I'm confused by Coq on its way dealing with existential quantification.
I have a predicate P and an assumption H
P : nat -> Prop
H : exists n, P n
while the current goal is (whatever)
(Some goal)
If I want to instantiate n in H, I will do
elim H.
However after the elimination, the current goal becomes
forall n, P n -> (Some goal)
It looks like Coq converts an existential quantifier into a universal one. I know that (forall a, P a -> Q a) -> ((exists a, P a) -> Q a) out of my limited knowledge on first-order logic. But the reverse proposition seems to be incorrect. If the 'forall' one and 'exists' one are not equivalent, why Coq would do such conversion?
Does 'elim' in Coq replace the goal with a harder to prove one? Or could anyone please show why ((exists a, P a) -> Q a) -> (forall a, P a -> Q a) holds in first-order logic?
Maybe the missing key is that the goal:
forall n, P n -> (Some goal)
is to be read as:
forall n, (P n -> (Some goal))
and not as:
(forall n, P n) -> (Some goal)
That is, the goal you are given just gives you an arbitrary n and a proof P n, which is indeed the proper way to eliminate an existential (you don't get to know the value of the witness since it could be any value that makes P true, you just get to know that there is a n and that P n holds).
On the contrary, the latter would provide you with a function that can build P n for any n you pass it, which is indeed a stronger statement than the one you have.