proposition: Lemma Qdiv_num_denom : forall q1 q2 p : Q, p <> Zero -> Qmult q1 (Qinv q2) = Qmult (Qmult q1 p) (Qinv (Qmult q2 p)).
proof: Proof. intros q1 [| qp| qp] p Hp; (simpl in |- *; ring) || (field; split; discriminate || assumption). Qed.
imports: Coq(FunInd) QArithSternBrocot(Field_Theory_Q)