703 lines
25 KiB
TeX
703 lines
25 KiB
TeX
\documentclass{report}
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\input{../../preamble}
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\makecode{../..}
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\externaldocument[S:]{Set}
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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 \nameref{ref:finite-sequence}
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$\ltuple{\epsilon_1}{\epsilon_n}$ of \nameref{ref:expression}s such that for
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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 \nameref{ref:finite-sequence} of symbols.
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\section{\defined{Finite Sequence}}%
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\hyperlabel{ref:finite-sequence}
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$S$ is a \textbf{finite sequence} (or \textbf{string}) of members of set $A$
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if and only if, for some positive integer $n$, we have
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$S = \ltuple{x_1}{x_n}$, where each $x_i \in A$.
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\section{\defined{Formula-Building Operations}}%
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\hyperlabel{ref:formula-building-operations}
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The \textbf{formula-building operations} (on expressions) are 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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\code*{Bookshelf/Enderton/Logic/Chapter\_1}
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{Enderton.Logic.Chapter\_1.Wff}
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\lean{Init/Prelude}
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{Not}
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\lean{Init/Prelude}
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{And}
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\lean{Init/Prelude}
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{Or}
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\lean{Init/Core}
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{Iff}
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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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\lean*{Init/Prelude}
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{Prod}
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\section{\defined{Well-Formed Formula}}%
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\hyperlabel{ref:well-formed-formula}
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A \textbf{well-formed formula} (wff) is an \nameref{ref:expression} that can
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be built up from the sentence symbols by applying some finite number of
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times the \nameref{ref:formula-building-operations}.
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\code*{Bookshelf/Enderton/Logic/Chapter\_1}
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{Enderton.Logic.Chapter\_1.Wff}
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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{induction}
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\hyperlabel{sec:lemma-0a-ih}
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If $\ltuple{x_1}{x_m} = \ltuple{y_1, \ldots, y_m}{y_{m+k}}$
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then $x_1 = \ltuple{y_1}{y_{k+1}}$.
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\end{induction}
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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-eq1}
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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-eq1} 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 \ihref{sec:lemma-0a-ih}, $P(m)$ 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{\unverified{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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\code{Bookshelf/Enderton/Logic/Chapter\_1}
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{Enderton.Logic.Chapter\_1.Wff.rec}
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\begin{proof}
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We note every \nameref{ref:well-formed-formula} can be characterized by a
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\nameref{ref:construction-sequence}.
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For natural number $m$, let $P(m)$ be the statement:
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\begin{induction}
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\hyperlabel{sub:induction-principle-1-ih}
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Every wff characterized by a construction sequence of length $m$ is in
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$S$.
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\end{induction}
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\noindent
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We proceed by strong induction on $m$.
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\paragraph{Base Case}%
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Let $\phi$ denote a wff characterized by a construction sequence of length
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$1$.
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Then it must be that $\phi$ is a single sentence symbol.
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By hypothesis, $S$ contains all the sentence symbols.
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Thus $P(1)$ holds true.
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\paragraph{Inductive Step}%
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Suppose $P(0)$, $P(1)$, $\ldots$, $P(m)$ holds true and let $\phi$ denote
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a wff characterized by a construction sequence of length $m + 1$.
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By definition of a construction sequence, one of the following holds:
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\begin{align}
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& \phi \text{ is a sentence symbol}
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& \label{sub:induction-principle-1-eq1} \\
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& \phi = \mathcal{E}_\neg(\epsilon_j)
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\text{ for some } j < m + 1
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& \label{sub:induction-principle-1-eq2} \\
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& \phi = \mathcal{E}_\square(\epsilon_j, \epsilon_k)
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\text{ for some } j < m + 1, k < m + 1
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& \label{sub:induction-principle-1-eq3}
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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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We consider each case in turn.
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\subparagraph{\eqref{sub:induction-principle-1-eq1}}%
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By hypothesis, all sentence symbols are in $S$.
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Thus $\phi \in S$.
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\subparagraph{\eqref{sub:induction-principle-1-eq2}}%
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Suppose $\phi = \mathcal{E}_\neg(\epsilon_j)$ for some $j < m + 1$.
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By \ihref{sub:induction-principle-1-ih}, $\epsilon_j$ is in $S$.
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By hypothesis, $S$ is closed under $\mathcal{E}_\neg$.
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Thus $\phi \in S$.
