intros l1 l2; try assumption.
unfold minuspf in |- *; case (minuspp (pair l1 l2)); simpl in |- *; auto.
Qed.
Hint Resolve minuspf_is_minusP.

(* Specifications: *)
Theorem minuspf_pO_is_eqP :
forall p q : list (Term A n),
eqP A eqA n (minuspf p q) (pO A n) -> eqP A eqA n p q.