Enderton (logic). Finish all but exercise 1.2.14.
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@ -118,6 +118,9 @@
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a function $$v \colon \mathcal{S} \rightarrow \{F, T\}$$ assigning either
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$T$ or $F$ to each symbol in $\mathcal{S}$.
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\suitdivider
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\noindent
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Let $\bar{\mathcal{S}}$ be the set of \nameref{ref:well-formed-formula}s that
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can be built up from $\mathcal{S}$ by the five
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\nameref{ref:formula-building-operations}.
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@ -246,6 +249,7 @@
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{Enderton.Logic.Chapter\_1.Wff.rec}
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\begin{proof}
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We note every 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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@ -1586,7 +1590,7 @@
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\end{proof}
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\subsection{\sorry{Exercise 1.2.10}}%
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\subsection{\unverified{Exercise 1.2.10}}%
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\hyperlabel{sub:exercise-1.2.10}
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Say that a set $\Sigma_1$ of wffs is \textit{equivalent} to a set $\Sigma_2$
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@ -1596,36 +1600,126 @@
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tautologically implied by the remaining members in $\Sigma$.
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Show that the following hold.
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\subsubsection{\sorry{Exercise 1.2.10a}}%
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\subsubsection{\unverified{Exercise 1.2.10a}}%
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\hyperlabel{ssub:exercise-1.2.10a}
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A finite set of wffs has an independent equivalent subset.
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\begin{proof}
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TODO
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For natural number $n$, let $P(n)$ be the statement:
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\begin{induction}
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\hyperlabel{sub:exercise-1.2.10a-ih}
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A set of wffs \nameref{S:ref:equinumerous} to $n$ has an independent
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equivalent subset.
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\end{induction}
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\noindent
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We proceed by induction on $n$.
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\paragraph{Base Case}%
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Consider a finite set of wffs equinumerous to $0$.
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This is simply the empty set.
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It is vacuously true that $\emptyset$ is independent.
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Thus $\emptyset \subseteq \emptyset$ is an independent equivalent subset
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meaning $P(0)$ is true.
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\paragraph{Inductive Step}%
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Suppose $P(n)$ holds true for some $n \geq 0$.
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That is, every finite set of wffs equinumerous to $n$ has an independent
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equivalent subset.
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Consider now set $\Sigma$ of wffs equinumerous to $n + 1$.
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There are two possibilities to consider:
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\subparagraph{Case 1}%
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Suppose $\Sigma$ is independent.
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Then $\Sigma \subseteq \Sigma$ is an independent equivalent subset.
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\subparagraph{Case 2}%
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Suppose $\Sigma$ is not independent.
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Then there exists some $\sigma \in \Sigma$ such that $\sigma$ is
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tautologically implied by the remaining members of $\Sigma$.
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Let $\Sigma_1 = \Sigma - \{\sigma\}$.
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By \ihref{sub:exercise-1.2.10a-ih}, $\Sigma_1$ has an independent
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equivalent subset $\Sigma_2$.
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Now let $\phi$ be an arbitrary wff.
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Then
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\begin{align*}
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\Sigma_2 \vDash \phi
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& \Rightarrow \Sigma_1 \vDash \phi
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& \text{def'n of equivalent} \\
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& \Rightarrow \Sigma_1; \sigma \vDash \phi
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& \sigma \text{ is redundant} \\
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& \Rightarrow \Sigma \vDash \phi.
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\end{align*}
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Likewise,
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\begin{align*}
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\Sigma \vDash \phi
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& \Rightarrow \Sigma_1; \sigma \vDash \phi \\
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& \Rightarrow \Sigma_1 \vDash \phi
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& \sigma \text{ is redundant} \\
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& \Rightarrow \Sigma_2 \vDash \phi.
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& \text{def'n of equivalent}
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\end{align*}
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Thus $\Sigma_2$ is an independent equivalent subset of $\Sigma$.
