\documentclass{report} \input{../../preamble} \makecode{../..} \externaldocument[S:]{Set} \begin{document} \header{A Mathematical Introduction to Logic}{Herbert B. Enderton} \tableofcontents \begingroup \renewcommand\thechapter{R} \chapter{Reference}% \hyperlabel{chap:reference} \section{\defined{Construction Sequence}}% \hyperlabel{ref:construction-sequence} A \textbf{construction sequence} is a \nameref{ref:finite-sequence} $\ltuple{\epsilon_1}{\epsilon_n}$ of \nameref{ref:expression}s 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} An \textbf{expression} is a \nameref{ref:finite-sequence} of symbols. \section{\defined{Finite Sequence}}% \hyperlabel{ref:finite-sequence} $S$ is a \textbf{finite sequence} (or \textbf{string}) of members of set $A$ if and only if, for some positive integer $n$, we have $S = \ltuple{x_1}{x_n}$, where each $x_i \in A$. \section{\defined{Formula-Building Operations}}% \hyperlabel{ref:formula-building-operations} The \textbf{formula-building operations} (on expressions) are 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*} \code*{Bookshelf/Enderton/Logic/Chapter\_1} {Enderton.Logic.Chapter\_1.WFF} \lean{Init/Prelude} {Not} \lean{Init/Prelude} {And} \lean{Init/Prelude} {Or} \lean{Init/Core} {Iff} \section{\defined{\texorpdfstring{$n$}{n}-tuple}}% \hyperlabel{ref:n-tuple} An \textbf{$n$-tuple} is recursively defined as $$\ltuple{x_1}{x_{n+1}} = \tuple{\ltuple{x_1}{x_n}, x_{n+1}}$$ for $n > 1$. We also define $\tuple{x} = x$. \lean*{Init/Prelude} {Prod} \section{\defined{Well-Formed Formula}}% \hyperlabel{ref:well-formed-formula} A \textbf{well-formed formula} (wff) is an \nameref{ref:expression} that can be built up from the sentence symbols by applying some finite number of times the \nameref{ref:formula-building-operations}. \code*{Bookshelf/Enderton/Logic/Chapter\_1} {Enderton.Logic.Chapter\_1.WFF} \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{\unverified{Lemma 0A}}% \hyperlabel{sec:lemma-0a} \begin{lemma}[0A] Assume that $\ltuple{x_1}{x_m} = \ltuple{y_1, \ldots, y_m}{y_{m+k}}$. Then $x_1 = \ltuple{y_1}{y_{k+1}}$. \end{lemma} \begin{proof} For natural number $m$, let $P(m)$ be the statement: \begin{induction} \hyperlabel{sec:lemma-0a-ih} If $\ltuple{x_1}{x_m} = \ltuple{y_1, \ldots, y_m}{y_{m+k}}$ then $x_1 = \ltuple{y_1}{y_{k+1}}$. \end{induction} \noindent We proceed by induction on $m$. \paragraph{Base Case}% Suppose $\tuple{x_1} = \ltuple{y_1}{y_{1 + k}}$. By definition of an \nameref{ref:n-tuple}, $\tuple{x_1} = x_1$. Thus $x_1 = \ltuple{y_1}{y_{k + 1}}$. Hence $P(1)$ holds true. \paragraph{Inductive Step}% Suppose for $m \geq 1$ that $P(m)$ is true and assume \begin{equation} \hyperlabel{sec:lemma-0a-eq1} \ltuple{x_1}{x_{m+1}} = \ltuple{y_1, \ldots, y_{m+1}}{y_{m+1+k}}. \end{equation} By definition of an \nameref{ref:n-tuple}, we can decompose \eqref{sec:lemma-0a-eq1} into the following two identities \begin{align*} x_{m+1} & = y_{m+1+k} \\ \ltuple{x_1}{x_m} & = \ltuple{y_1}{y_{m+k}}. \end{align*} By \ihref{sec:lemma-0a-ih}, $P(m)$ implies $x_1 = \ltuple{y_1}{y_{k+1}}$. Hence $P(m+1)$ holds true. \paragraph{Conclusion}% By induction, $P(m)$ holds true for all $m \geq 1$. \end{proof} \chapter{Sentential Logic}% \hyperlabel{chap:sentential-logic} \section{The Language of Sentential Logic}% \hyperlabel{sec:language-sentential-logic} \subsection{\unverified{Induction Principle}}% \hyperlabel{sub:induction-principle-1} \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} \begin{proof} We note every \nameref{ref:well-formed-formula} can be characterized by a \nameref{ref:construction-sequence}. For natural