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Which one of the following is NOT logically equivalent to $¬∃x(∀ y (α)∧∀z(β ))$ ?

1. $∀ x(∃ z(¬β )→∀ y(α))$
2. $∀x(∀ z(β )→∃ y(¬α))$
3. $∀x(∀ y(α)→∃z(¬β ))$
4. $∀x(∃ y(¬α)→∃z(¬β ))$

It's simply based on Demorgan's law

$\sim \exists x(\forall y(\alpha )\wedge \forall z(\beta ))$

// Simply Replace $\forall$ with $\exists$ and vice versa , $\vee$ with $\wedge$ and vice versa , $P(t)$ with $\sim$$P(t) and vice versa. \equiv \forall x(\exists y(\sim\alpha )\vee\exists z(\sim(\beta ))) // like \overline{a+b}\equiv \overline{a} .\overline{b} this \overline{\forall xP(t)} \equiv \exists x\sim P(t) \equiv \forall x(\forall z(\beta )\rightarrow \exists y(\sim(\alpha ))) -------option B matched Let suppose \forall z(\beta ) is 'a' and \exists y(\sim\alpha ) is 'b' So option B is like a->b and option C is \sim b -> \sim a , both are equivalent so hence C is also equivalent. Option A is \sim a -> \sim b which is not equivalent to a->b so out Option D is b->\sim a which is also not equivalent so out. hope you like my approach ## 2 Answers Best answer A useful rule:$$\forall x (\alpha) = \neg \exists (x) (\neg \alpha)$$i.e.; If some property$\alpha$is true for all$x$, then it is equivalent ot say that no$x$exists such that property$\alpha$does not hold for it. Starting with choices: 1.$\forall x (\exists z (\neg \beta) \to \forall y (\alpha)) \implies \forall x (\neg \exists z (\neg \beta) \vee \forall y (\alpha)) \implies \forall x (\forall z ( \beta) \vee \forall y (\alpha)) \implies \neg \exists x \neg (\forall z ( \beta) \vee \forall y (\alpha)) \implies \neg \exists x  (\neg \forall z ( \beta) \wedge \neg \forall y (\alpha)) $So, A is not matching with the logical statement in question. 2.$\forall x (\forall z (\beta) \to \exists y (\neg \alpha)) \implies \forall x (\neg \forall z (\beta) \vee \exists y (\neg \alpha)) \implies \neg \exists x \neg(\neg \forall z (\beta) \vee \exists y (\neg \alpha)) \implies \neg \exists x (\forall z (\beta) \wedge \neg \exists y (\neg \alpha)) \implies \neg \exists x (\forall z (\beta) \wedge \forall y (\alpha)) $Hence, matches with the given statement. 3.$\forall x (\forall y (\alpha) \to \exists z (\neg \beta)) \implies \forall x (\neg \forall y (\alpha) \vee \exists z (\neg \beta)) \implies \neg \exists x \neg(\neg \forall y (\alpha) \vee \exists z (\neg \beta)) \implies \neg \exists x (\forall y (\alpha) \wedge \neg \exists z (\neg \beta)) \implies \neg \exists x (\forall y (\alpha) \wedge \forall z (\beta)) $Hence, matches with the given statement. 4.$\forall x (\exists y (\neg \alpha) \to \exists z (\neg \beta)) \implies \forall x (\neg \exists y (\neg \alpha) \vee \exists z (\neg \beta)) \implies \forall x (\forall y ( \alpha) \vee \exists z (\neg \beta)) \implies \neg \exists x \neg (\forall y ( \alpha) \vee \exists  z (\neg \beta)) \implies \neg \exists x  (\neg \forall y ( \alpha) \wedge \neg \exists z (\neg \beta)) \implies \neg \exists x  (\neg \forall y ( \alpha) \wedge \forall z (\beta)) \$

So, D is not matching with the logical statement in question.

Thus both (A) and (D) are not logically equivalent to the given statement.
In GATE 2013 marks were given to all for this question

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### ∃x(∀z(β)∨∀y(α)) as  ∃x(∀y(α)∨∀z(β)) ?

→ Yes we can write that way. @ankitrazzagmail.com

→ Option D should be as per commented by @Tesla!

D.∀x(∃y(¬α)→∃z(β))

its ∃z (¬β) in D  option.

option A

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Thanks a lotss ..Sir
Good method of solving through variables p and q.