303 lines
9.6 KiB
TeX
303 lines
9.6 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 \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{\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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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 \textbf{formula-building operations} (on expressions) defined by
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the 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{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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\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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\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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