proposition: Lemma QSORaddon : SORaddon 0 1 Qplus Qmult Qminus Qopp Qeq Qle 0 1 Qplus Qmult Qminus Qopp Qeq_bool Qle_bool (fun x => x) (fun x => x) (pow_N 1 Qmult).
proof: Proof. constructor. - constructor ; intros ; try reflexivity. apply Qeq_bool_eq; auto. - constructor. reflexivity. - intros x y. apply Qeq_bool_neq ; auto. - apply Qle_bool_imp_le. Qed.
imports: OrderedRing RingMicromega Refl QArith Qfield