How to prove (forall x, P x /\ Q x) -> (forall x, P x)

You can do it more quickly by just applying H, but this script should be more clear.

Lemma foo : forall (A:Type) (P Q: A-> Prop), (forall x, P x /\ Q x) -> (forall x, P x).
intros.
destruct (H x). 
exact H0.
Qed.

Try

elim (H x).

Actually, I figured this one out when I found this:

Mathematics for Computer Scientists 2

In lesson 5 he solves the exact same problem and uses "cut (P x /\ Q x)" which re-writes the goal from "P x" to "P x /\ Q x -> P x". From there you can do some manipulations and when the goal is just "P x /\ Q x" you can apply "forall x : P x /\ Q x" and the rest is straightforward.