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_sym :
forall (X : Typ1) (A : Rel X) (a : X) (Y : Typ1) (B : Rel Y) (b : Y),
EQV X A a Y B b -> EQV Y B b X A a.