Enderton (logic). Continue working through section 1.1.
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@ -63,7 +63,7 @@
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\end{align*}
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\code*{Bookshelf/Enderton/Logic/Chapter\_1}
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{Enderton.Logic.Chapter\_1.WFF}
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{Enderton.Logic.Chapter\_1.Wff}
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\lean{Init/Prelude}
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{Not}
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@ -96,7 +96,7 @@
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times the \nameref{ref:formula-building-operations}.
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\code*{Bookshelf/Enderton/Logic/Chapter\_1}
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{Enderton.Logic.Chapter\_1.WFF}
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{Enderton.Logic.Chapter\_1.Wff}
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\endgroup
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@ -169,6 +169,9 @@
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wffs.
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\end{theorem}
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\code{Bookshelf/Enderton/Logic/Chapter\_1}
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{Enderton.Logic.Chapter\_1.Wff.rec}
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\begin{proof}
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We note every \nameref{ref:well-formed-formula} can be characterized by a
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\nameref{ref:construction-sequence}.
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@ -244,6 +247,109 @@
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\end{proof}
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\subsection{\unverified{Balanced Parentheses}}%
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\hyperlabel{sub:balanced-parentheses}
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\begin{lemma}
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All well-formed formulas have an equal number of left and right parentheses.
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\end{lemma}
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\begin{proof}
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Define $$S = \{ \phi \mid
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\phi \text{ is a wff with a balanced number of parentheses} \}.$$
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We prove that (i) all the sentence symbols are members of $S$ and (ii)
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$S$ is closed under the five \nameref{ref:formula-building-operations}.
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We then conclude with (iii) the proof of the theorem statement.
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\paragraph{(i)}%
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\hyperlabel{par:balanced-parentheses-i}
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By definition, well-formed formulas comprising a single sentence symbol
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do not have any parentheses.
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Thus all sentence symbols are members of $S$.
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\paragraph{(ii)}%
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\hyperlabel{par:balanced-parentheses-ii}
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Let $\alpha, \beta \in S$.
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By definition, $\mathcal{E}_{\neg}(\alpha) = (\neg\alpha)$.
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Thus one additional left and right parenthesis is introduced.
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Since $\alpha$ is assumed to have an equal number of left and right
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parentheses, $\mathcal{E}_{\neg}(\alpha) \in S$.
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Likewise,
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$\mathcal{E}_{\square}(\alpha, \beta) = (\alpha \mathop{\square} \beta)$
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where $\square$ is one of the binary connectives $\land$, $\lor$,
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$\Rightarrow$, $\Leftrightarrow$.
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Again, an additional left and right parenthesis is introduced.
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Since $\alpha$ and $\beta$ are assumed to have a balanced number of
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parentheses, $\mathcal{E}_{\square}(\alpha, \beta) \in S$.
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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:balanced-parentheses-i} and
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\nameref{par:balanced-parentheses-ii}, the
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\nameref{sub:induction-principle-1} implies $S$ is the set of all wffs.
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Thus all well-formed formulas have an equal number of left and right
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parentheses.
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\end{proof}
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\subsection{\verified{Parentheses Count}}%
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\hyperlabel{sub:parentheses-count}
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\begin{lemma}
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Let $\phi$ be a well-formed formula and $c$ be the number of places at which
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a sentential connective symbol exists.
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Then there is $2c$ parentheses in $\phi$.
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\end{lemma}
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\code{Enderton.Logic.Chapter\_1}
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{Enderton.Logic.Chapter\_1.paren\_count\_double\_sentential\_count}
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\begin{proof}
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Define $$S = \{ \phi \mid
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\phi \text{ is a wff with } 2c \text{ parentheses} \}.$$
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We prove that (i) all the sentence symbols are members of $S$ and (ii)
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$S$ is closed under the five \nameref{ref:formula-building-operations}.
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We then conclude with (iii) the proof of the theorem statement.
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\paragraph{(i)}%
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\hyperlabel{par:parentheses-count-i}
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A sentence symbol, by itself, has no sentential connectives.
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Likewise, it has 0 parentheses.
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Thus $S$ contains every sentence symbol.
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\paragraph{(ii)}%
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\hyperlabel{par:parentheses-count-ii}
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Let $\alpha, \beta \in S$.
