Enderton (logic). 1.2.6-1.2.9.
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@ -272,13 +272,13 @@
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By definition of a construction sequence, one of the following holds:
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\begin{align}
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& \phi \text{ is a sentence symbol}
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& \label{sub:induction-principle-1-eq1} \\
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& \hyperlabel{sub:induction-principle-1-eq1} \\
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& \phi = \mathcal{E}_\neg(\epsilon_j)
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\text{ for some } j < m + 1
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& \label{sub:induction-principle-1-eq2} \\
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& \hyperlabel{sub:induction-principle-1-eq2} \\
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& \phi = \mathcal{E}_\square(\epsilon_j, \epsilon_k)
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\text{ for some } j < m + 1, k < m + 1
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& \label{sub:induction-principle-1-eq3}
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& \hyperlabel{sub:induction-principle-1-eq3}
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\end{align}
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where $\square$ is one of the binary connectives $\land$, $\lor$,
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$\Rightarrow$, $\Leftrightarrow$.
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@ -1300,7 +1300,7 @@
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\end{proof}
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\subsection{\sorry{Exercise 1.2.6b}}%
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\subsection{\unverified{Exercise 1.2.6b}}%
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\hyperlabel{sub:exercise-1.2.6b}
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Let $\mathcal{S}$ be a set of sentence symbols that includes those in $\Sigma$
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@ -1315,10 +1315,32 @@
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are a great many -- uncountably many -- of them.)
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\begin{proof}
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TODO
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Let $\mathcal{S}$ be a set of sentence symbols that includes those in
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$\Sigma$ and $\tau$ (and possibly more).
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Let $\mathcal{S}' \subseteq \mathcal{S}$ be the set containing precisely
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the sentence symbols found in $\Sigma$ and $\tau$.
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Let $v$ be a truth assignment for $S'$ that satisfies every member of
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$\Sigma$ and $w$ be an arbitrary extension of $v$ for $S$.
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By construction, $v$ and $w$ agree on all the sentence symbols found in both
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$\Sigma$ and $\tau$.
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\nameref{sub:exercise-1.2.6a} then implies $\bar{v}(\tau) = \bar{w}(\tau)$
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and $\bar{v}(\sigma) = \bar{w}(\sigma)$ for all $\sigma \in \Sigma$.
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Thus $v$ satisfies every member of $\Sigma$ if and only if $w$ satisfies
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every member of $\Sigma$.
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Likewise, $v$ satisfies $\tau$ if and only if $w$ satisfies $\tau$.
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Hence, by definition of \nameref{ref:tautological-implication},
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$\Sigma \vDash \tau$ if and only if every truth assignment for the
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sentence symbols in $S'$ that satisfies every member of $\Sigma$ also
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satisfies $\tau$ if and only if every truth assignment for the sentence
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symbols in $S$ that satisfies every member of $\Sigma$ also satisfies
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$\tau$.
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\end{proof}
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\subsection{\sorry{Exercise 1.2.7}}%
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\subsection{\unverified{Exercise 1.2.7}}%
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\hyperlabel{sub:exercise-1.2.7}
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You are in a land inhabited by people who either always tell the truth or
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@ -1331,29 +1353,103 @@
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\textit{Suggestion}: Make a table.
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\begin{proof}
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TODO
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Consider the self-referential question,
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"Would you respond 'yes' to the question,
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'Should I take the left road to get to the capital?'"
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Let $I$ denote whether the inhabitant is truthful, and $L$ denote whether
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the left road actually goes to the capital.
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We have $R$ denote the answer given by the inhabitant in the following
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"truth table":
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$$\begin{array}{s|s|e}
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I & L & R \\
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\hline
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T & T & \text{Yes} \\
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T & F & \text{No} \\
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F & T & \text{Yes} \\
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F & F & \text{No}
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\end{array}$$
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Regardless of the inhabitant's honesty, we receive the answer "Yes" if and
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only if the left road actually goes to the capital.
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\end{proof}
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\subsection{\sorry{Exercise 1.2.8}}%
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\subsection{\unverified{Exercise 1.2.8}}%
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\hyperlabel{sub:exercise-1.2.8}
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(Substitution) Consider a suquence $\alpha_1, \alpha_2, \ldots$ of wffs.
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(Substitution) Consider a sequence $\alpha_1, \alpha_2, \ldots$ of wffs.
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For each wff $\phi$ let $\phi^*$ be the result of replacing the sentence
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symbol $A_n$ by $\alpha_n$ for each $n$.
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\subsubsection{\sorry{Exercise 1.2.8a}}%
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\subsubsection{\unverified{Exercise 1.2.8a}}%
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\hyperlabel{ssub:exercise-1.2.8a}
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Let $v$ be a truth assignment for the set of all sentence symbols; define $u$
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to be the truth assignment for which $u(A_n) = \bar{v}(\alpha_n)$.
