I been learning predicates and quantifiers, and now I am going through this question that suppose to take the negation of $(\forall_{x} \in U)(P(x) \implies Q(x))$. So worked it out from what I understood and got this:
$\neg[(\forall_{x} \in U)(P(x) \implies Q(x))]$ =
$(\exists_{x} \in U)\neg(P(x) \implies Q(x))$ =
$(\exists_{x} \in U)(\neg P(x) \implies \neg Q(x))$.
Is this correctly done? Any help would be appreciated.
The negation of $P\implies Q$ is not $\neg P\implies \neg Q$. It is, on the other hand, $P \wedge \neg Q$. Otherwise, you seem to have it right.
To put this in terms of a concrete example, the negation of "every dog has a tail" is "there is a dog without a tail". In this case, $P(x)$ means "$x$ is a dog", and $Q(x)$ means "$x$ has a tail."