Proof fun X A a Y B b R H1 H2 H3 E e => e R H1 H2 H3.
(** It would be useless to define the corresponding elimination rule whose behaviour is obtained by using [Apply H] instead of [Elim H] in the following proof scripts.  We first check that bisimilarity is an equivalence relation: *)

Lemma EQV_refl : forall (X : Typ1) (A : Rel X) (a : X), EQV X A a X A a.