Reputation: 43
For example I have these two hypotheses (one is negation of other)
H : forall e : R, e > 0 -> exists p : X, B e x p -> ~ F p
H0 : exists e : R, e > 0 -> forall p : X, B e x p -> F p
And goal
False
How to prove it?
Upvotes: 3
Views: 384
Reputation: 6047
You can't, because H0
is not the negation of H
. The correct statement would be
Definition R := nat.
Parameter X: Type.
Parameter T: Type.
Parameter x: T.
Parameter B : R -> T -> X -> Prop.
Parameter F : X -> Prop.
Lemma foobar: forall (H: forall e : R, e > 0 -> exists p : X, B e x p -> ~ F p)
(H0: exists e: R, e > 0 /\ forall p: X, B e x p /\ F p), False.
Proof.
intros H H0.
destruct H0 as [e [he hforall]].
destruct (H e he) as [p hp].
destruct (hforall p) as [hB hF].
exact (hp hB hF).
Qed.
Upvotes: 2