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$\forall x: P(x) \rightarrow Q(x) \vee R(x) $

Is it for some $x$, if not $P(x)$, then not $Q(x)$ and not $R(x)$?

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Note that the negation of

$\forall x \phi(x)$


$\exists x \neg \phi(x)$.

Then we use that

$\neg(\phi \rightarrow \psi) = \phi \land \neg \psi$

and De Morgan's laws to get

$\exists x [ P(x) \land \neg Q(x) \land \neg R(x) ]$

as the negation we want.

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There exists $x$ such that $P(x)$, not $Q(x)$ and not $R(x)$.

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Right answer. I would format it as $\exists x [P(x) \wedge \neg (Q(x) \wedge R(x))]$ One could use DeMorgan on the right clause. – Ross Millikan Nov 9 '12 at 4:49
There's a tiny typo, it should be $\exists x[P(x)\land\neg (Q(x)\lor R(x))]$ ;). – ՃՃՃ Nov 9 '12 at 4:52
right. Thanks.... – Ross Millikan Nov 9 '12 at 4:58

$$\neg \forall x:(P(x)\rightarrow Q(x)) \vee R(x)\equiv \exists x:\neg(\neg P(x)\vee Q(x))\wedge (\neg R(x))\equiv\exists x:(P(x) \wedge (\neg Q(x)))\wedge (\neg R(x))$$

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