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By definition, $\mathcal{E}_{\neg}(\alpha) = (\neg \alpha)$.
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Then $\mathcal{E}_{\neg}(\alpha)$ introduces two additional parentheses
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and one additional sentential connective symbol.
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Thus $\mathcal{E}_{\neg}(\alpha) \in S$.
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Likewise,
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$\mathcal{E}_{\square}(\alpha, \beta) = (\alpha \mathop{\square} \beta)$
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where $\square$ is one of the binary connectives $\land$, $\lor$,
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$\Rightarrow$, $\Leftrightarrow$.
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$\mathcal{E}_{\square}(\alpha, \beta)$ also introduces two additional
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parentheses and one additional connective symbol.
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Thus $\mathcal{E}_{\square}(\alpha, \beta) \in S$.
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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:parentheses-count-i} and
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\nameref{par:parentheses-count-ii}, the
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\nameref{sub:induction-principle-1} implies $S$ is the set of all wffs.
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Thus every wff has $2c$ parentheses in $\phi$, where $c$ denotes the
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number of places at which a sentential connective symbol exists.
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\end{proof}
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\section{Exercises 1}%
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\hyperlabel{sec:exercises-1}
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@ -258,7 +364,7 @@
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\begin{answer}
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We begin first with the English sentences:
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\begin{enumerate}[i]
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\begin{enumerate}[(i)]
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\item He can juggle beach balls, bowling pins, and hackysacks unless
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he is tired, in which case he can only juggle beach balls.
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\item
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@ -327,21 +433,21 @@
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\hyperlabel{par:exercise-1.1.2-ii}
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Define $L$ to be the length function mapping arbitrary wff to its length.
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Let $\phi, \psi \in S$.
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Then $L(\phi)$ and $L(\psi)$ each evaluate to 1, 4, 5, or a value larger
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than 6.
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Let $\alpha, \beta \in S$.
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Then $L(\alpha)$ and $L(\beta)$ each evaluate to 1, 4, 5, or a value
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larger than 6.
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By definition, $\mathcal{E}_{\neg}(\phi) = (\neg \phi)$.
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Thus $L(\mathcal{E}_{\neg}(\phi)) = L(\phi) + 3$.
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Enumerating through the possible values of $L(\phi)$ shows
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$\mathcal{E}_{\neg}(\phi) \in S$.
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By definition, $\mathcal{E}_{\neg}(\alpha) = (\neg \alpha)$.
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Thus $L(\mathcal{E}_{\neg}(\alpha)) = L(\alpha) + 3$.
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Enumerating through the possible values of $L(\alpha)$ shows
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$\mathcal{E}_{\neg}(\alpha) \in S$.
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Likewise,
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$\mathcal{E}_{\square}(\phi, \psi) = (\phi \mathop{\square} \psi)$
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$\mathcal{E}_{\square}(\alpha, \beta) = (\alpha \mathop{\square} \beta)$
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where $\square$ is one of the binary connectives $\land$, $\lor$,
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$\Rightarrow$, $\Leftrightarrow$.
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Thus $L(\mathcal{E}_{\square}(\phi, \psi)) = L(\phi) + L(\psi) + 3$.
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Again, enumerating through the possible values of $L(\phi)$ and $L(\psi)$
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shows $\mathcal{E}_{\square}(\phi, \psi) \in S$.
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Thus $L(\mathcal{E}_{\square}(\alpha, \beta)) = L(\alpha) + L(\beta) + 3$.
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Again, enumerating through the possible values of $L(\alpha)$ and
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$L(\beta)$ shows $\mathcal{E}_{\square}(\alpha, \beta) \in S$.
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Hence $S$ is closed under the five formula-building operations.
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@ -368,7 +474,7 @@
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\end{proof}
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\subsection{\pending{Exercise 1.1.3}}%
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\subsection{\verified{Exercise 1.1.3}}%
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\hyperlabel{sub:exercise-1.1.3}
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Let $\alpha$ be a wff; let $c$ be the number of places at which binary
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$s = 2$.)
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Show by using the induction principle that $s = c + 1$.
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\code*{Enderton.Logic.Chapter\_1}
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{Enderton.Logic.Chapter\_1.exercise\_1\_1\_3}
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\begin{proof}
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Define
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$\mathcal{E}_{\neg}(\alpha)$ does not change.
