bookshelf/Bookshelf/Enderton/Logic.tex

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\documentclass{report}
\input{../../preamble}
\makecode{../..}
\begin{document}
\header{A Mathematical Introduction to Logic}{Herbert B. Enderton}
\tableofcontents
\begingroup
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\renewcommand\thechapter{R}
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\chapter{Reference}%
\hyperlabel{chap:reference}
\section{\defined{Construction Sequence}}%
\hyperlabel{ref:construction-sequence}
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A \textbf{construction sequence} is a finite sequence
$\langle \epsilon_1, \ldots, \epsilon_n \rangle$ of expressions such that
for each $i \leq n$ we have at least one of
\begin{align*}
& \epsilon_i \text{ is a sentence symbol} \\
& \epsilon_i = \mathcal{E}_\neg(\epsilon_j) \text{ for some } j < i \\
& \epsilon_i = \mathcal{E}_\square(\epsilon_j, \epsilon_k)
\text{ for some } j < i, k < i
\end{align*}
where $\square$ is one of the binary connectives $\land$, $\lor$,
$\Rightarrow$, $\Leftrightarrow$.
\section{\defined{Expression}}%
\hyperlabel{ref:expression}
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An \textbf{expression} is a finite sequence of symbols.
\section{\defined{Well-Formed Formula}}%
\hyperlabel{ref:well-formed-formula}
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An \nameref{ref:expression} that can be built up from the sentence symbols by
applying some finite number of times the
\textbf{formula-building operations} (on expressions) defined by the
equations:
\begin{align*}
\mathcal{E}_{\neg}(\alpha)
& = (\neg \alpha) \\
\mathcal{E}_{\land}(\alpha, \beta)
& = (\alpha \land \beta) \\
\mathcal{E}_{\lor}(\alpha, \beta)
& = (\alpha \lor \beta) \\
\mathcal{E}_{\Rightarrow}(\alpha, \beta)
& = (\alpha \Rightarrow \beta) \\
\mathcal{E}_{\Leftrightarrow}(\alpha, \beta)
& = (\alpha \Leftrightarrow \beta)
\end{align*}
\endgroup
% Reset counter to mirror Enderton's book.
\setcounter{chapter}{0}
\addtocounter{chapter}{-1}
\chapter{Useful Facts About Sets}%
\hyperlabel{chap:useful-facts-about-sets}
\section{\sorry{Lemma 0A}}%
\hyperlabel{sec:lemma-0a}
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\begin{lemma}[0A]
Assume that $\langle x_1, \ldots, x_m \rangle =
\langle y_1, \ldots, y_m, \ldots, y_{m+k} \rangle$.
Then $x_1 = \langle y_1, \ldots, y_{k+1} \rangle$.
\end{lemma}
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\begin{proof}
TODO
\end{proof}
\chapter{Sentential Logic}%
\hyperlabel{chap:sentential-logic}
\section{The Language of Sentential Logic}%
\hyperlabel{sec:language-sentential-logic}
\subsection{\sorry{Induction Principle}}%
\hyperlabel{sub:induction-principle-1}
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\begin{theorem}
If $S$ is a set of wffs containing all the sentence symbols and closed under
all five formula-building operations, then $S$ is the set of \textit{all}
wffs.
\end{theorem}
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\begin{proof}
TODO
\end{proof}
\section{Exercises 1}%
\hyperlabel{sec:exercises-1}
\subsection{\sorry{Exercise 1.1.1}}%
\hyperlabel{sub:exercise-1.1.1}
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Give three sentences in English together with translations into our formal
language.
The sentences shoudl be chosen so as to have an interesting structure, and the
translations should each contain 15 or more symbols.
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\begin{answer}
TODO
\end{answer}
\subsection{\sorry{Exercise 1.1.2}}%
\hyperlabel{sub:exercise-1.1.2}
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Show that there are no wffs of length 2, 3, or 6, but that any other positive
length is possible.
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\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 1.1.3}}%
\hyperlabel{sub:exercise-1.1.3}
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Let $\alpha$ be a wff; let $c$ be the number of places at which binary
connective symbols $(\land, \lor, \Rightarrow, \Leftrightarrow)$ occur in
$\alpha$; let $s$ be the number of places at which sentence symbols occur in
$\alpha$.
(For example, if $\alpha$ is $(A \Rightarrow (\neg A))$ then $c = 1$ and
$s = 2$.)
Show by using the induction principle that $s = c + 1$.
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\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 1.1.4}}%
\hyperlabel{sub:exercise-1.1.4}
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Assume we have a construction sequence ending in $\phi$, where $\phi$ does not
contain the symbol $A_4$.
Suppose we delete all the expressions in the construction sequence that
contain $A_4$.
Show that the result is still a legal construction sequence.
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\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 1.1.5}}%
\hyperlabel{sub:exercise-1.1.5}
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Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
\begin{enumerate}[(a)]
\item Show that the length of $\alpha$ (i.e., the number of symbols in the
string) is odd.
\item Show that more than a quarter of the symbols are sentence symbols.
\end{enumerate}
\textit{Suggestion}: Apply induction to show that the length is of the form
$4k + 1$ and the number of sentence symbols is $k + 1$.
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\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 1.1.6}}%
\hyperlabel{sub:exercise-1.1.6}
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Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
\begin{enumerate}[(a)]
\item Show that the length of $\alpha$ (i.e., the number of symbols in the
string) is odd.
\item Show that more than a quarter of the symbols are sentence symbols.
\end{enumerate}
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\begin{proof}
TODO
\end{proof}
\end{document}