The Coq mailing list

["Matthew Brecknell" <coq-club AT brecknell.org>]

Taral wrote:
> The point is that f is parametric in var. Try this instead:
> 
> Inductive test (var : Type) : Type :=
> | A : var -> test var
> | B : test var.
> 
> Goal forall T var (f : forall var, T -> test var) x y v, f var x = A
> var v -> exists v', f var y = A var v'.

Hi Taral,

I haven't got as far as proving your goal, but here's a more basic
result you might find useful as a starting point:

Inductive mytest (var : Type) : Type :=
  | mk_test : var -> mytest var.

Lemma p0: forall (f: forall (fvar: Type), mytest fvar), False.
Proof (fun f => match (f Empty_set) with mk_test v => Empty_set_ind (fun
e => False) v end).

(Inspired by Coq'Art section 6.6 on empty types, which I just happened
to read yesterday).

The corresponding lemma for your inductive type would be:

Lemma p1: forall (f: forall (fvar: Type), test fvar) (var: Type), f var
= B var.
Proof ???.

After that, it should just be a matter of adding in the T parameter,
then your goal would be vacuously true.

Cheers,
Matthew