proposition: Lemma ofnat_mul : forall n m, [n*m] == [n]*[m].
proof: Proof. induction n; simpl; intros. - symmetry. apply mul_0_l. - rewrite Nat.add_comm. rewrite ofnat_add, mul_succ_l. now f_equiv. Qed.
imports: NumPrelude NZAxioms NZBase NZOrder NZAddOrder PeanoNat