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\subparagraph{Subconclusion}%
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The above two cases are exhaustive.
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Hence $P(n + 1)$ holds true.
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\paragraph{Conclusion}%
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By induction, it follows $P(n)$ holds true for all $n \geq 0$.
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That is, every set of wffs equinumerous to a natural number has an
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independent equivalent subset.
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In other words, every finite set of wffs has an independent equivalent
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subset.
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\end{proof}
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\subsubsection{\sorry{Exercise 1.2.10b}}%
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\subsubsection{\unverified{Exercise 1.2.10b}}%
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\hyperlabel{ssub:exercise-1.2.10b}
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An infinite set need not have an independent equivalent subset.
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\begin{proof}
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TODO
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Let $$S = \{A_1 \land \cdots \land A_n \mid n \in \omega\}$$ be an infinite
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set of wffs.
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For the sake of contradiction, suppose $S$ has an independent equivalent
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subset $S'$.
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There are two cases to consider:
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\paragraph{Case 1}%
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Suppose $S' = \emptyset$.
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Then it trivally follows $S'$ is not equivalent to $S$, a contradiction.
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\paragraph{Case 1}%
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Suppose $S' \neq \emptyset$.
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By the \nameref{S:sub:well-ordering-natural-numbers}, there exists a least
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$n \in \mathbb{N}$ such that $\phi = A_1 \land \cdots \land A_n$ is in
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$S'$.
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It cannot be that another element of $S'$ exists since such an element
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would tautologically imply $\phi$, contradicting independence.
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Thus $S' = \{\phi\}$.
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But $\{\phi\}$ cannot be equivalent to $S$ since it has no information
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about sentence symbol e.g. $A_{n+1}$, another contradiction.
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\paragraph{Conclusion}%
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The above two cases are exhaustive and both yield contradictions.
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It must be that $S$ does not have an independent equivalent subset.
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\end{proof}
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\subsubsection{\sorry{Exercise 1.2.10c}}%
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\hyperlabel{ssub:exercise-1.2.10c}
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Let $\Sigma = \{\sigma_0, \sigma_1, \ldots\}$; show that there is an
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independent equivalent set $\Sigma'$.
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(By part (b), we cannot hope to have $\Sigma' \subseteq \Sigma$ in general.)
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 1.2.11}}%
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\subsection{\unverified{Exercise 1.2.11}}%
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\hyperlabel{sub:exercise-1.2.11}
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Show that a truth assignment $v$ satisfies the wff
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@ -1636,10 +1730,88 @@
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parentheses is not crucial.)
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\begin{proof}
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TODO
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Define $\sigma_n$ recursively as follows:
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$\sigma_0 = (A_1 \Leftrightarrow A_2)$ and
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$\sigma_{n+1} = (\sigma_n \Leftrightarrow A_{n+3})$.
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For natural number $n$, let $P(n)$ be the statement:
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\begin{induction}
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\hyperlabel{sub:exercise-1.2.11-ih}
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Truth assignment $v$ satisfies $\sigma_n$ if and only if $v(A_i) = F$
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for an even number of $i$'s, $1 \leq i \leq n + 2$.
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\end{induction}
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\noindent
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We proceed by induction on $n$.
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\paragraph{Base Case}%
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Let $n = 0$.
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Then $\sigma_n = \sigma_0 = (A_1 \Leftrightarrow A_2)$.
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We proceed by truth table:
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$$\begin{array}{s|e|s}
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(A_1 & \Leftrightarrow & A_2) \\
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\hline
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T & T & T \\
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T & F & F \\
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F & F & T \\
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F & T & F
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\end{array}$$
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Here we see $A_1 \Leftrightarrow A_2$ is true if and only if both $A_1$
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and $A_2$ are true or neither $A_1$ nor $A_2$ are true.
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Thus $P(0)$ holds true.
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\paragraph{Inductive Step}%
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Suppose $P(n)$ holds true for some $n \geq 0$.