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\subparagraph{\eqref{sub:induction-principle-1-eq3}}%
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Suppose $\phi = \mathcal{E}_\square(\epsilon_j, \epsilon_k)$ for some
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$j < m + 1, k < m + 1$,
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By \ihref{sub:induction-principle-1-ih}, $\epsilon_j$ and $\epsilon_k$
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is in $S$.
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By hypothesis, $S$ is closed under $\mathcal{E}_\square$ for all
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possible candidates of $\square$.
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Thus $\phi \in S$.
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\subparagraph{Subconclusion}%
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Since the above three cases are exhaustive, $P(m + 1)$ holds.
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\paragraph{Conclusion}%
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By strong induction, $P(m)$ holds true for all natural numbers $m \geq 1$.
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Since every well-formed formula is characterized by a construction
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sequence, the set of all wffs is a subset of $S$.
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Likewise, it obviously holds that $S$ is a subset of all wffs.
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Thus $S$ is precisely the set of all wffs.
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\end{proof}
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\subsection{\unverified{Balanced Parentheses}}%
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\hyperlabel{sub:balanced-parentheses}
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\begin{lemma}
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All well-formed formulas have an equal number of left and right parentheses.
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\end{lemma}
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\begin{proof}
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Define $$S = \{ \phi \mid
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\phi \text{ is a wff with a balanced number of parentheses} \}.$$
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We prove that (i) all the sentence symbols are members of $S$ and (ii)
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$S$ is closed under the five \nameref{ref:formula-building-operations}.
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We then conclude with (iii) the proof of the theorem statement.
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\paragraph{(i)}%
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\hyperlabel{par:balanced-parentheses-i}
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By definition, well-formed formulas comprising a single sentence symbol
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do not have any parentheses.
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Thus all sentence symbols are members of $S$.
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\paragraph{(ii)}%
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\hyperlabel{par:balanced-parentheses-ii}
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Let $\alpha, \beta \in S$.
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By definition, $\mathcal{E}_{\neg}(\alpha) = (\neg\alpha)$.
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Thus one additional left and right parenthesis is introduced.
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Since $\alpha$ is assumed to have an equal number of left and right
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parentheses, $\mathcal{E}_{\neg}(\alpha) \in S$.
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Likewise,
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$\mathcal{E}_{\square}(\alpha, \beta) = (\alpha \mathop{\square} \beta)$
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where $\square$ is one of the binary connectives $\land$, $\lor$,
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$\Rightarrow$, $\Leftrightarrow$.
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Again, an additional left and right parenthesis is introduced.
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Since $\alpha$ and $\beta$ are assumed to have a balanced number of
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parentheses, $\mathcal{E}_{\square}(\alpha, \beta) \in S$.
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Hence $S$ is closed under the five formula-building operations.
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\paragraph{(iii)}%
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By \nameref{par:balanced-parentheses-i} and
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\nameref{par:balanced-parentheses-ii}, the
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\nameref{sub:induction-principle-1} implies $S$ is the set of all wffs.
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Thus all well-formed formulas have an equal number of left and right
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parentheses.
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\end{proof}
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\subsection{\verified{Parentheses Count}}%
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\hyperlabel{sub:parentheses-count}
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\begin{lemma}
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Let $\phi$ be a well-formed formula and $c$ be the number of places at which
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a sentential connective symbol exists.
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Then there is $2c$ parentheses in $\phi$.
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\end{lemma}
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\code{Enderton.Logic.Chapter\_1}
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{Enderton.Logic.Chapter\_1.paren\_count\_double\_sentential\_count}
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\begin{proof}
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Define $$S = \{ \phi \mid
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\phi \text{ is a wff with } 2c \text{ parentheses} \}.$$
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We prove that (i) all the sentence symbols are members of $S$ and (ii)
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$S$ is closed under the five \nameref{ref:formula-building-operations}.
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We then conclude with (iii) the proof of the theorem statement.
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\paragraph{(i)}%
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\hyperlabel{par:parentheses-count-i}
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A sentence symbol, by itself, has no sentential connectives.
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Likewise, it has 0 parentheses.
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Thus $S$ contains every sentence symbol.
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\paragraph{(ii)}%
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\hyperlabel{par:parentheses-count-ii}
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Let $\alpha, \beta \in S$.
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By definition, $\mathcal{E}_{\neg}(\alpha) = (\neg \alpha)$.
