What's the negative statement of

Question

$\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 $R(x)$?

Answer 1

Note that the negation of

$\forall x \phi(x)$

is

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

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.

Answer 2

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

Answer 3

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