number $m$, let $P(m)$ be the statement: \begin{induction} \hyperlabel{sub:induction-principle-1-ih} Every wff characterized by a construction sequence of length $m$ is in $S$. \end{induction} \noindent We proceed by strong induction on $m$. \paragraph{Base Case}% Let $\phi$ denote a wff characterized by a construction sequence of length $1$. Then it must be that $\phi$ is a single sentence symbol. By hypothesis, $S$ contains all the sentence symbols. Thus $P(1)$ holds true. \paragraph{Inductive Step}% Suppose $P(0)$, $P(1)$, $\ldots$, $P(m)$ holds true and let $\phi$ denote a wff characterized by a construction sequence of length $m + 1$. By definition of a construction sequence, one of the following holds: \begin{align} & \phi \text{ is a sentence symbol} & \label{sub:induction-principle-1-eq1} \\ & \phi = \mathcal{E}_\neg(\epsilon_j) \text{ for some } j < m + 1 & \label{sub:induction-principle-1-eq2} \\ & \phi = \mathcal{E}_\square(\epsilon_j, \epsilon_k) \text{ for some } j < m + 1, k < m + 1 & \label{sub:induction-principle-1-eq3} \end{align} where $\square$ is one of the binary connectives $\land$, $\lor$, $\Rightarrow$, $\Leftrightarrow$. We consider each case in turn. \subparagraph{\eqref{sub:induction-principle-1-eq1}}% By hypothesis, all sentence symbols are in $S$. Thus $\phi \in S$. \subparagraph{\eqref{sub:induction-principle-1-eq2}}% Suppose $\phi = \mathcal{E}_\neg(\epsilon_j)$ for some $j < m + 1$. By \ihref{sub:induction-principle-1-ih}, $\epsilon_j$ is in $S$. By hypothesis, $S$ is closed under $\mathcal{E}_\neg$. Thus $\phi \in S$. \subparagraph{\eqref{sub:induction-principle-1-eq3}}% Suppose $\phi = \mathcal{E}_\square(\epsilon_j, \epsilon_k)$ for some $j < m + 1, k < m + 1$, By \ihref{sub:induction-principle-1-ih}, $\epsilon_j$ and $\epsilon_k$ is in $S$. By hypothesis, $S$ is closed under $\mathcal{E}_\square$ for all possible candidates of $\square$. Thus $\phi \in S$. \subparagraph{Subconclusion}% Since the above three cases are exhaustive, $P(m + 1)$ holds. \paragraph{Conclusion}% By strong induction, $P(m)$ holds true for all natural numbers $m \geq 1$. Since every well-formed formula is characterized by a construction sequence, the set of all wffs is a subset of $S$. Likewise, it obviously holds that $S$ is a subset of all wffs. Thus $S$ is precisely the set of all wffs. \end{proof} \section{Exercises 1}% \hyperlabel{sec:exercises-1} \subsection{\unverified{Exercise 1.1.1}}% \hyperlabel{sub:exercise-1.1.1} Give three sentences in English together with translations into our formal language. The sentences should be chosen so as to have an interesting structure, and the translations should each contain 15 or more symbols. \begin{answer} We begin first with the English sentences: \begin{enumerate}[i] \item He can juggle beach balls, bowling pins, and hackysacks unless he is tired, in which case he can only juggle beach balls. \item If Lauren goes to the moves with Sam, he will watch Barbie and eat popcorn, but if Lauren does not, he will watch Oppenheimer and eat gummy worms. \item Trees produce oxygen if they are alive and well, able to pull nutrients from the earth, and receive ample water. \end{enumerate} \paragraph{(i)}% We use the following translation: "To juggle beach balls" (B), "to juggle bowling pins" (P), "to juggle hackysacks" (H), and "he is tired" (T). This yields the following translation: $$(B \land ((\neg T) \Rightarrow (P \land H))).$$ \paragraph{(ii)}% We use the following translation: "Lauren goes to the movies" (L), "Sam will watch Oppenheimer" (O), "Sam will watch "Barbie" (B), "Sam will eat popcorn" (P), and "Sam will eay gummy worms" (G). This yields the following translation: $$(((L \land B) \land P) \lor (((\neg L) \land O) \land G)).