intros a b nZb p q r H' H'0 H'1 H'2; try assumption.
apply
(eqp_trans _ _ _ _ _ _ _ _ _ cs n)
with
(y := spminusf a b nZb
(pluspf (A:=A) A0 (eqA:=eqA) plusA eqA_dec (n:=n) (ltM:=ltM)
ltM_dec q p) r); [ auto | idtac ].
apply
(eqp_trans _ _ _ _ _ _ _ _ _ cs n)
with
(y := pluspf (A:=A) A0 (eqA:=eqA) plusA eqA_dec (n:=n) (ltM:=ltM) ltM_dec
(spminusf a b nZb q r) p); [ auto | idtac ].
apply (eqp_sym _ _ _ _ _ _ _ _ _ cs n); auto.
Qed.
Hint Resolve spminusf_plusTerm_r.

Theorem divP_minusTerm_comp :
forall a b c : Term A n,
divP A A0 eqA multA divA n a c ->
divP A A0 eqA multA divA n b c ->
eqT a b ->
~ zeroP (A:=A) A0 eqA (n:=n) (minusTerm (A:=A) minusA (n:=n) a b) ->
divP A A0 eqA multA divA n (minusTerm (A:=A) minusA (n:=n) a b) c.