Is $\lnot\forall x(Px \lor Qx) \equiv \lnot(\forall x(Px \lor Qx))?$

Question

Is $\lnot\forall x(Px \lor Qx) \equiv \lnot(\forall x(Px \lor Qx))? $

I'm trying to figure out how to apply the distributive law for universal quantifiers when the universal quantifier is negated, and the book that I'm studying doesn't specify.

If they're equivalent, then applying the law should result in $\lnot((\forall x Px) \lor (\forall x \lor Qx))$.

If they're not equivalent, maybe in $\lnot\forall x Px \lor \lnot \forall Qx$.

Answer 1

Yes, they are equivalent, and they are both equivalent to

$$\exists x \textrm{ such that }(\lnot Px) \wedge (\lnot Qx)$$

Answer 2

Of course they are ... what else would the $\neg$ in $\lnot(\forall x(Px \lor Qx))$ be negating other than $\forall x(Px \lor Qx)$? In other words, the extra parentheses simply do no work at all and can be dropped, resulting in $\lnot \forall x(Px \lor Qx)$

So, these statement are not just equivalent, they are really just one and the same.