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Then $\mathcal{E}_{\neg}(\alpha)$ introduces two additional parentheses
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and one additional sentential connective symbol.
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Thus $\mathcal{E}_{\neg}(\alpha) \in S$.
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Likewise,
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$\mathcal{E}_{\square}(\alpha, \beta) = (\alpha \mathop{\square} \beta)$
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where $\square$ is one of the binary connectives $\land$, $\lor$,
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$\Rightarrow$, $\Leftrightarrow$.
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$\mathcal{E}_{\square}(\alpha, \beta)$ also introduces two additional
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parentheses and one additional connective symbol.
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Thus $\mathcal{E}_{\square}(\alpha, \beta) \in S$.
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Hence $S$ is closed under the five formula-building operations.
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\paragraph{(iii)}%
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By \nameref{par:parentheses-count-i} and
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\nameref{par:parentheses-count-ii}, the
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\nameref{sub:induction-principle-1} implies $S$ is the set of all wffs.
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Thus every wff has $2c$ parentheses in $\phi$, where $c$ denotes the
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number of places at which a sentential connective symbol exists.
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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{\unverified{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 should 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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We begin first with the English sentences:
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\begin{enumerate}[(i)]
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\item He can juggle beach balls, bowling pins, and hackysacks unless
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he is tired, in which case he can only juggle beach balls.
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\item
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If Lauren goes to the moves with Sam, he will watch Barbie and
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eat popcorn, but if Lauren does not, he will watch Oppenheimer and
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eat gummy worms.
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\item
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Trees produce oxygen if they are alive and well, able to pull
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nutrients from the earth, and receive ample water.
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\end{enumerate}
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\paragraph{(i)}%
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We use the following translation: "To juggle beach balls" (B),
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"to juggle bowling pins" (P), "to juggle hackysacks" (H), and
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"he is tired" (T).
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This yields the following translation:
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$$(B \land ((\neg T) \Rightarrow (P \land H))).$$
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\paragraph{(ii)}%
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We use the following translation: "Lauren goes to the movies" (L),
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"Sam will watch Oppenheimer" (O), "Sam will watch "Barbie" (B),
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"Sam will eat popcorn" (P), and "Sam will eay gummy worms" (G).
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This yields the following translation:
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$$(((L \land B) \land P) \lor (((\neg L) \land O) \land G)).$$
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\paragraph{(iii)}%
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We use the following translation: "Trees produce oxygen" (O),
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"the tree is alive" (A), "the tree is well" (W), "can pull nutrients
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from the earth" (N), and "receives ample water" (R).
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This yields the following translation:
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$$(O \iff (((A \land W) \land N) \land R)).$$
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\end{answer}
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\subsection{\pending{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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\code*{Enderton.Logic.Chapter\_1}
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{Enderton.Logic.Chapter\_1.exercise\_1\_1\_2\_i}
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\code{Enderton.Logic.Chapter\_1}
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{Enderton.Logic.Chapter\_1.exercise\_1\_1\_2\_ii}
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\begin{proof}
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Define $$S = \{ \phi \mid
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\phi \text{ is a wff and the length of } \phi
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\text{ is not } 2, 3, \text{or } 6. \}.$$
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We prove that (i) all the sentence symbols are members of $S$ and (ii)
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$S$ is closed under the five \nameref{ref:formula-building-operations}.
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We then conclude with (iii) the proof of the theorem statement.
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\paragraph{(i)}%
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\hyperlabel{par:exercise-1.1.2-i}
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Sentence symbols, by definition, have length 1.
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Thus every sentence symbol is a member of $S$.
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\paragraph{(ii)}%
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\hyperlabel{par:exercise-1.1.2-ii}
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Define $L$ to be the length function mapping arbitrary wff to its length.
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Let $\alpha, \beta \in S$.
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Then $L(\alpha)$ and $L(\beta)$ each evaluate to 1, 4, 5, or a value
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larger than 6.
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By definition, $\mathcal{E}_{\neg}(\alpha) = (\neg \alpha)$.
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Thus $L(\mathcal{E}_{\neg}(\alpha)) = L(\alpha) + 3$.
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Enumerating through the possible values of $L(\alpha)$ shows
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$\mathcal{E}_{\neg}(\alpha) \in S$.
