From c458eca8ae7b1f3ef3a3f4ab84be76e7156b3301 Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Mon, 21 Aug 2023 14:51:49 -0600 Subject: [PATCH] Enderton (logic). 1.2.6-1.2.9. --- Bookshelf/Enderton/Logic.tex | 239 ++++++++++++++++++++++++++++++++--- 1 file changed, 222 insertions(+), 17 deletions(-) diff --git a/Bookshelf/Enderton/Logic.tex b/Bookshelf/Enderton/Logic.tex index 71cc037..e7de261 100644 --- a/Bookshelf/Enderton/Logic.tex +++ b/Bookshelf/Enderton/Logic.tex @@ -272,13 +272,13 @@ 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} \\ + & \hyperlabel{sub:induction-principle-1-eq1} \\ & \phi = \mathcal{E}_\neg(\epsilon_j) \text{ for some } j < m + 1 - & \label{sub:induction-principle-1-eq2} \\ + & \hyperlabel{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} + & \hyperlabel{sub:induction-principle-1-eq3} \end{align} where $\square$ is one of the binary connectives $\land$, $\lor$, $\Rightarrow$, $\Leftrightarrow$. @@ -1300,7 +1300,7 @@ \end{proof} -\subsection{\sorry{Exercise 1.2.6b}}% +\subsection{\unverified{Exercise 1.2.6b}}% \hyperlabel{sub:exercise-1.2.6b} Let $\mathcal{S}$ be a set of sentence symbols that includes those in $\Sigma$ @@ -1315,10 +1315,32 @@ are a great many -- uncountably many -- of them.) \begin{proof} - TODO + + Let $\mathcal{S}$ be a set of sentence symbols that includes those in + $\Sigma$ and $\tau$ (and possibly more). + Let $\mathcal{S}' \subseteq \mathcal{S}$ be the set containing precisely + the sentence symbols found in $\Sigma$ and $\tau$. + + Let $v$ be a truth assignment for $S'$ that satisfies every member of + $\Sigma$ and $w$ be an arbitrary extension of $v$ for $S$. + By construction, $v$ and $w$ agree on all the sentence symbols found in both + $\Sigma$ and $\tau$. + \nameref{sub:exercise-1.2.6a} then implies $\bar{v}(\tau) = \bar{w}(\tau)$ + and $\bar{v}(\sigma) = \bar{w}(\sigma)$ for all $\sigma \in \Sigma$. + Thus $v$ satisfies every member of $\Sigma$ if and only if $w$ satisfies + every member of $\Sigma$. + Likewise, $v$ satisfies $\tau$ if and only if $w$ satisfies $\tau$. + + Hence, by definition of \nameref{ref:tautological-implication}, + $\Sigma \vDash \tau$ if and only if every truth assignment for the + sentence symbols in $S'$ that satisfies every member of $\Sigma$ also + satisfies $\tau$ if and only if every truth assignment for the sentence + symbols in $S$ that satisfies every member of $\Sigma$ also satisfies + $\tau$. + \end{proof} -\subsection{\sorry{Exercise 1.2.7}}% +\subsection{\unverified{Exercise 1.2.7}}% \hyperlabel{sub:exercise-1.2.7} You are in a land inhabited by people who either always tell the truth or @@ -1331,29 +1353,103 @@ \textit{Suggestion}: Make a table. \begin{proof} - TODO + Consider the self-referential question, + "Would you respond 'yes' to the question, + 'Should I take the left road to get to the capital?'" + Let $I$ denote whether the inhabitant is truthful, and $L$ denote whether + the left road actually goes to the capital. + We have $R$ denote the answer given by the inhabitant in the following + "truth table": + $$\begin{array}{s|s|e} + I & L & R \\ + \hline + T & T & \text{Yes} \\ + T & F & \text{No} \\ + F & T & \text{Yes} \\ + F & F & \text{No} + \end{array}$$ + Regardless of the inhabitant's honesty, we receive the answer "Yes" if and + only if the left road actually goes to the capital. \end{proof} -\subsection{\sorry{Exercise 1.2.8}}% +\subsection{\unverified{Exercise 1.2.8}}% \hyperlabel{sub:exercise-1.2.8} - (Substitution) Consider a suquence $\alpha_1, \alpha_2, \ldots$ of wffs. + (Substitution) Consider a sequence $\alpha_1, \alpha_2, \ldots$ of wffs. For each wff $\phi$ let $\phi^*$ be the result of replacing the sentence symbol $A_n$ by $\alpha_n$ for each $n$. -\subsubsection{\sorry{Exercise 1.2.8a}}% +\subsubsection{\unverified{Exercise 1.2.8a}}% \hyperlabel{ssub:exercise-1.2.8a} Let $v$ be a truth assignment for the set of all sentence symbols; define $u$ to be the truth assignment for which $u(A_n) = \bar{v}(\alpha_n)$. Show that $\bar{u}(\phi) = \bar{v}(\phi^*)$. - Use the induction principle. + Use the \nameref{sub:induction-principle-1}. \begin{proof} - TODO + + Let $$S = \{\phi \mid + \phi \text{ is a wff such that } \bar{u}(\phi) = \bar{v}(\phi^*)\}.