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Thus $\mathcal{E}_{\neg}(\alpha) \in S$.
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Likewise,
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$\mathcal{E}_{\square}(\phi, \psi) = (\phi \mathop{\square} \psi)$
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$\mathcal{E}_{\square}(\alpha, \beta) = (\alpha \mathop{\square} \beta)$
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where $\square$ is one of the binary connectives $\land$, $\lor$,
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$\Rightarrow$, $\Leftrightarrow$.
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Therefore $\mathcal{E}_{\square}(\phi, \psi)$ has $s_\alpha + s_\beta$
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Therefore $\mathcal{E}_{\square}(\alpha, \beta)$ has $s_\alpha + s_\beta$
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sentence symbols and $c_\alpha + c_\beta + 1$ binary connective symbols.
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But \eqref{sub:exercise-1.1.3-eq1} implies
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\begin{align*}
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& = (c_\alpha + 1) + (c_\beta + 1) \\
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& = (c_\alpha + c_\beta + 1) + 1,
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\end{align*}
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meaning $\mathcal{E}_{\square}(\phi, \psi) \in S$.
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meaning $\mathcal{E}_{\square}(\alpha, \beta) \in S$.
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Hence $S$ is closed under the five formula-building operations.
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\end{proof}
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\subsection{\sorry{Exercise 1.1.5}}%
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\subsection{\pending{Exercise 1.1.5}}%
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\hyperlabel{sub:exercise-1.1.5}
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Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
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\subsubsection{\sorry{Exercise 1.1.5a}}%
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\hyperlabel{ssub:exercise-1.1.5-a}
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\subsubsection{\verified{Exercise 1.1.5a}}%
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\hyperlabel{ssub:exercise-1.1.5.a}
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Show that the length of $\alpha$ (i.e., the number of symbols in the string)
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is odd.
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\textit{Suggestion}: Apply induction to show that the length is of the form
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$4k + 1$.
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\code*{Enderton.Logic.Chapter\_1}
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{Enderton.Logic.Chapter\_1.exercise\_1\_1\_5\_a}
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\begin{proof}
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TODO
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Define $L$ to be the length function mapping arbitrary
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\nameref{ref:well-formed-formula} to its length and let
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\begin{equation}
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\hyperlabel{ssub:exercise-1.1.5.a-eq1}
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S = \{\phi \mid
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\phi \text{ is a wff containing } \neg \text{ or }
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\exists k \in \mathbb{N}, L(\phi) = 4k + 1\}.
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\end{equation}
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We prove that (i) all the sentence symbols are members of $S$ and (ii)
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$S$ is closed under the five \nameref{ref:formula-building-operations}.
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We then conclude with (iii) the proof of the theorem statement.
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\paragraph{(i)}%
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\hyperlabel{par:exercise-1.1.5.a-i}
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Every sentence symbol has length 1 by definition.
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That is, every sentence symbol has length $(4)(0) + 1$.
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Hence $S$ contains every sentence symbol.
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\paragraph{(ii)}%
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\hyperlabel{par:exercise-1.1.5.a-ii}
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Let $\alpha, \beta \in S$.
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Then there exists some $k_\alpha$ and $k_\beta$ such that
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$L(\alpha) = 4k_\alpha + 1$ and $L(\beta) = 4k_\beta + 1$.
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Clearly $S$ is closed under $\mathcal{E}_{\neg}$.
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Next consider
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$\mathcal{E}_{\square}(\alpha, \beta) = (\alpha \mathop{\square} \beta)$
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where $\square$ is one of the binary connectives $\land$, $\lor$,
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$\Rightarrow$, $\Leftrightarrow$.
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Then
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\begin{align*}
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L(\alpha, \beta)
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& = L(\alpha) + L(\beta) + 3 \\
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& = (4k_\alpha + 1) + (4k_\beta + 1) + 3 \\
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& = 4k_\alpha + 4k_\beta + 4 + 1 \\
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& = 4(k_\alpha + k_\beta + 1) + 1.
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\end{align*}
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Therefore, there exists a $k \in \mathbb{N}$, namely
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$k = k_\alpha + k_\beta + 1$, such that
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$L(\mathcal{E}_{\square}(\alpha, \beta)) = 4k + 1$.