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Likewise,
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$\mathcal{E}_{\square}(\alpha, \beta) = (\alpha \mathop{\square} \beta)$
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where $\square$ is one of the binary connectives $\land$, $\lor$,
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$\Rightarrow$, $\Leftrightarrow$.
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Thus $L(\mathcal{E}_{\square}(\alpha, \beta)) = L(\alpha) + L(\beta) + 3$.
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Again, enumerating through the possible values of $L(\alpha)$ and
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$L(\beta)$ shows $\mathcal{E}_{\square}(\alpha, \beta) \in S$.
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Hence $S$ is closed under the five formula-building operations.
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\paragraph{(iii)}%
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By \nameref{par:exercise-1.1.2-i} and \nameref{par:exercise-1.1.2-ii}, the
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\nameref{sub:induction-principle-1} implies $S$ is the set of all wffs.
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It remains to be shown that a wff of any positive length excluding 2, 3,
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and 6 are possible.
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Let $\phi_1 = A_1$, $\phi_2 = (A_1 \land A_2)$, and
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$\phi_3 = ((A_1 \land A_2) \land A_3)$.
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Note these are wffs of lengths 1, 5, and 9 respectively.
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Then $n$ repeated applications of $\mathcal{E}_{\neg}$ yields wffs of
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length $1 + 3n$, $5 + 3n$, and $9 + 3n$ respectively.
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But
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\begin{align*}
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& \{ 1 + 3n \mid n \in \mathbb{N} \}, \\
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& \{ 5 + 3n \mid n \in \mathbb{N} \}, \text{ and } \\
|
|
& \{ 9 + 3n \mid n \in \mathbb{N} \}
|
|
\end{align*}
|
|
form a \nameref{S:ref:partition} of set $\mathbb{N} - \{ 2, 3, 6 \}$.
|
|
Thus a wff of any other positive length besides 2, 3, and 6 is possible.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\verified{Exercise 1.1.3}}%
|
|
\hyperlabel{sub:exercise-1.1.3}
|
|
|
|
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 $s = c + 1$.
|
|
|
|
\code*{Enderton.Logic.Chapter\_1}
|
|
{Enderton.Logic.Chapter\_1.exercise\_1\_1\_3}
|
|
|
|
\begin{proof}
|
|
|
|
Define
|
|
\begin{equation}
|
|
\hyperlabel{sub:exercise-1.1.3-eq1}
|
|
S = \{\phi \mid \phi \text{ is a wff such that } s = c + 1\}.
|
|
\end{equation}
|
|
We prove that (i) all the sentence symbols are members of $S$ and (ii)
|
|
$S$ is closed under the five \nameref{ref:formula-building-operations}.
|
|
We then conclude with (iii) the proof of the theorem statement.
|
|
|
|
\paragraph{(i)}%
|
|
\hyperlabel{par:exercise-1.1.3-i}
|
|
|
|
Let $\phi = A_n$ be an arbitrary sentence symbol.
|
|
The number of places at which sentence symbols occur in $\phi$ is 1.
|
|
The number of places at which binary connective symbols occur in $\phi$ is
|
|
0.
|
|
Hence $\phi \in S$.
|
|
|
|
\paragraph{(ii)}%
|
|
\hyperlabel{par:exercise-1.1.3-ii}
|
|
|
|
Let $\alpha, \beta \in S$.
|
|
Denote the number of places at which sentence symbols occur in each as
|
|
$s_\alpha$ and $s_\beta$ respectively.
|
|
Likewise, denote the number of places at which binary connective symbols
|
|
occur as $c_\alpha$ and $c_\beta$.
|
|
|
|
By definition, $\mathcal{E}_{\neg}(\alpha) = (\neg\alpha)$.
|
|
The number of sentence and binary connective symbols in
|
|
$\mathcal{E}_{\neg}(\alpha)$ does not change.
|
|
Thus $\mathcal{E}_{\neg}(\alpha) \in S$.
|
|
Likewise,
|
|
$\mathcal{E}_{\square}(\alpha, \beta) = (\alpha \mathop{\square} \beta)$
|
|
where $\square$ is one of the binary connectives $\land$, $\lor$,
|
|
$\Rightarrow$, $\Leftrightarrow$.
|
|
Therefore $\mathcal{E}_{\square}(\alpha, \beta)$ has $s_\alpha + s_\beta$
|
|
sentence symbols and $c_\alpha + c_\beta + 1$ binary connective symbols.
