From 8f15ec7d6dbf2fb367fdfd346471cb4282b82155 Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Tue, 8 Aug 2023 14:50:48 -0600 Subject: [PATCH] Enderton (logic). Draft up chapter 0 and chapter 1 theorems/exercises. --- Bookshelf/Enderton/Logic.tex | 165 +++++++++++++++++++++++++++++++++++ 1 file changed, 165 insertions(+) diff --git a/Bookshelf/Enderton/Logic.tex b/Bookshelf/Enderton/Logic.tex index 0262d80..1cf57f6 100644 --- a/Bookshelf/Enderton/Logic.tex +++ b/Bookshelf/Enderton/Logic.tex @@ -15,6 +15,45 @@ \chapter{Reference}% \hyperlabel{chap:reference} +\section{\defined{Construction Sequence}}% +\hyperlabel{ref:construction-sequence} + +A \textbf{construction sequence} is a finite sequence + $\langle \epsilon_1, \ldots, \epsilon_n \rangle$ of expressions such that for + each $i \leq n$ we have at least one of + \begin{align*} + & \epsilon_i \text{ is a sentence symbol} \\ + & \epsilon_i = \mathcal{E}_\neg(\epsilon_j) \text{ for some } j < i \\ + & \epsilon_i = \mathcal{E}_\square(\epsilon_j, \epsilon_k) + \text{ for some } j < i, k < i + \end{align*} + where $\square$ is one of the binary connectives $\land$, $\lor$, + $\Rightarrow$, $\Leftrightarrow$. + +\section{\defined{Expression}}% +\hyperlabel{ref:expression} + +An \textbf{expression} is a finite sequence of symbols. + +\section{\defined{Well-Formed Formula}}% +\hyperlabel{ref:well-formed-formula} + +An \nameref{ref:expression} that can be built up from the sentence symbols by + applying some finite number of times the \textbf{formula-building operations} + (on expressions) defined by the equations: + \begin{align*} + \mathcal{E}_{\neg}(\alpha) + & = (\neg \alpha) \\ + \mathcal{E}_{\land}(\alpha, \beta) + & = (\alpha \land \beta) \\ + \mathcal{E}_{\lor}(\alpha, \beta) + & = (\alpha \lor \beta) \\ + \mathcal{E}_{\Rightarrow}(\alpha, \beta) + & = (\alpha \Rightarrow \beta) \\ + \mathcal{E}_{\Leftrightarrow}(\alpha, \beta) + & = (\alpha \Leftrightarrow \beta) + \end{align*} + \endgroup % Reset counter to mirror Enderton's book. @@ -26,8 +65,134 @@ \section{\sorry{Lemma 0A}}% \hyperlabel{sec:lemma-0a} +\begin{lemma}[0A] + Assume that $\langle x_1, \ldots, x_m \rangle = \langle y_1, \ldots, y_m, \ldots, y_{m+k} \rangle$. Then $x_1 = \langle y_1, \ldots, y_{k+1} \rangle$. +\end{lemma} + +\begin{proof} + + TODO + +\end{proof} + +\chapter{Sentential Logic}% +\hyperlabel{chap:sentential-logic} + +\section{The Language of Sentential Logic}% +\hyperlabel{sec:language-sentential-logic} + +\subsection{\sorry{Induction Principle}}% +\hyperlabel{sub:induction-principle-1} + +\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} + + TODO + +\end{proof} + +\section{Exercises 1}% +\hyperlabel{sec:exercises-1} + +\subsection{\sorry{Exercise 1.1}}% +\hyperlabel{sub:exercise-1.1} + +Give three sentences in English together with translations into our formal + language. +The sentences shoudl be chosen so as to have an interesting structure, and the + translations should each contain 15 or more symbols. + +\begin{answer} + + TODO + +\end{answer} + +\subsection{\sorry{Exercise 1.2}}% +\hyperlabel{sub:exercise-1.2} + +Show that there are no wffs of length 2, 3, or 6, but that any other positive + length is possible. + +\begin{answer} + + TODO + +\end{answer} + +\subsection{\sorry{Exercise 1.3}}% +\hyperlabel{sub:exercise-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 exmaple, if $\alpha$ is $(A \Rightarrow (\neg A))$ then $c = 1$ and $s = 2$.) +Show by using the induction principle that $s = c + 1$. + +\begin{answer} + + TODO + +\end{answer} + +\subsection{\sorry{Exercise 1.4}}% +\hyperlabel{sub:exercise-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{answer} + + TODO + +\end{answer} + +\subsection{\sorry{Exercise 1.5}}% +\hyperlabel{sub:exercise-1.5} + +Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$. +\begin{enumerate}[(a)] + \item Show that the length of $\alpha$ (i.e., the number of symbols in the + string) is odd. + \item Show that more than a quarter of the symbols are sentence symbols. +\end{enumerate} +\textit{Suggestion}: Apply induction to show that the length is of the form + $4k + 1$ and the number of sentence symbols is $k + 1$. + +\begin{answer} + + TODO + +\end{answer} + +\subsection{\sorry{Exercise 1.6}}% +\hyperlabel{sub:exercise-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{answer} + + TODO + +\end{answer} + \end{document}