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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-1.1.5.a-i} and \nameref{par:exercise-1.1.5.a-ii},
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the \nameref{sub:induction-principle-1} indicates $S$ is the set of all
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wffs.
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Thus all well-formed formulas not containing symbol $\neg$ has length
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$4k + 1$ for some $k \in \mathbb{N}$.
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Therefore these well-formed formulas have odd length.
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\end{proof}
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\subsubsection{\sorry{Exercise 1.1.6a}}%
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\hyperlabel{ssub:exercise-1.1.6-a}
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\subsubsection{\pending{Exercise 1.1.5b}}%
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\hyperlabel{ssub:exercise-1.1.5-b}
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Show that more than a quarter of the symbols are sentence symbols.
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\textit{Suggestion}: Apply induction to show that the number of sentence
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symbols is $k + 1$.
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symbols is of the form $k + 1$.
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\code*{Enderton.Logic.Chapter\_1}
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{Enderton.Logic.Chapter\_1.exercise\_1\_1\_5\_b}
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 1.1.6}}%
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\hyperlabel{sub:exercise-1.1.6}
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Let $\phi$ be a \nameref{ref:well-formed-formula}.
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By \nameref{sub:exercise-1.1.3}, the number of sentence symbols of $\phi$ is
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$k + 1$, where $k$ is the number of places at which binary connective
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symbols occur in $\phi$.
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By \nameref{sub:parentheses-count}, the number of parentheses in $\phi$ is
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$2k$.
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Thus $\phi$ has length $(k + 1) + k + 2k = 4k + 1$.
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But $$\frac{k + 1}{4k + 1} > \frac{k + 1}{4k + 4} = \frac{1}{4}.$$
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Hence more than a quarter of the symbols of $\phi$ are sentence symbols.
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Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
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\begin{enumerate}[(a)]
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\item Show that the length of $\alpha$ (i.e., the number of symbols in the
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string) is odd.
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\item Show that more than a quarter of the symbols are sentence symbols.
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\end{enumerate}
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\begin{proof}
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TODO
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\end{proof}
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\end{document}
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@ -1,6 +1,9 @@
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import Common.Logic.Basic
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import Common.Nat.Basic
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import Mathlib.Algebra.Parity
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import Mathlib.Data.Nat.Basic
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import Mathlib.Tactic.NormNum
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import Mathlib.Tactic.Ring
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/-! # Enderton.Logic.Chapter_1
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@ -23,7 +26,7 @@ inductive Wff where
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namespace Wff
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/--
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Returns the length of the expression, including symbols.
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Returns the length of the expression, i.e. a count of all symbols..
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-/
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def length : Wff → ℕ
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| Wff.SS _ => 1
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@ -34,7 +37,7 @@ def length : Wff → ℕ
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| Wff.Iff e₁ e₂ => length e₁ + length e₂ + 3
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/--
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Every well-formed formula has a positive length.
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Every `Wff` has a positive length.
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-/
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theorem length_gt_zero (φ : Wff)
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: length φ > 0 := by
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@ -47,16 +50,122 @@ theorem length_gt_zero (φ : Wff)
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| Cond _ _
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| Iff _ _ => simp
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/--
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The number of sentence symbols found in the provided `Wff`.
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-/
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def sentenceSymbolCount : Wff → ℕ
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| Wff.SS _ => 1
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| Wff.Not e => sentenceSymbolCount e
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| Wff.And e₁ e₂
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| Wff.Or e₁ e₂
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| Wff.Cond e₁ e₂
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| Wff.Iff e₁ e₂ => sentenceSymbolCount e₁ + sentenceSymbolCount e₂
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/--
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The number of sentential connective symbols found in the provided `Wff`.
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-/
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def sententialSymbolCount : Wff → ℕ
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| Wff.SS _ => 0
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| Wff.Not e => sententialSymbolCount e + 1
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| Wff.And e₁ e₂
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| Wff.Or e₁ e₂
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| Wff.Cond e₁ e₂
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| Wff.Iff e₁ e₂ => sententialSymbolCount e₁ + sententialSymbolCount e₂ + 1
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/--
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The number of binary connective symbols found in the provided `Wff`.
