Proof spec_Qquadratic_Qpositive_to_Q_dL_dL.

Lemma outside_square_correct_1 :
forall (a b c d : Z) (p1 p2 : Qpositive),
(2 < outside_square a b c d)%Z ->
~ (b = 0%Z /\ c = 0%Z /\ d = 0%Z) ->
Qlt Zero
(Qplus
(Qplus
(Qplus (Qmult (Qmult a (Qpos p1)) (Qpos p2)) (Qmult b (Qpos p1)))
(Qmult c (Qpos p2))) d).