$$ \paragraph{(iii)}% We use the following translation: "Trees produce oxygen" (O), "the tree is alive" (A), "the tree is well" (W), "can pull nutrients from the earth" (N), and "receives ample water" (R). This yields the following translation: $$(O \iff (((A \land W) \land N) \land R)).$$ \end{answer} \subsection{\pending{Exercise 1.1.2}}% \hyperlabel{sub:exercise-1.1.2} Show that there are no wffs of length 2, 3, or 6, but that any other positive length is possible. \code*{Enderton.Logic.Chapter\_1} {Enderton.Logic.Chapter\_1.exercise\_1\_1\_2\_i} \code{Enderton.Logic.Chapter\_1} {Enderton.Logic.Chapter\_1.exercise\_1\_1\_2\_ii} \begin{proof} Define $$S = \{ \phi \mid \phi \text{ is a wff and the length of } \phi \text{ is not } 2, 3, \text{or } 6. \}.$$ 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-1.1.2-i} Sentence symbols, by definition, have length 1. Thus every sentence symbol is a member of $S$. \paragraph{(ii)}% \hyperlabel{par:exercise-1.1.2-ii} Define $L$ to be the length function mapping arbitrary wff to its length. Let $\phi, \psi \in S$. Then $L(\phi)$ and $L(\psi)$ each evaluate to 1, 4, 5, or a value larger than 6. By definition, $\mathcal{E}_{\neg}(\phi) = (\neg \phi)$. Thus $L(\mathcal{E}_{\neg}(\phi)) = L(\phi) + 3$. Enumerating through the possible values of $L(\phi)$ shows $\mathcal{E}_{\neg}(\phi) \in S$. Likewise, $\mathcal{E}_{\square}(\phi, \psi) = (\phi \mathop{\square} \psi)$ where $\square$ is one of the binary connectives $\land$, $\lor$, $\Rightarrow$, $\Leftrightarrow$. Thus $L(\mathcal{E}_{\square}(\phi, \psi)) = L(\phi) + L(\psi) + 3$. Again, enumerating through the possible values of $L(\phi)$ and $L(\psi)$ shows $\mathcal{E}_{\square}(\phi, \psi) \in S$. Hence $S$ is closed under the five formula-building operations. \paragraph{(iii)}% By \nameref{par:exercise-1.1.2-i} and \nameref{par:exercise-1.1.2-ii}, the \nameref{sub:induction-principle-1} implies $S$ is the set of all wffs. It remains to be shown that a wff of any positive length excluding 2, 3, and 6 are possible. Let $\phi_1 = A_1$, $\phi_2 = (A_1 \land A_2)$, and $\phi_3 = ((A_1 \land A_2) \land A_3)$. Note these are wffs of lengths 1, 5, and 9 respectively. Then $n$ repeated applications of $\mathcal{E}_{\neg}$ yields wffs of length $1 + 3n$, $5 + 3n$, and $9 + 3n$ respectively. But \begin{align*} & \{ 1 + 3n \mid n \in \mathbb{N} \}, \\ & \{ 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{\pending{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 $s = c + 1$. \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, $\mathcal{E}_{\square}(\phi, \psi) = (\phi \mathop{\square} \psi)$ where $\square$ is one of the binary connectives $\land$, $\lor$, $\Rightarrow$, $\Leftrightarrow$. Therefore $\mathcal{E}_{\square}(\phi, \psi)$ 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 $\mathcal{E}_{\square}(\phi, \psi) \in S$. 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{\sorry{Exercise 1.1.5}}% \hyperlabel{sub:exercise-1.1.5} Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$. \subsubsection{\sorry{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 $4k + 1$. \begin{proof} TODO \end{proof} \subsubsection{\sorry{Exercise 1.1.6a}}% \hyperlabel{ssub:exercise-1.1.6-a} 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 $k + 1$. \begin{proof} TODO \end{proof} \subsection{\sorry{Exercise 1.1.6}}% \hyperlabel{sub:exercise-1.1.6} 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} \begin{proof} TODO \end{proof} \end{document}