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-/
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def binarySymbolCount : Wff → ℕ
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| Wff.SS _ => 0
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| Wff.Not e => binarySymbolCount e
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| Wff.And e₁ e₂
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| Wff.Or e₁ e₂
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| Wff.Cond e₁ e₂
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| Wff.Iff e₁ e₂ => binarySymbolCount e₁ + binarySymbolCount e₂ + 1
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/--
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The number of parentheses found in the provided `Wff`.
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-/
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def parenCount : Wff → ℕ
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| Wff.SS _ => 0
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| Wff.Not e => 2 + parenCount e
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| Wff.And e₁ e₂
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| Wff.Or e₁ e₂
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| Wff.Cond e₁ e₂
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| Wff.Iff e₁ e₂ => 2 + parenCount e₁ + parenCount e₂
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/--
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Whether or not the `Wff` contains a `¬`.
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-/
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def hasNotSymbol : Wff → Prop
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| Wff.SS _ => False
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| Wff.Not _ => True
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| Wff.And e₁ e₂
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| Wff.Or e₁ e₂
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| Wff.Cond e₁ e₂
|
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| Wff.Iff e₁ e₂ => hasNotSymbol e₁ ∨ hasNotSymbol e₂
|
||||
|
||||
/--
|
||||
If a `Wff` does not contain the `¬` symbol, it has the same number of sentential
|
||||
connective symbols as it does binary connective symbols. In other words, the
|
||||
negation symbol is the only non-binary sentential connective.
|
||||
-/
|
||||
lemma no_neg_sentential_count_eq_binary_count {φ : Wff} (h : ¬φ.hasNotSymbol)
|
||||
: φ.sententialSymbolCount = φ.binarySymbolCount := by
|
||||
sorry
|
||||
|
||||
/-- #### Parentheses Count
|
||||
|
||||
Let `φ` be a well-formed formula and `c` be the number of places at which a
|
||||
sentential connective symbol exists. Then there is `2c` parentheses in `φ`.
|
||||
-/
|
||||
theorem paren_count_double_sentential_count (φ : Wff)
|
||||
: φ.parenCount = 2 * φ.sententialSymbolCount := by
|
||||
induction φ with
|
||||
| SS _ =>
|
||||
unfold parenCount sententialSymbolCount
|
||||
simp
|
||||
| Not e ih =>
|
||||
unfold parenCount sententialSymbolCount
|
||||
rw [ih]
|
||||
ring
|
||||
| And e₁ e₂ ih₁ ih₂
|
||||
| Or e₁ e₂ ih₁ ih₂
|
||||
| Cond e₁ e₂ ih₁ ih₂
|
||||
| Iff e₁ e₂ ih₁ ih₂ =>
|
||||
unfold parenCount sententialSymbolCount
|
||||
rw [ih₁, ih₂]
|
||||
ring
|
||||
|
||||
/--
|
||||
The length of a `Wff` corresponds to the summation of sentence symbols,
|
||||
sentential connective symbols, and parentheses.
|
||||
-/
|
||||
theorem length_eq_sum_symbol_count (φ : Wff)
|
||||
: φ.length = φ.sentenceSymbolCount +
|
||||
φ.sententialSymbolCount +
|
||||
φ.parenCount := by
|
||||
induction φ with
|
||||
| SS _ =>
|
||||
unfold length sentenceSymbolCount sententialSymbolCount parenCount
|
||||
simp
|
||||
| Not e ih =>
|
||||
unfold length sentenceSymbolCount sententialSymbolCount parenCount
|
||||
rw [ih]
|
||||
ac_rfl
|
||||
| And e₁ e₂ ih₁ ih₂
|
||||
| Or e₁ e₂ ih₁ ih₂
|
||||
| Cond e₁ e₂ ih₁ ih₂
|
||||
| Iff e₁ e₂ ih₁ ih₂ =>
|
||||
unfold length sentenceSymbolCount sententialSymbolCount parenCount
|
||||
rw [ih₁, ih₂]
|
||||
ac_rfl
|
||||
|
||||
end Wff
|
||||
|
||||
/-! #### Exercise 1.1.2
|
||||
|
||||
Show that there are no wffs of length `2`, `3`, or `6`, but that any other
|
||||
positive length is possible.
|
||||
/--
|
||||
An enumeration of values `m` and `n` can take on in equation `m + n = 3`.
