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Re: [Coq-Club] fun <-> forall


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  • From: Burak Ekici <ekcburak AT hotmail.com>
  • To: Clément Pit-Claudel <cpitclaudel AT gmail.com>, "coq-club AT inria.fr" <coq-club AT inria.fr>
  • Subject: Re: [Coq-Club] fun <-> forall
  • Date: Thu, 26 Oct 2017 15:11:10 +0000
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Hi Clément,

Thanks a lot! That's exactly what I was looking for...

Best,



Burak.



On 26-10-2017 18:08, Clément Pit-Claudel wrote:
> On 2017-10-26 10:25, Burak Ekici wrote:
>> However, could not make it work whatever I did...
> Does the following work for you?
>
> pose proof (fun x y z => eq_ind_r (fun f => f x y z = _ x y z) eq_refl
> H1c) as H';
> cbv beta in H'.
>
> Here's a complete example:
>
> Axiom catC: Type.
> Axiom obj: Type -> Type.
> Axiom arrow: forall (catC: Type), obj catC -> obj catC -> Type.
> Axiom fmap : forall {C: Type} (a b: obj C), arrow C b a -> Type.
> Axiom foo : forall {C a b}, arrow C b a -> arrow C b a.
>
> Goal ((fun (a b : obj catC) (f : arrow catC b a) => fmap a b (foo f)) =
> (fun (a b : obj catC) (f : arrow catC b a) => fmap a b f) -> True).
> Proof.
> intros.
> pose proof (fun x y z => eq_ind_r (fun f => f x y z = _ x y z) eq_refl
> H) as H';
> cbv beta in H'.
>




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