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In classical propositional logic $\neg(A \to B) \equiv A \land \neg B$. In predicate logic, therefore, do we have $\neg (\forall x (\phi(x) \to \psi(x)) \equiv \forall x (\phi(x) \land \neg \psi(x))$?

Or, instead, would $\neg (\forall x (\phi(x) \to \psi(x)) \equiv \exists x (\phi(x) \land \neg \psi(x))$?

Or do we have, in fact, $\neg (\forall x (\phi(x) \to \psi(x)) \equiv \forall x (\phi(x) \land \neg \psi(x))\equiv \exists x (\phi(x) \land \neg \psi(x))$. If this is the case, then why do we have $\forall x (\phi(x) \land \neg \psi(x))\equiv \exists x (\phi(x) \land \neg \psi(x))$ ?

More generally, I don't understand how we understand a formula like (1), as opposed to one like (2):

$$(1)\;\;\forall x (\phi(x) \land \neg \psi(x))$$

$$(2)\; \;\forall x (\phi(x) \to \neg \psi(x))$$


Edit: It is still unclear to me what the precise difference between (1) and (2) is (and their variants $sans$ negation). (1) requires the truth of $\phi(x)$ for all substitution instances, whereas (2) does not. Is that all?

But then, what might be a natural sentence of English which expresses (1)? Would it be something like (1')?

(1') Every man dances and he doesn't play football

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Take it step by step, first apply quantifier duality, and next negate the predicate.

$$\begin{align}&\lnot\forall x~\big(\phi(x)\to\psi(x)\big)\\[1ex]\equiv ~&\exists x~\lnot\big(\phi(x)\to\psi(x)\big)\\[1ex]\equiv ~&\exists x~\big(\phi(x)\land\lnot\psi(x)\big)\end{align}$$

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  • $\begingroup$ Ah, yes. What about the difference between (1) and (2)? $\endgroup$ – Edward.Lin Dec 11 '18 at 10:40
  • $\begingroup$ I have revised my question $\endgroup$ – Edward.Lin Dec 11 '18 at 10:48
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In general $\neg\forall x\;P(x)$ and $\exists x\;\neg P(x)$ are the equivalent.

In words: "not all $x$ have property $P$" is the same as "some $x$ exists that has not property $P$".

So if $P(x)$ is $\phi(x)\to\psi(x)$ then it appears that $\neg\forall x\;[\phi(x)\to\psi(x)]$ and $\exists x\;\neg [\phi(x)\to\psi(x)]$ are equivalent.

Since $\neg [\phi(x)\to\psi(x)]$ and $\phi(x)\wedge\neg\psi(x)$ are equivalent, there is also equivalence with $\exists x\;[\phi(x)\wedge\neg\psi(x)]$

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