Enderton (logic). Draft up chapter 0 and chapter 1 theorems/exercises.

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}