T1
用等值演算、构造指派等方式判断公式的永真性
\[(\forall xP(x)\rightarrow \exist xQ(x))\rightarrow\exist x (P(x)\rightarrow Q(x) ) \](1)
\[(\forall x P(x)\rightarrow \forall x Q(x))\rightarrow \forall x (P(x)\rightarrow Q(x)) \](2)
T2
\[\begin{aligned} & \forall x(G(x)\lor H(x))\\ \models & \neg\neg\forall x(G(x)\lor H(x)) & 1\\ \models & \neg\exists x\neg(G(x)\lor H(x)) & 2\\ \models & \neg\exists x(\neg G(x)\land\neg H(x)) & 3\\ \models & \neg(\exists x\neg G(x)\land \exists x\neg H(x)) & 4\\ \models & \neg\exists x\neg G(x) \lor \neg\exists x\neg H(x) & 5\\ \models & \forall x G(x)\lor \forall x H(x) & 6 \end{aligned} \]以下哪一步出现错误?
A: 3 | B: 4 | C: 5 | D: 6 | E: 没有错误 |
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T3
前束范式\(A\)的无\(\forall\)前束范式\(A'\)的递归定义如下
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若\(A\)不含\(\forall\)量词,则\(A'\)是$ A$
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若\(A\)是\(\forall y B\),则\(A'\)是\((B^y_a)'\),其中\(a\)是\(B\)所不包含的常元
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若\(A\)是\(\exists x_1\exists x_2\cdots \exists x_n\forall yB\),则\(A'\)是\(\left(\exists x_1\exists x_2\cdots \exists x_nB^y_{f(x_1,x_2,\cdots x_n)}\right)'\),其中\(f\)是\(B\)所不包含的函词
标签:Week,exists,models,neg,Problems,lor,forall,rightarrow From: https://www.cnblogs.com/fallqs/p/18157094证明前束范式\(A\)是永真式当且仅当\(A'\)是永真式,其中,\(A'\)是\(A\)的无\(\forall\)前束范式