proposition: Lemma ofnat_le : forall n m : nat, [n]<=[m] <-> (n<=m)%nat.
proof: Proof. intros. rewrite lt_eq_cases, ofnat_lt, ofnat_eq. split. - destruct 1; subst; auto. apply Nat.lt_le_incl; assumption. - apply Nat.lt_eq_cases. Qed.
imports: NumPrelude NZAxioms NZBase NZOrder NZAddOrder PeanoNat