|
|
But \eqref{sub:exercise-1.1.3-eq1} implies
|
|
\begin{align*}
|
|
s_\alpha + s_\beta
|
|
& = (c_\alpha + 1) + (c_\beta + 1) \\
|
|
& = (c_\alpha + c_\beta + 1) + 1,
|
|
\end{align*}
|
|
meaning $\mathcal{E}_{\square}(\alpha, \beta) \in S$.
|
|
|
|
Hence $S$ is closed under the five formula-building operations.
|
|
|
|
\paragraph{(iii)}%
|
|
\hyperlabel{par:exercise-1.1.3-iii}
|
|
|
|
By \nameref{par:exercise-1.1.3-i} and \nameref{par:exercise-1.1.3-ii}, the
|
|
\nameref{sub:induction-principle-1} indicates $S$ is the set of all
|
|
wffs.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Exercise 1.1.4}}%
|
|
\hyperlabel{sub:exercise-1.1.4}
|
|
|
|
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.
|
|
|
|
\begin{proof}
|
|
|
|
|
|
Let $S$ denote a \nameref{ref:construction-sequence}
|
|
$\ltuple{\epsilon_1}{\epsilon_n}$ such that $\epsilon_n = \phi$.
|
|
Let $S' = \ltuple{\epsilon_{i_1}}{\epsilon_{i_m}}$ denote the construction
|
|
sequence resulting from deleting all expressions in $S$ containing $A_4$.
|
|
Fix $1 \leq j \leq m$.
|
|
Then there exists some $1 \leq k \leq n$ such that
|
|
$\epsilon_{i_j} = \epsilon_k$.
|
|
By definition of a construction sequence, there are three cases to consider:
|
|
|
|
\paragraph{Case 1}%
|
|
|
|
Suppose $\epsilon_k$ is a sentence symbol.
|
|
Then $\epsilon_{i_j}$ is also sentence symbol.
|
|
|
|
\paragraph{Case 2}%
|
|
|
|
Suppose $\epsilon_k = \mathcal{E}_{\neg}(\epsilon_a)$ for some $a < k$.
|
|
It must be that $A_4$ is not found in $\epsilon_a$, else an immediate
|
|
contradiction is raised.
|
|
Therefore $\epsilon_a$ is a member of $S'$ that precedes $\epsilon_{i_j}$.
|
|
Hence $\epsilon_{i_j} = \mathcal{E}_{\neg}(\epsilon_{i_a})$ for some
|
|
$a < j$.
|
|
|
|
\paragraph{Case 3}%
|
|
|
|
Suppose $\epsilon_k = \mathcal{E}_{\square}(\epsilon_a, \epsilon_b)$ for
|
|
some $a, b < k$ where $\square$ is one of the binary connectives
|
|
$\land$, $\lor$, $\Rightarrow$, $\Leftrightarrow$.
|
|
It must be that $A_4$ is found in neither $\epsilon_a$ nor $\epsilon_b$,
|
|
else an immediate contradiction is raised.
|
|
Therefore $\epsilon_a$ and $\epsilon_b$ is a member of $S'$, both of which
|
|
precede $\epsilon_{i_j}$.
|
|
Hence
|
|
$\epsilon_{i_j} = \mathcal{E}_{\square}(\epsilon_{i_a}, \epsilon_{i_b})$
|
|
for some $a, b < j$.
|
|
|
|
\paragraph{Conclusion}%
|
|
|
|
Since the above cases are exhaustive and apply to an arbitrary member of
|
|
$S'$, it must be that every member of $S'$ is valid.
|
|
Hence $S'$ is still a legal construction sequence.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\verified{Exercise 1.1.5}}%
|
|
\hyperlabel{sub:exercise-1.1.5}
|
|
|
|
Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
|
|
|
|
\subsubsection{\verified{Exercise 1.1.5a}}%
|
|
\hyperlabel{ssub:exercise-1.1.5.a}
|
|
|
|
Show that the length of $\alpha$ (i.e., the number of symbols in the string)
|
|
is odd.
|
|
\textit{Suggestion}: Apply induction to show that the length is of the form
|
|
$4k + 1$.
