80 lines
3.4 KiB
TeX
80 lines
3.4 KiB
TeX
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\documentclass[全部作业]{subfiles}
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\pagestyle{fancyplain}
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\fancyhead{}
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\fancyhead[C]{\mysignature}
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\setcounter{chapter}{3}
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\setcounter{section}{2}
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\begin{document}
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\section{范式和联结词的功能完备集}
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\begin{enumerate}
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\item 通过等值演算求$p \to (p\land (q \to p))$的主析取范式和主合取范式。
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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p \to (p \land (q \to p))&=\lnot p\lor (p\land (\lnot q\lor p))=\lnot p\lor ((p\lor 0)\land (p\lor \lnot q)) \\
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&=\lnot p \lor (p\lor (0 \land \lnot q)) = \lnot p \lor p = 1 \\
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&=(p\land q)\lor (p\land \lnot q)\lor (\lnot p\land q)\lor (\lnot p\land \lnot q) \qquad (主析取范式)\\
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\end{aligned}
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$$
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$$
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因为原式为永真式，所以无主合取范式。
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$$
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\end{zhongwen}
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\end{proof}
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\item 证明$\{ \lnot ,\to \}$是功能完备集。
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\begin{proof}
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\begin{zhongwen}
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$$
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\begin{aligned}
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&\lnot p=\lnot p \\
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&p\lor q=\lnot p \to q \\
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&\because \{ \lnot ,\lor \}是功能完备集 \\
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&\therefore \{ \lnot ,\to \}是功能完备集 \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\end{enumerate}
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\section{命题逻辑的推理理论}
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\begin{enumerate}
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\item 证明$p \to (q \to s),q,p\lor \lnot r \Rightarrow r \to s$。
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\begin{proof}
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\begin{zhongwen}
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\begin{enumerate}[label=\mycircle{\arabic{enumii}},leftmargin=\linewidth/4,rightmargin=\linewidth/4]
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\item $r$\hfill 附加前提引入
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\item $p\lor \lnot r$ \hfill 前提引入
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\item $p$\hfill \mycircle{1}\mycircle{2}析取三段论
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\item $p \to (q \to s)$\hfill 前提引入
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\item $q \to s$\hfill \mycircle{3}\mycircle{4}假言推理
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\item $q$ \hfill 前提引入
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\item $s$ \hfill \mycircle{5}\mycircle{6}假言推理
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\end{enumerate}
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\end{zhongwen}
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\end{proof}
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\item 构造下列推理的形式证明：
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“今天下午没有出太阳并且今天比昨天冷。只有今天下午出太阳，我们才去游泳。若我们不去游泳，则我们乘独木舟游览。若我们乘独木舟游览，则我们在黄昏时回家。 所以，我们在黄昏时回家。”
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\begin{proof}
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\begin{zhongwen}
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$$
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\begin{aligned}
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令&p:今天下午出太阳，q:今天比昨天冷，r:我们去游泳， \\
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&s:我们乘独木舟游览，t:我们在黄昏时回家 \\
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&则需要证明：\lnot p \land q,r \to p,\lnot r \to s, s \to t \Rightarrow t \\
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\end{aligned}
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$$
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\begin{enumerate}[label=\mycircle{\arabic{enumii}},leftmargin=\linewidth/4,rightmargin=\linewidth/4]
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\item $\lnot p\land q$\hfill 前提引入
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\item $\lnot p$\hfill \mycircle{1}化简
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\item $r \to p$\hfill 前提引入
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\item $\lnot r$\hfill \mycircle{2}\mycircle{3}拒取式
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\item $\lnot r \to s$\hfill 前提引入
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\item $s$\hfill \mycircle{4}\mycircle{5}假言推理
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\item $s \to t$\hfill 前提引入
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\item $t$\hfill \mycircle{6}\mycircle{7}假言推理
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\end{enumerate}
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\end{zhongwen}
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\end{proof}
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\end{enumerate}
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\end{document}
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