225 lines
6.5 KiB
TeX
225 lines
6.5 KiB
TeX
\documentclass{report}
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\input{../../preamble}
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\makecode{../..}
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\begin{document}
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\header{A Mathematical Introduction to Logic}{Herbert B. Enderton}
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\tableofcontents
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\begingroup
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\renewcommand\thechapter{R}
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\chapter{Reference}%
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\hyperlabel{chap:reference}
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\section{\defined{Construction Sequence}}%
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\hyperlabel{ref:construction-sequence}
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A \textbf{construction sequence} is a finite sequence
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$\langle \epsilon_1, \ldots, \epsilon_n \rangle$ of expressions such that
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for each $i \leq n$ we have at least one of
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\begin{align*}
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& \epsilon_i \text{ is a sentence symbol} \\
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& \epsilon_i = \mathcal{E}_\neg(\epsilon_j) \text{ for some } j < i \\
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& \epsilon_i = \mathcal{E}_\square(\epsilon_j, \epsilon_k)
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\text{ for some } j < i, k < i
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\end{align*}
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where $\square$ is one of the binary connectives $\land$, $\lor$,
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$\Rightarrow$, $\Leftrightarrow$.
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\section{\defined{Expression}}%
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\hyperlabel{ref:expression}
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An \textbf{expression} is a finite sequence of symbols.
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\section{\defined{\texorpdfstring{$n$}{n}-tuple}}%
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\hyperlabel{ref:n-tuple}
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An \textbf{$n$-tuple} is recursively defined as
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$$\ltuple{x_1}{x_{n+1}} = \tuple{\ltuple{x_1}{x_n}, x_{n+1}}$$
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for $n > 1$.
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We also define $\tuple{x} = x$.
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\section{\defined{Well-Formed Formula}}%
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\hyperlabel{ref:well-formed-formula}
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An \nameref{ref:expression} that can be built up from the sentence symbols by
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applying some finite number of times the
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\textbf{formula-building operations} (on expressions) defined by the
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equations:
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\begin{align*}
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\mathcal{E}_{\neg}(\alpha)
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& = (\neg \alpha) \\
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\mathcal{E}_{\land}(\alpha, \beta)
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& = (\alpha \land \beta) \\
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\mathcal{E}_{\lor}(\alpha, \beta)
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& = (\alpha \lor \beta) \\
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\mathcal{E}_{\Rightarrow}(\alpha, \beta)
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& = (\alpha \Rightarrow \beta) \\
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\mathcal{E}_{\Leftrightarrow}(\alpha, \beta)
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& = (\alpha \Leftrightarrow \beta)
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\end{align*}
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\endgroup
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% Reset counter to mirror Enderton's book.
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\setcounter{chapter}{0}
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\addtocounter{chapter}{-1}
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\chapter{Useful Facts About Sets}%
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\hyperlabel{chap:useful-facts-about-sets}
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\section{\unverified{Lemma 0A}}%
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\hyperlabel{sec:lemma-0a}
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\begin{lemma}[0A]
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Assume that $\ltuple{x_1}{x_m} = \ltuple{y_1, \ldots, y_m}{y_{m+k}}$.
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Then $x_1 = \ltuple{y_1}{y_{k+1}}$.
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\end{lemma}
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\begin{proof}
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For natural number $m$, let $P(m)$ be the statement:
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\begin{assumption}
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\hyperlabel{sec:lemma-0a-eq1}
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\text{If } \ltuple{x_1}{x_m} = \ltuple{y_1, \ldots, y_m}{y_{m+k}}
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\text{ then } x_1 = \ltuple{y_1}{y_{k+1}}.
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\end{assumption}
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\noindent
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We proceed by induction on $m$.
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\paragraph{Base Case}%
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Suppose $\tuple{x_1} = \ltuple{y_1}{y_{1 + k}}$.
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By definition of an \nameref{ref:n-tuple}, $\tuple{x_1} = x_1$.
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Thus $x_1 = \ltuple{y_1}{y_{k + 1}}$.
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Hence $P(1)$ holds true.
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\paragraph{Inductive Step}%
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Suppose for $m \geq 1$ that $P(m)$ is true and assume
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\begin{equation}
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\hyperlabel{sec:lemma-0a-eq2}
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\ltuple{x_1}{x_{m+1}} = \ltuple{y_1, \ldots, y_{m+1}}{y_{m+1+k}}.
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\end{equation}
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By definition of an \nameref{ref:n-tuple}, we can decompose
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\eqref{sec:lemma-0a-eq2} into the following two identities
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\begin{align*}
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x_{m+1} & = y_{m+1+k} \\
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\ltuple{x_1}{x_m} & = \ltuple{y_1}{y_{m+k}}.
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\end{align*}
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By \eqref{sec:lemma-0a-eq1}, $P(m)$ implies $x_1 = \ltuple{y_1}{y_{k+1}}$.
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Hence $P(m+1)$ holds true.
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\paragraph{Conclusion}%
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By induction, $P(m)$ holds true for all $m \geq 1$.
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\end{proof}
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\chapter{Sentential Logic}%
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\hyperlabel{chap:sentential-logic}
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\section{The Language of Sentential Logic}%
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\hyperlabel{sec:language-sentential-logic}
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\subsection{\sorry{Induction Principle}}%
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\hyperlabel{sub:induction-principle-1}
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\begin{theorem}
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If $S$ is a set of wffs containing all the sentence symbols and closed under
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all five formula-building operations, then $S$ is the set of \textit{all}
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wffs.
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\end{theorem}
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\begin{proof}
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TODO
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\end{proof}
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\section{Exercises 1}%
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\hyperlabel{sec:exercises-1}
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\subsection{\sorry{Exercise 1.1.1}}%
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\hyperlabel{sub:exercise-1.1.1}
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Give three sentences in English together with translations into our formal
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language.
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The sentences shoudl be chosen so as to have an interesting structure, and the
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translations should each contain 15 or more symbols.
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\begin{answer}
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TODO
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\end{answer}
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\subsection{\sorry{Exercise 1.1.2}}%
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\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
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length is possible.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 1.1.3}}%
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\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
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connective symbols $(\land, \lor, \Rightarrow, \Leftrightarrow)$ occur in
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$\alpha$; let $s$ be the number of places at which sentence symbols occur in
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$\alpha$.
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(For example, if $\alpha$ is $(A \Rightarrow (\neg A))$ then $c = 1$ and
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$s = 2$.)
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Show by using the induction principle that $s = c + 1$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 1.1.4}}%
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\hyperlabel{sub:exercise-1.1.4}
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Assume we have a construction sequence ending in $\phi$, where $\phi$ does not
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contain the symbol $A_4$.
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Suppose we delete all the expressions in the construction sequence that
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contain $A_4$.
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Show that the result is still a legal construction sequence.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 1.1.5}}%
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\hyperlabel{sub:exercise-1.1.5}
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Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
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\begin{enumerate}[(a)]
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\item Show that the length of $\alpha$ (i.e., the number of symbols in the
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string) is odd.
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\item Show that more than a quarter of the symbols are sentence symbols.
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\end{enumerate}
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\textit{Suggestion}: Apply induction to show that the length is of the form
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$4k + 1$ and the number of sentence symbols is $k + 1$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 1.1.6}}%
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\hyperlabel{sub:exercise-1.1.6}
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Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
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\begin{enumerate}[(a)]
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\item Show that the length of $\alpha$ (i.e., the number of symbols in the
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string) is odd.
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\item Show that more than a quarter of the symbols are sentence symbols.
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\end{enumerate}
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\begin{proof}
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TODO
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\end{proof}
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\end{document}
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