$$ + We prove that (i) $S$ contains the set of all sentence symbols and (ii) $S$ + is closed under the five \nameref{ref:formula-building-operations}. + Afterward we prove that (iii) our theorem statement holds. + + \paragraph{(i)}% + \hyperlabel{par:exercise-1.2.8-i} + + Let $\phi = A_n$ be an arbitrary sentence symbol. + By definition, $u(A_n) = \bar{v}(\alpha_n)$. + Then $$\bar{u}(\phi) + = \bar{u}(A_n) \\ + = u(A_n) \\ + = \bar{v}(\alpha_n) \\ + = \bar{v}(\phi^*).$$ + Hence every sentence symbol is in $S$. + + \paragraph{(ii)}% + \hyperlabel{par:exercise-1.2.8-ii} + + Let $\beta, \gamma \in S$. + That is, $\bar{u}(\beta) = \bar{v}(\beta^*)$ and + $\bar{u}(\gamma) = \bar{v}(\gamma^*)$. + By definition, $\mathcal{E}_{\neg}(\beta) = (\neg\beta)$. + Therefore + \begin{align*} + \bar{u}(\mathcal{E}_{\neg}(\beta)) + & = (\neg\bar{u}(\beta)) \\ + & = (\neg\bar{v}(\beta^*)) \\ + & = \bar{v}((\neg\beta^*)) \\ + & = \bar{v}((\neg\beta)^*) \\ + & = \bar{v}(\mathcal{E}_{\neg}(\beta)^*). + \end{align*} + Likewise, + $\mathcal{E}_{\square}(\beta, \gamma) = + (\beta \mathop{\square} \gamma)$ + where $\square$ is one of the binary connectives $\land$, $\lor$, + $\Rightarrow$, $\Leftrightarrow$. + Therefore + \begin{align*} + \bar{u}(\mathcal{E}_{\square}(\beta, \gamma)) + & = \bar{u}(\beta) \mathop{\square} \bar{u}(\gamma) \\ + & = \bar{v}(\beta^*) \mathop{\square} \bar{v}(\gamma^*) \\ + & = \bar{v}((\beta^* \mathop{\square} \gamma^*)) \\ + & = \bar{v}((\beta \mathop{\square} \gamma)^*) \\ + & = \bar{v}(\mathcal{E}_{\square}(\beta, \gamma)^*). + \end{align*} + Hence $S$ is closed under the five formula-building operations. + + \paragraph{(iii)}% + + By \nameref{par:exercise-1.2.8-i} and \nameref{par:exercise-1.2.8-ii}, + the \nameref{sub:induction-principle-1} implies $S$ is the set of all + wffs. + Thus for any well-formed formula $\phi$, + $\bar{u}(\phi) = \bar{v}(\phi^*)$. + \end{proof} -\subsubsection{\sorry{Exercise 1.2.8b}}% +\subsubsection{\unverified{Exercise 1.2.8b}}% \hyperlabel{ssub:exercise-1.2.8b} Show that if $\phi$ is a tautology, then so is $\phi^*$. @@ -1364,21 +1460,130 @@ tautology, for any wffs $\alpha$ and $\beta$.) \begin{proof} - TODO + Suppose $\phi$ is a tautology and let $S$ be the set of all sentence + symbols. + Let $v$ be a truth assignment for $S$ and define $u$ to be the truth + assignment for which $u(A_n) = \bar{v}(\alpha_n)$. + By \nameref{ssub:exercise-1.2.8a}, $\bar{u}(\phi) = \bar{v}(\phi^*)$. + Since $\phi$ is a tautology, $\bar{u}(\phi)$ is true meaning + $\bar{v}(\phi^*)$ is also true. + Since $v$ is an arbitrary truth assignment, it follows that every truth + assignment for $S$ satisfies $\phi^*$. + By \nameref{sub:exercise-1.2.6b}, $\vDash \phi^*$, i.e. $\phi^*$ is a + tautology. \end{proof} -\subsection{\sorry{Exercise 1.2.9}}% +\subsection{\unverified{Exercise 1.2.9}}% \hyperlabel{sub:exercise-1.2.9} (Duality) Let $\alpha$ be a wff whose only connective symbols are $\land$, $\lor$, and $\neg$. Let $\alpha^*$ be the result of interchanging $\land$ and $\lor$ and replacing each sentence symbol by its negation. - Show that $\alpha^*$ is tautologically equivalent to $(\neg \alpha)$. + Show that $\alpha^*$ is tautologically equivalent to $(\neg\alpha)$. Use the \nameref{sub:induction-principle-1}. + \code*{Common/Logic/Basic} + {not\_and\_de\_morgan} + + \code{Common/Logic/Basic} + {not\_or\_de\_morgan} + \begin{proof} - TODO + + Let + \begin{align*} + S = \{ \alpha \mid + & \alpha \text{ is a wff containing a } \Rightarrow \text{ or } \\ + & \alpha \text{ is a wff containing a } \Leftrightarrow \text{ or } \\ + & \alpha^* \vDash \Dashv (\neg\alpha) \}. + \end{align*} + We prove that (i) $S$ contains the set of all sentence symbols and (ii) $S$ + is closed under the five \nameref{ref:formula-building-operations}. + Afterward we prove that (iii) our theorem statement holds. + + \paragraph{(i)}% + \hyperlabel{par:exercise-1.2.9-i} + + Let $\alpha = A_n$ be an arbitrary sentence symbol. + By definition, $$\alpha^* = A_n^* = (\neg A_n) = (\neg\alpha).$$ + Hence every sentence symbol is in $S$. + + \paragraph{(ii)}% + \hyperlabel{par:exercise-1.2.9-ii} + + Let $\alpha, \beta \in S$. + Suppose $\alpha$ contains a $\Rightarrow$ or $\Leftrightarrow$ symbol. + Then $\mathcal{E}_{\neg}(\alpha)$ also does. + The same holds for $\beta$. + Furthermore, if either $\alpha$ or $\beta$ contains a $\Rightarrow$ or + $\Leftrightarrow$ symbol, then so does + $\mathcal{E}_{\square}(\alpha, \beta)$ + where $\square$ is one of the binary connectives $\land$, $\lor$, + $\Rightarrow$, $\Leftrightarrow$. + In any of these above cases, it is trivial to see each of the five-formula + building operations take a wff from $S$ and produce another wff in $S$. + + Now, suppose neither $\alpha$ nor $\beta$ contain a $\Rightarrow$ or + $\Leftrightarrow$ symbol. + Then it must be that $\alpha^* \vDash\Dashv (\neg\alpha)$ and + $\beta^* \vDash\Dashv (\neg\beta)$. + Consider first $\mathcal{E}_{\neg}(\alpha) = (\neg\alpha)$. + By definition, + $$\mathcal{E}_{\neg}(\alpha)^* + = (\neg\alpha^*) + \vDash\Dashv (\neg(\neg\alpha)) + = (\neg(\mathcal{E}_{\neg}(\alpha))).$$ + Therefore $\mathcal{E}_{\neg}(\alpha) \in S$. + + Likewise, consider $\mathcal{E}_{\square}(\alpha, \beta)$ where $\square$ + is one of the binary connectives $\land$, $\lor$, $\Rightarrow$, + $\Leftrightarrow$. + It trivially follows that $\mathcal{E}_{\Rightarrow}(\alpha, \beta) \in S$ + and $\mathcal{E}_{\Leftrightarrow}(\alpha, \beta) \in S$. + We cover the remaining two cases in turn: + + \subparagraph{Case 1}% + + Suppose $\square = \land$. + Then + \begin{align*} + \mathcal{E}_{\land}(\alpha, \beta)^* + & = (\alpha \land \beta)^* \\ + & = (\alpha^* \lor \beta^*) \\ + & \vDash\Dashv ((\neg\alpha) \lor (\neg\beta)) \\ + & \vDash\Dashv \neg(\alpha \land \beta) \\ + & = (\neg(\mathcal{E}_{\land}(\alpha, \beta))), + \end{align*} + where the last tautological equivalence follows from De Morgan's laws. + + \subparagraph{Case 2}% + + Suppose $\square = \lor$. + Then + \begin{align*} + \mathcal{E}_{\lor}(\alpha, \beta)^* + & = (\alpha \lor \beta)^* \\ + & = (\alpha^* \land \beta^*) \\ + & \vDash\Dashv ((\neg\alpha) \land (\neg\beta)) \\ + & \vDash\Dashv \neg(\alpha \lor \beta) \\ + & = (\neg(\mathcal{E}_{\lor}(\alpha, \beta))), + \end{align*} + where the last tautological equivalence follows from De Morgan's laws. + + \subparagraph{Subconclusion}% + + The foregoing analysis shows that $S$ is indeed closed under the five + formula-building operations. + + \paragraph{(iii)}% + + By \nameref{par:exercise-1.2.9-i} and \nameref{par:exercise-1.2.9-ii}, + the \nameref{sub:induction-principle-1} implies $S$ is the set of all + wffs. + Thus for any well-formed formula $\alpha$ whose only connective symbols + are $\land$, $\lor$, and $\neg$, $\alpha^* \vDash\Dashv (\neg\alpha)$. + \end{proof} \subsection{\sorry{Exercise 1.2.10}}%