Quantifiers and their negations

Question

I was revising logic and I realised that I had a very basic lack of understanding about quantifiers and their negations.

As I understand it ¬$\forall$$x$ $P$($x$) is equivalent to $\exists$$x$ ¬$P$($x$).

So far so simple. But what about quantifiers over a particular set.

Again, as I understand it $\forall$$x$ ($x \in X$) can be thought of as $\forall$$x$ $P$($x$) where $P$($x$) $=$ $x \in X$.

Yet the negation of $\forall$$x$ ($x \in X$) $P$($x$) is $\exists$$x$ ($x \in X$) ¬$P$($x$) rather than $\exists$$x$ ($x \notin X$) $P$($x$).

Are the two equivalent? What am I missing?

Answer 1

$\forall x \in X, P(x)$ is shorthand for $\forall x, (x \in X \implies P(x))$,
and $\exists x \in X, P(x)$ is shorthand for $\exists x, (x \in X \land P(x))$.

(your " $\forall x (x \in X) P(x)$ " isn't a well-formed formula and doesn't mean anything)

Notice that one uses an implication or the other uses a logical and.

Then the following are equivalent :

$\neg \forall x \in X, P(x) \\ \neg \forall x, (x \in X \implies P(x)) \\ \exists x, \neg (x \in X \implies P(x)) \\ \exists x, (x \in X \land \neg P(x)) \\ \exists x \in X, \neg P(x) $

So as expected the connectives $\forall x \in X$ and $\exists x \in X$ are still dual to each other.

Answer 2

With $X=\{a,b,\cdots z\}$, by De Morgan,

$$\lnot(P(a)\land P(b)\land\cdots P(z))=\lnot P(a)\lor\lnot P(b)\lor\cdots\lnot P(z).$$