Question

Which one of the following is NOT logically equivalent to$\neg \exists x\left(\forall y\right(\alpha )\wedge \forall z(\beta \left)\right)$

• A. $\forall x\left(\exists z\right(\neg \beta )\to \forall y(\alpha \left)\right)$
• B. $\forall x\left(\forall z\right(\beta )\to \exists y(\neg \alpha \left)\right)$
• C. $\forall x\left(\forall y\right(\alpha )\to \exists z(\neg \beta \left)\right)$
• D. $\forall x\left(\exists y\right(\neg \alpha )\vee \exists z(\neg \beta \left)\right)$

Answer 1

The correct option is A $\forall x\left(\exists z\right(\neg \beta )\to \forall y(\alpha \left)\right)$

Let,$\forall y\left(\alpha \right)=P,\forall z\left(\beta \right)=Q$
Then, $\exists y\left(\neg \alpha \right)=\neg Pand\exists z\left(\neg \beta \right)=\neg Q$
Given, $\neg \exists \left(\forall y\right(\alpha )\wedge \forall z(\beta \left)\right)$
$=\neg \exists x(p\wedge Q)=\neg \forall x(\neg p\vee \neg Q)$
$=\forall x(p\to \neg Q)$
(a) $\forall x\left(\exists z\right(\neg \beta )\to \forall y(\alpha )=\forall x(\neg Q\to P\left)\right)$
(b) $\forall x\left(\forall z\right(\beta )\to \exists y(\neg \alpha \left)\right)=\forall x(Q\to \neg P)=\forall x(p\to \neg Q)$
(c) $\forall x\left(\forall y\right(\alpha )\to \exists z(\neg \beta \left)\right)=\forall x(P\to \neg Q)$
(d) $\forall x\left(\exists y\right(\neg \alpha )\vee \exists z(\neg \beta \left)\right)=\forall x(\neg P\vee \neg Q)$
$=\forall x(\neg P\vee \neg Q)$
$\therefore$ Only (a) is not logically equivalent to $\forall x(\neg P\to \neg Q)$