$\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 $R(x)$?
$\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 $R(x)$?
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.
There exists $x$ such that $P(x)$, not $Q(x)$ and not $R(x)$.
$\neg \forall x:(P(x)\rightarrow Q(x)) \vee R(x)\equiv \exists x:\neg(\neg P(x)\vee Q(x))\wedge (\neg R(x))\equiv\exists x:(P(x) \wedge (\neg Q(x)))\wedge (\neg R(x))$