proposition: Theorem Nge_out : forall n m, n >= m -> (nat_of_N n >= nat_of_N m)%nat.
proof: Proof. unfold N.ge; intros ?? H. rewrite nat_of_Ncompare in H. apply nat_compare_ge; assumption. Qed.
imports: Coq.Bool.Bool Coq.Classes.Morphisms Coq.Arith.Arith Coq.micromega.Lia Coq.NArith.NArith Coq.ZArith.ZArith