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Show that $\bar{u}(\phi) = \bar{v}(\phi^*)$.
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Use the induction principle.
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Use the \nameref{sub:induction-principle-1}.
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\begin{proof}
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TODO
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Let $$S = \{\phi \mid
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\phi \text{ is a wff such that } \bar{u}(\phi) = \bar{v}(\phi^*)\}.$$
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We prove that (i) $S$ contains the set of all sentence symbols and (ii) $S$
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is closed under the five \nameref{ref:formula-building-operations}.
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Afterward we prove that (iii) our theorem statement holds.
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\paragraph{(i)}%
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\hyperlabel{par:exercise-1.2.8-i}
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Let $\phi = A_n$ be an arbitrary sentence symbol.
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By definition, $u(A_n) = \bar{v}(\alpha_n)$.
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Then $$\bar{u}(\phi)
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= \bar{u}(A_n) \\
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= u(A_n) \\
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= \bar{v}(\alpha_n) \\
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= \bar{v}(\phi^*).$$
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Hence every sentence symbol is in $S$.
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\paragraph{(ii)}%
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\hyperlabel{par:exercise-1.2.8-ii}
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Let $\beta, \gamma \in S$.
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That is, $\bar{u}(\beta) = \bar{v}(\beta^*)$ and
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$\bar{u}(\gamma) = \bar{v}(\gamma^*)$.
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By definition, $\mathcal{E}_{\neg}(\beta) = (\neg\beta)$.
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Therefore
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\begin{align*}
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\bar{u}(\mathcal{E}_{\neg}(\beta))
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& = (\neg\bar{u}(\beta)) \\
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& = (\neg\bar{v}(\beta^*)) \\
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& = \bar{v}((\neg\beta^*)) \\
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& = \bar{v}((\neg\beta)^*) \\
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& = \bar{v}(\mathcal{E}_{\neg}(\beta)^*).
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\end{align*}
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Likewise,
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$\mathcal{E}_{\square}(\beta, \gamma) =
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(\beta \mathop{\square} \gamma)$
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where $\square$ is one of the binary connectives $\land$, $\lor$,
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$\Rightarrow$, $\Leftrightarrow$.
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Therefore
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\begin{align*}
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\bar{u}(\mathcal{E}_{\square}(\beta, \gamma))
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& = \bar{u}(\beta) \mathop{\square} \bar{u}(\gamma) \\
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& = \bar{v}(\beta^*) \mathop{\square} \bar{v}(\gamma^*) \\
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& = \bar{v}((\beta^* \mathop{\square} \gamma^*)) \\
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& = \bar{v}((\beta \mathop{\square} \gamma)^*) \\
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& = \bar{v}(\mathcal{E}_{\square}(\beta, \gamma)^*).
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\end{align*}
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Hence $S$ is closed under the five formula-building operations.
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\paragraph{(iii)}%
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By \nameref{par:exercise-1.2.8-i} and \nameref{par:exercise-1.2.8-ii},
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the \nameref{sub:induction-principle-1} implies $S$ is the set of all
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wffs.
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Thus for any well-formed formula $\phi$,
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$\bar{u}(\phi) = \bar{v}(\phi^*)$.
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\end{proof}
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\subsubsection{\sorry{Exercise 1.2.8b}}%
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\subsubsection{\unverified{Exercise 1.2.8b}}%
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\hyperlabel{ssub:exercise-1.2.8b}
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Show that if $\phi$ is a tautology, then so is $\phi^*$.
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@ -1364,10 +1460,20 @@
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tautology, for any wffs $\alpha$ and $\beta$.)
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\begin{proof}
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TODO
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Suppose $\phi$ is a tautology and let $S$ be the set of all sentence
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symbols.
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Let $v$ be a truth assignment for $S$ and define $u$ to be the truth
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assignment for which $u(A_n) = \bar{v}(\alpha_n)$.
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By \nameref{ssub:exercise-1.2.8a}, $\bar{u}(\phi) = \bar{v}(\phi^*)$.
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Since $\phi$ is a tautology, $\bar{u}(\phi)$ is true meaning
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$\bar{v}(\phi^*)$ is also true.
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Since $v$ is an arbitrary truth assignment, it follows that every truth
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assignment for $S$ satisfies $\phi^*$.
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By \nameref{sub:exercise-1.2.6b}, $\vDash \phi^*$, i.e. $\phi^*$ is a
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tautology.
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\end{proof}
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\subsection{\sorry{Exercise 1.2.9}}%
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\subsection{\unverified{Exercise 1.2.9}}%
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\hyperlabel{sub:exercise-1.2.9}
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(Duality) Let $\alpha$ be a wff whose only connective symbols are $\land$,
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@ -1377,8 +1483,107 @@
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Show that $\alpha^*$ is tautologically equivalent to $(\neg\alpha)$.
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Use the \nameref{sub:induction-principle-1}.
