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Re: [Coq-Club] What tactic can be used to prove 'forall' to '~ exists'?


Chronological Thread 
  • From: Catalin Hritcu <catalin.hritcu AT gmail.com>
  • To: Coq Club <coq-club AT inria.fr>
  • Subject: Re: [Coq-Club] What tactic can be used to prove 'forall' to '~ exists'?
  • Date: Sat, 23 Aug 2014 09:03:10 +0200

On Sat, Aug 23, 2014 at 8:55 AM, ....
<nixiejun AT 163.com>
wrote:
> Dear everyone:
> I'mc confused with the following subgoal:
> Theorem T1(P:Prop->Prop): forall B:Prop,~P B->~(exists A:Prop,P A).
> Is it possible to prove?

No, because it's wrong. If you put the parentheses the right way, then yes:

Theorem T1_fixed (P:Prop->Prop): (forall B:Prop,~P B)->~(exists A:Prop,P A).
Proof. intros H [x Hc]. eapply H. eassumption. Qed.

Hope this helps,
Catalin



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