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Consider now $$\sigma_{n+1} = (\sigma_k \Leftrightarrow A_{n+3}).$$
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Let $v$ be a truth assignment for $A_1, \ldots, A_{n+3}$.
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There are two cases to consider:
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\subparagraph{Case 1}%
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Suppose $v(A_i) = F$ for an even number of $i$'s, $1 \leq i \leq n + 2$.
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By \ihref{sub:exercise-1.2.11-ih}, $v$ satisfies $\sigma_n$.
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We now have the following truth table:
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$$\begin{array}{s|e|s}
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(\sigma_n & \Leftrightarrow & A_{n+3}) \\
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\hline
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T & T & T \\
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T & F & F \\
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\end{array}$$
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In this case, it follows $v$ satisfies $\sigma_{n+1}$ if and only if
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$v(A_i) = F$ for an even number of $i$'s, $1 \leq i \leq n + 3$.
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\subparagraph{Case 2}%
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Suppose $v(A_i) = F$ for an odd number of $i$'s, $1 \leq i \leq n + 2$.
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By \ihref{sub:exercise-1.2.11-ih}, $v$ does not satisfy $\sigma_n$.
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We now have the following truth table:
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$$\begin{array}{s|e|s}
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(\sigma_n & \Leftrightarrow & A_{n+3}) \\
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\hline
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F & F & T \\
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F & T & F \\
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\end{array}$$
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In this case, it follows $v$ satisfies $\sigma_{n+1}$ if and only if
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$v(A_i) = F$ for an even number of $i$'s, $1 \leq i \leq n + 3$.
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\subparagraph{Subconclusion}%
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The above two cases are exhaustive.
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Hence $P(n + 1)$ holds true.
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\paragraph{Conclusion}%
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By induction, it follows $P(n)$ holds true for all $n \geq 0$.
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That is, truth assignment $v$ satisfies
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$$(\cdots (A_1 \Leftrightarrow A_2)
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\Leftrightarrow \cdots \Leftrightarrow A_n)$$
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if and only if $v(A_i) = F$ for an even number of $i$'s,
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$1 \leq i \leq n$.
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\end{proof}
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\subsection{\sorry{Exercise 1.2.12}}%
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\subsection{\unverified{Exercise 1.2.12}}%
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\hyperlabel{sub:exercise-1.2.12}
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There are three suspects for a murder: Adams, Brown, and Clark.
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@ -1647,17 +1819,20 @@
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But Clark hated him."
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Brown states "I didn't do it. I didn't even know the guy. Besides I was out of
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town all that week."
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Clark says "I didn't do it. I saw both ADams and Brown downtown with the
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Clark says "I didn't do it. I saw both Adams and Brown downtown with the
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victim that day; one of them must have done it."
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Assume that the two innocent men are telling the truth, but that the guilty
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man might not be.
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Who did it?
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\begin{proof}
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TODO
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It must be that Brown is the guilty one.
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Adam claims the victim was an old acquaintance of Brown's.
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Clark claims Brown was downtown with the victim that day.
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Brown's testimony conflicts with both of these statements.
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\end{proof}
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\subsection{\sorry{Exercise 1.2.13}}%
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\subsection{\unverified{Exercise 1.2.13}}%
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\hyperlabel{sub:exercise-1.2.13}
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An advertisement for a tennis magazine states, "If I'm not playing tennis,
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truth assignments.)
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\begin{proof}
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TODO
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Let $P$ denote playing tennis, $W$ denote watching tennis, and $R$ denote
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reading about tennis.
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These statements can be translated as:
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\begin{enumerate}[(a)]
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\item $\neg P \Rightarrow W$.
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\item $\neg W \Rightarrow R$.
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\end{enumerate}
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Thus either the speaker is playing tennis, or, if not, he is watching
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tennis.
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Since we assume the speaker cannot do more than one of these activities at
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a time, reading is never a possibility.