|
|
|
|
\code*{Enderton.Logic.Chapter\_1}
|
|
{Enderton.Logic.Chapter\_1.exercise\_1\_1\_5\_a}
|
|
|
|
\begin{proof}
|
|
|
|
Define $L$ to be the length function mapping arbitrary
|
|
\nameref{ref:well-formed-formula} to its length and let
|
|
\begin{equation}
|
|
\hyperlabel{ssub:exercise-1.1.5.a-eq1}
|
|
S = \{\phi \mid
|
|
\phi \text{ is a wff containing } \neg \text{ or }
|
|
\exists k \in \mathbb{N}, L(\phi) = 4k + 1\}.
|
|
\end{equation}
|
|
We prove that (i) all the sentence symbols are members of $S$ and (ii)
|
|
$S$ is closed under the five \nameref{ref:formula-building-operations}.
|
|
We then conclude with (iii) the proof of the theorem statement.
|
|
|
|
\paragraph{(i)}%
|
|
\hyperlabel{par:exercise-1.1.5.a-i}
|
|
|
|
Every sentence symbol has length 1 by definition.
|
|
That is, every sentence symbol has length $(4)(0) + 1$.
|
|
Hence $S$ contains every sentence symbol.
|
|
|
|
\paragraph{(ii)}%
|
|
\hyperlabel{par:exercise-1.1.5.a-ii}
|
|
|
|
Let $\alpha, \beta \in S$.
|
|
Then there exists some $k_\alpha$ and $k_\beta$ such that
|
|
$L(\alpha) = 4k_\alpha + 1$ and $L(\beta) = 4k_\beta + 1$.
|
|
Clearly $S$ is closed under $\mathcal{E}_{\neg}$.
|
|
Next consider
|
|
$\mathcal{E}_{\square}(\alpha, \beta) = (\alpha \mathop{\square} \beta)$
|
|
where $\square$ is one of the binary connectives $\land$, $\lor$,
|
|
$\Rightarrow$, $\Leftrightarrow$.
|
|
Then
|
|
\begin{align*}
|
|
L(\alpha, \beta)
|
|
& = L(\alpha) + L(\beta) + 3 \\
|
|
& = (4k_\alpha + 1) + (4k_\beta + 1) + 3 \\
|
|
& = 4k_\alpha + 4k_\beta + 4 + 1 \\
|
|
& = 4(k_\alpha + k_\beta + 1) + 1.
|
|
\end{align*}
|
|
Therefore, there exists a $k \in \mathbb{N}$, namely
|
|
$k = k_\alpha + k_\beta + 1$, such that
|
|
$L(\mathcal{E}_{\square}(\alpha, \beta)) = 4k + 1$.
|
|
|
|
Hence $S$ is closed under the five formula-building operations.
|
|
|
|
\paragraph{(iii)}%
|
|
|
|
By \nameref{par:exercise-1.1.5.a-i} and \nameref{par:exercise-1.1.5.a-ii},
|
|
the \nameref{sub:induction-principle-1} indicates $S$ is the set of all
|
|
wffs.
|
|
Thus all well-formed formulas not containing symbol $\neg$ has length
|
|
$4k + 1$ for some $k \in \mathbb{N}$.
|
|
Therefore these well-formed formulas have odd length.
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\verified{Exercise 1.1.5b}}%
|
|
\hyperlabel{ssub:exercise-1.1.5-b}
|
|
|
|
Show that more than a quarter of the symbols are sentence symbols.
|
|
\textit{Suggestion}: Apply induction to show that the number of sentence
|
|
symbols is of the form $k + 1$.
|
|
|
|
\code*{Enderton.Logic.Chapter\_1}
|
|
{Enderton.Logic.Chapter\_1.exercise\_1\_1\_5\_b}
|
|
|
|
\begin{proof}
|
|
|
|
Let $\phi$ be a \nameref{ref:well-formed-formula}.
|
|
By \nameref{sub:exercise-1.1.3}, the number of sentence symbols of $\phi$ is
|
|
$k + 1$, where $k$ is the number of places at which binary connective
|
|
symbols occur in $\phi$.
|
|
By \nameref{sub:parentheses-count}, the number of parentheses in $\phi$ is
|
|
$2k$.
|
|
Thus $\phi$ has length $(k + 1) + k + 2k = 4k + 1$.
|
|
But $$\frac{k + 1}{4k + 1} > \frac{k + 1}{4k + 4} = \frac{1}{4}.$$
|
|
Hence more than a quarter of the symbols of $\phi$ are sentence symbols.
|
|
|
|
\end{proof}
|
|
|
|
\end{document}
|