(** NB: without the [step_good] hypothesis, we have proved that [strong_rec a f 0] is [f (fun _ => a) 0]. Now we can prove that the first argument of [f] is arbitrary in this case... *)

Theorem strong_rec_0_any : forall (a : A)(any : N.t->A),
Aeq (strong_rec a f 0) (f any 0).