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\end{proof}
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\subsection{\sorry{Exercise 1.2.14}}%
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@ -1684,10 +1869,11 @@
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\nameref{sub:induction-principle-1} to show that $\bar{v}_1 = \bar{v}_2$.
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\begin{proof}
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The conditions 0-5 can be found at \nameref{ref:truth-assignment}.
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 1.2.15}}%
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\subsection{\verified{Exercise 1.2.15}}%
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\hyperlabel{sub:exercise-1.2.15}
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Of the following three formulas, which tautologically implies which?
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\item $(((\neg A) \lor B) \land (A \lor (\neg B)))$
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\end{enumerate}
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\code{Bookshelf/Enderton/Logic/Chapter\_1}
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{Enderton.Logic.Chapter\_1.exercise\_1\_2\_15\_i}
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\code{Bookshelf/Enderton/Logic/Chapter\_1}
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{Enderton.Logic.Chapter\_1.exercise\_1\_2\_15\_ii}
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\begin{proof}
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TODO
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All three are tautologically equivalent.
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We prove that (i) (a) is tautologically equivalent to (b) and (ii) (a) is
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tautologically equivalent to (c). It then immediately follows that (b) is
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tautologically equivalent to (c).
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\paragraph{(i)}%
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By \nameref{sub:exercise-1.2.4}, $(a) \vDash\Dashv (b)$ if and only if
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$\vDash ((a) \Leftrightarrow (b))$.
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We now construct the corresponding truth table:
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$$\begin{array}{s|c|s|e|c|s|c|s|c|c|s|c|s}
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(A & \Leftrightarrow & B) & \Leftrightarrow &
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(\neg & ((A & \Rightarrow & B) &
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\Rightarrow & (\neg & (B & \Rightarrow & A)))) \\
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\hline
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T & T & T & T & T & T & T & T & F & F & T & T & T \\
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T & F & F & T & F & T & F & F & T & F & F & T & T \\
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F & F & T & T & F & F & T & T & T & T & T & F & F \\
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F & T & F & T & T & F & T & F & F & F & F & T & F
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\end{array}$$
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Therefore (a) and (b) are tautologically equivalent.
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\paragraph{(ii)}%
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By \nameref{sub:exercise-1.2.4}, $(a) \vDash\Dashv (c)$ if and only if
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$\vDash ((a) \Leftrightarrow (c))$.
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We now construct the corresponding truth table:
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$$\begin{array}{s|c|s|e|c|s|c|s|c|s|c|c|s}
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(A & \Leftrightarrow & B) & \Leftrightarrow &
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(((\neg & A) & \lor & B) & \land & (A & \lor & (\neg & B))) \\
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\hline
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T & T & T & T & F & T & T & T & T & T & T & F & T \\
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T & F & F & T & F & T & F & F & F & T & T & T & F \\
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F & F & T & T & T & F & T & T & F & F & F & F & T \\
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F & T & F & T & T & F & T & F & T & F & T & T & F
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\end{array}$$
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\end{proof}
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\end{document}
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@ -519,4 +519,20 @@ theorem exercise_1_2_6b
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: (False ∨ True) ∧ ¬ False := by
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simp
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/-! #### Exercise 1.2.15
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Of the following three formulas, which tautologically implies which?
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(a) `(A ↔ B)`
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(b) `(¬((A → B) →(¬(B → A))))`
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(c) `(((¬ A) ∨ B) ∧ (A ∨ (¬ B)))`
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-/
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theorem exercise_1_2_15_i (A B : Prop)
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: (A ↔ B) ↔ (¬((A → B) → (¬(B → A)))) := by
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tauto
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theorem exercise_1_2_15_ii (A B : Prop)
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: (A ↔ B) ↔ (((¬ A) ∨ B) ∧ (A ∨ (¬ B))) := by
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tauto
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end Enderton.Logic.Chapter_1
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