proposition: Lemma count_occ_elt_eq l1 l2 x y : x = y -> count_occ (l1 ++ x :: l2) y = S (count_occ (l1 ++ l2) y).
proof: Proof. intros ->. rewrite ? count_occ_app; cbn. destruct (eq_dec y y) as [Heq | Hneq]; [ apply Nat.add_succ_r | now contradiction Hneq ]. Qed.
imports: PeanoNat