Are these equivalent?

Question

$\forall x \in D, (P(x) \Rightarrow Q(x))$ is equivalent to $(\forall x \in D \cap P,Q(x))$.

However, is this also equivalent to $(\forall x\in D)( P(x)\land Q(x))$?

If not, what's the difference? Thank you.

Answer 1

No.

$\forall x \in D, (P(x)\rightarrow Q(x)) \not\equiv \forall x\in D(P(x) \land Q(x))$.

$P(x) = x$ is a pink unicorn and

$Q(x) = x$ can fly.

Then it's certainly true that $\forall x\in D(P(x) \rightarrow Q(x))$, but not $\forall x\in D(P(x) \land Q(x))$.

In the first case, $P(x) \rightarrow Q(x)$ holds for all $x$ because for every x, $P(x)$ is false, and therefore the implication is true, vacuously.

In the second case, we have a proposition asserting that "Every x in the domain is both a pink unicorn and can fly." That's a different statement altogether, which happens not to be true: it never holds.