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: *)

(* Specifications: *)
Lemma EQV_refl : forall (X : Typ1) (A : Rel X) (a : X), EQV X A a X A a.
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.
Lemma EQV_trans :
forall (X : Typ1) (A : Rel X) (a : X) (Y : Typ1) (B : Rel Y)
(b : Y) (Z : Typ1) (C : Rel Z) (c : Z),
EQV X A a Y B b -> EQV Y B b Z C c -> EQV X A a Z C c.