proposition: Lemma divide_lcm_iff : forall n m, 0<=m -> ((n|m) <-> lcm n m == m).
proof: Proof. intros n m Hn. split. - now apply divide_lcm_eq_r. - intros EQ. rewrite <- EQ. apply divide_lcm_l. Qed.
imports: ZAxioms ZMulOrder ZSgnAbs ZGcd ZDivTrunc ZDivFloor