|
||||
-/
|
||||
|
||||
section Exercise_1_1_2
|
||||
|
||||
private lemma eq_3_by_cases (m n : ℕ) (h : m + n = 3)
|
||||
: m = 0 ∧ n = 3 ∨
|
||||
m = 1 ∧ n = 2 ∨
|
||||
|
|
@ -93,10 +202,16 @@ private lemma eq_3_by_cases (m n : ℕ) (h : m + n = 3)
|
|||
right; right; right
|
||||
exact ⟨m_eq_3, h⟩
|
||||
|
||||
/-! #### Exercise 1.1.2
|
||||
|
||||
Show that there are no wffs of length `2`, `3`, or `6`, but that any other
|
||||
positive length is possible.
|
||||
-/
|
||||
|
||||
theorem exercise_1_1_2_i (φ : Wff)
|
||||
: φ.length ≠ 2 ∧ φ.length ≠ 3 ∧ φ.length ≠ 6 := by
|
||||
induction φ with
|
||||
| SS c =>
|
||||
| SS _ =>
|
||||
unfold Wff.length
|
||||
simp
|
||||
| Not e ih =>
|
||||
|
|
@ -140,45 +255,95 @@ theorem exercise_1_1_2_i (φ : Wff)
|
|||
intro h₃
|
||||
exact absurd h₃.left ih₁.right.left
|
||||
|
||||
theorem exercise_1_1_2_ii (n : ℕ) (h : n ≠ 2 ∧ n ≠ 3 ∧ n ≠ 6)
|
||||
theorem exercise_1_1_2_ii (n : ℕ) (h : n ∈ Set.univ \ {2, 3, 6})
|
||||
: ∃ φ : Wff, φ.length = n := by
|
||||
let φ₁ := Wff.SS 1
|
||||
let φ₂ := Wff.And φ₁ (Wff.SS 2)
|
||||
let φ₃ := Wff.And φ₂ (Wff.SS 3)
|
||||
sorry
|
||||
|
||||
end Exercise_1_1_2
|
||||
|
||||
/-! #### Exercise 1.1.3
|
||||
/-- #### Exercise 1.1.3
|
||||
|
||||
Let `α` be a wff; let `c` be the number of places at which binary connective
|
||||
symbols (`∧`, `∨`, `→`, `↔`) occur in `α`; let `s` be the number of places at
|
||||
which sentence symbols occur in `α`. (For example, if `α` is `(A → (¬ A))` then
|
||||
`c = 1` and `s = 2`.) Show by using the induction principle that `s = c + 1`.
|
||||
-/
|
||||
|
||||
section Exercise_1_1_3
|
||||
|
||||
private def binary_symbol_count : Wff → ℕ
|
||||
| Wff.SS _ => 0
|
||||
| Wff.Not e => binary_symbol_count e
|
||||
| Wff.And e₁ e₂
|
||||
| Wff.Or e₁ e₂
|
||||
| Wff.Cond e₁ e₂
|
||||
| Wff.Iff e₁ e₂ => binary_symbol_count e₁ + binary_symbol_count e₂ + 1
|
||||
|
||||
private def sentence_symbol_count : Wff → ℕ
|
||||
| Wff.SS _ => 1
|
||||
| Wff.Not e => sentence_symbol_count e
|
||||
| Wff.And e₁ e₂
|
||||
| Wff.Or e₁ e₂
|
||||
| Wff.Cond e₁ e₂
|
||||
| Wff.Iff e₁ e₂ => sentence_symbol_count e₁ + sentence_symbol_count e₂
|
||||
|
||||
theorem exercise_1_1_3 (φ : Wff)
|
||||
: sentence_symbol_count φ = binary_symbol_count φ + 1 := by
|
||||
sorry
|
||||
: φ.sentenceSymbolCount = φ.binarySymbolCount + 1 := by
|
||||
induction φ with
|
||||
| SS _ =>
|
||||
unfold Wff.sentenceSymbolCount Wff.binarySymbolCount
|
||||
simp
|
||||
| Not e ih =>
|
||||
unfold Wff.sentenceSymbolCount Wff.binarySymbolCount
|
||||
exact ih
|
||||
| And e₁ e₂ ih₁ ih₂
|
||||
| Or e₁ e₂ ih₁ ih₂
|
||||
| Cond e₁ e₂ ih₁ ih₂
|
||||
| Iff e₁ e₂ ih₁ ih₂ =>
|
||||
unfold Wff.sentenceSymbolCount Wff.binarySymbolCount
|
||||
rw [ih₁, ih₂]
|
||||
ring
|
||||
|
||||
end Exercise_1_1_3
|
||||
/-- #### Exercise 1.1.5 (a)
|
||||
|
||||
Suppose that `α` is a wff not containing the negation symbol `¬`. Show that the
|
||||
length of `α` (i.e., the number of symbols in the string) is odd.
