Proof
fun X A a E _ _ H3 =>
H3 (eq_refl (opt X) (none X)) (eq_refl (opt X) (some X a)).

Lemma SUCC_compat :
forall (X : Typ1) (A : Rel X) (a : X) (Y : Typ1) (B : Rel Y) (b : Y),
EQV X A a Y B b ->
EQV (opt X) (SUCC X A a) (none X) (opt Y) (SUCC Y B b) (none Y).