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\code*{Common/Logic/Basic}
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{not\_and\_de\_morgan}
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\code{Common/Logic/Basic}
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{not\_or\_de\_morgan}
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\begin{proof}
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TODO
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Let
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\begin{align*}
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S = \{ \alpha \mid
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& \alpha \text{ is a wff containing a } \Rightarrow \text{ or } \\
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& \alpha \text{ is a wff containing a } \Leftrightarrow \text{ or } \\
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& \alpha^* \vDash \Dashv (\neg\alpha) \}.
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\end{align*}
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We prove that (i) $S$ contains the set of all sentence symbols and (ii) $S$
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is closed under the five \nameref{ref:formula-building-operations}.
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Afterward we prove that (iii) our theorem statement holds.
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\paragraph{(i)}%
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\hyperlabel{par:exercise-1.2.9-i}
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Let $\alpha = A_n$ be an arbitrary sentence symbol.
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By definition, $$\alpha^* = A_n^* = (\neg A_n) = (\neg\alpha).$$
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Hence every sentence symbol is in $S$.
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\paragraph{(ii)}%
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\hyperlabel{par:exercise-1.2.9-ii}
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Let $\alpha, \beta \in S$.
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Suppose $\alpha$ contains a $\Rightarrow$ or $\Leftrightarrow$ symbol.
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Then $\mathcal{E}_{\neg}(\alpha)$ also does.
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The same holds for $\beta$.
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Furthermore, if either $\alpha$ or $\beta$ contains a $\Rightarrow$ or
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$\Leftrightarrow$ symbol, then so does
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$\mathcal{E}_{\square}(\alpha, \beta)$
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where $\square$ is one of the binary connectives $\land$, $\lor$,
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$\Rightarrow$, $\Leftrightarrow$.
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In any of these above cases, it is trivial to see each of the five-formula
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building operations take a wff from $S$ and produce another wff in $S$.
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Now, suppose neither $\alpha$ nor $\beta$ contain a $\Rightarrow$ or
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$\Leftrightarrow$ symbol.
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Then it must be that $\alpha^* \vDash\Dashv (\neg\alpha)$ and
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$\beta^* \vDash\Dashv (\neg\beta)$.
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Consider first $\mathcal{E}_{\neg}(\alpha) = (\neg\alpha)$.
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By definition,
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$$\mathcal{E}_{\neg}(\alpha)^*
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= (\neg\alpha^*)
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\vDash\Dashv (\neg(\neg\alpha))
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= (\neg(\mathcal{E}_{\neg}(\alpha))).$$
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Therefore $\mathcal{E}_{\neg}(\alpha) \in S$.
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Likewise, consider $\mathcal{E}_{\square}(\alpha, \beta)$ where $\square$
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is one of the binary connectives $\land$, $\lor$, $\Rightarrow$,
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$\Leftrightarrow$.
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It trivially follows that $\mathcal{E}_{\Rightarrow}(\alpha, \beta) \in S$
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and $\mathcal{E}_{\Leftrightarrow}(\alpha, \beta) \in S$.
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We cover the remaining two cases in turn:
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\subparagraph{Case 1}%
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Suppose $\square = \land$.
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Then
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\begin{align*}
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\mathcal{E}_{\land}(\alpha, \beta)^*
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& = (\alpha \land \beta)^* \\
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& = (\alpha^* \lor \beta^*) \\
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& \vDash\Dashv ((\neg\alpha) \lor (\neg\beta)) \\
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& \vDash\Dashv \neg(\alpha \land \beta) \\
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& = (\neg(\mathcal{E}_{\land}(\alpha, \beta))),
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\end{align*}
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where the last tautological equivalence follows from De Morgan's laws.
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\subparagraph{Case 2}%
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Suppose $\square = \lor$.
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Then
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\begin{align*}
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\mathcal{E}_{\lor}(\alpha, \beta)^*
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& = (\alpha \lor \beta)^* \\
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& = (\alpha^* \land \beta^*) \\
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& \vDash\Dashv ((\neg\alpha) \land (\neg\beta)) \\
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& \vDash\Dashv \neg(\alpha \lor \beta) \\
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& = (\neg(\mathcal{E}_{\lor}(\alpha, \beta))),
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\end{align*}
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where the last tautological equivalence follows from De Morgan's laws.
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\subparagraph{Subconclusion}%
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The foregoing analysis shows that $S$ is indeed closed under the five
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formula-building operations.
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\paragraph{(iii)}%
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By \nameref{par:exercise-1.2.9-i} and \nameref{par:exercise-1.2.9-ii},
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the \nameref{sub:induction-principle-1} implies $S$ is the set of all
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wffs.
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Thus for any well-formed formula $\alpha$ whose only connective symbols
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are $\land$, $\lor$, and $\neg$, $\alpha^* \vDash\Dashv (\neg\alpha)$.
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\end{proof}
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\subsection{\sorry{Exercise 1.2.10}}%
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