proposition: Lemma Qpower_1_le: forall (q : Q) (n : Z), (1<=q)%Q -> (0<=n)%Z -> (1 <= q ^ n)%Q.
proof: Proof. intros q n Hq Hn. destruct n. - apply Qle_refl. - apply Qpower_1_le_pos; assumption. - discriminate (Z.le_lt_trans _ _ _ Hn (Pos2Z.neg_is_neg p)). Qed.
imports: Zpow_facts Qfield Qreduction