(* We need an assumption saying that for every n, the step function (f h n) calls h only on the segment [0 ... n - 1]. This means that if h1 and h2 coincide on values < n, then (f h1 n) coincides with (f h2 n) *)
Hypothesis step_good :
forall (n : N.t) (h1 h2 : N.t -> A),
(forall m : N.t, m < n -> Aeq (h1 m) (h2 m)) -> Aeq (f h1 n) (f h2 n).

Lemma strong_rec0_more_steps : forall a k n m, m < n ->
Aeq (strong_rec0 a f n m) (strong_rec0 a f (n+k) m).