c)$ is !!a{sentence},
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$\Sat{M}{!B(c)}[s]$ by
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$\Sat{M'}{!B(c)}[s]$ by
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\olref[syn][ass]{prop:sentence-sat-true}. $\Sat{M'}{!B(x)}[s]$ iff
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$\Sat{M'}{!B(c)}$ by \olref[syn][ext]{prop:ext-formulas} (recall that
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$\Sat{M'}{!B(c)}[s]$ by \olref[syn][ext]{prop:ext-formulas} (recall that
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$!B(c)$ is just $\Subst{!B(x)}{c}{x}$). So,
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$\Sat{M'}{!B(x)}[s]$. Since $c$ does not occur in~$!B(x)$, by
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\olref[syn][ext]{prop:extensionality}, $\Sat{M}{!B(x)}[s]$. But $s$

content/model-theory/interpolation/definability.tex

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By Craig's Interpolation Theorem there is !!a{sentence} $!C(c_1,\dots, c_n)$
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not containing $P$ or $P'$ such that:
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\begin{align*}
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!D(P) \land \Atom{P}{c_1, \dots, c_n} & \Entails !C(c_1,\dots, c_n); &
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!D(P) \land \Atom{P}{c_1, \dots, c_n} & \Entails !C(c_1,\dots, c_n); \\
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!C(c_1,\dots, c_n) & \Entails !D(P') \to \Atom{P'}{c_1, \dots, c_n}.
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\end{align*}
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From the former of these two entailments we have: $!D(P) \Entails

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