|
||||
|
||||
*Suggestion*: Apply induction to show that the length is of the form `4k + 1`.
|
||||
-/
|
||||
theorem exercise_1_1_5_a (α : Wff) (hα : ¬α.hasNotSymbol)
|
||||
: Odd α.length := by
|
||||
suffices ∃ k : ℕ, α.length = 4 * k + 1 by
|
||||
have ⟨k, hk⟩ := this
|
||||
unfold Odd
|
||||
exact ⟨2 * k, by rw [hk, ← mul_assoc]; norm_num⟩
|
||||
induction α with
|
||||
| SS _ =>
|
||||
refine ⟨0, ?_⟩
|
||||
unfold Wff.length
|
||||
simp
|
||||
| Not e _ =>
|
||||
unfold Wff.hasNotSymbol at hα
|
||||
exfalso
|
||||
exact hα trivial
|
||||
| And e₁ e₂ ih₁ ih₂
|
||||
| Or e₁ e₂ ih₁ ih₂
|
||||
| Cond e₁ e₂ ih₁ ih₂
|
||||
| Iff e₁ e₂ ih₁ ih₂ =>
|
||||
unfold Wff.hasNotSymbol at hα
|
||||
rw [not_or_de_morgan] at hα
|
||||
have ⟨k₁, hk₁⟩ := ih₁ hα.left
|
||||
have ⟨k₂, hk₂⟩ := ih₂ hα.right
|
||||
refine ⟨k₁ + k₂ + 1, ?_⟩
|
||||
unfold Wff.length
|
||||
rw [hk₁, hk₂]
|
||||
ring
|
||||
|
||||
/-- #### Exercise 1.1.5 (b)
|
||||
|
||||
Suppose that `α` is a wff not containing the negation symbol `¬`. Show that more
|
||||
than a quarter of the symbols are sentence symbols.
|
||||
|
||||
*Suggestion*: Apply induction to show that the number of sentence symbols is
|
||||
`k + 1`.
|
||||
-/
|
||||
theorem exercise_1_1_5_b (α : Wff) (hα : ¬α.hasNotSymbol)
|
||||
: α.sentenceSymbolCount > α.length / 4 := by
|
||||
have h₁ := α.sentenceSymbolCount
|
||||
have h₂ := α.sententialSymbolCount
|
||||
have h₃ := α.parenCount
|
||||
rw [
|
||||
α.length_eq_sum_symbol_count,
|
||||
Wff.paren_count_double_sentential_count α,
|
||||
Wff.no_neg_sentential_count_eq_binary_count hα,
|
||||
exercise_1_1_3 α
|
||||
]
|
||||
generalize Wff.binarySymbolCount α = k
|
||||
calc k + 1
|
||||
_ = 4 * (k + 1) / 4 := Eq.symm $ Nat.mul_div_cancel_left (k + 1) (by simp)
|
||||
_ = (k + 4 + k + 2 * k) / 4 := by ring_nf
|
||||
_ > (k + 1 + k + 2 * k) / 4 := by sorry
|
||||
|
||||
end Enderton.Logic.Chapter_1
|
||||
|
|
|
|||
|
|
@ -1,5 +1,6 @@
|
|||
import Bookshelf.Enderton.Set.Chapter_1
|
||||
import Bookshelf.Enderton.Set.Chapter_2
|
||||
import Bookshelf.Enderton.Set.Chapter_3
|
||||
import Bookshelf.Enderton.Set.Chapter_4
|
||||
import Bookshelf.Enderton.Set.OrderedPair
|
||||
import Bookshelf.Enderton.Set.Relation
|
||||
Loading…
Reference in New Issue