(** NB : We would like to have [pred n == n] for the initial element, but nothing forces that. For instance we can have -3 as initial point, and P(-3) = 2. A bit odd indeed, but legal according to [NZDomainSig]. We can hence have [n == (P^k) m] without [exists k', m == (S^k') n]. *)
(** We need decidability of [eq] (or classical reasoning) for this: *)
Hypothesis eq_decidable : forall n m, n==m \/ n~=m.

Lemma succ_pred_approx : forall n, ~initial n -> S (P n) == n.
End SuccPred.