intros a b H' H'0; try assumption.
lapply (ppc_is_ppcm a b);
[ intros H'3; lapply H'3; clear H'3; [ intros H'4 | idtac ] | idtac ];
auto; auto.
inversion H'4; auto.
Qed.
Hint Resolve divP_ppcl divP_ppcr.

Theorem divTerm_compo :
forall (a b c : Term A n) (nZb : ~ zeroP (A:=A) A0 eqA (n:=n) b)
(nZc : ~ zeroP (A:=A) A0 eqA (n:=n) c),
divP a b ->
divP b c ->
eqTerm (A:=A) eqA (n:=n) (divTerm a nZc)
(multTerm (A:=A) multA (n:=n) (divTerm a nZb) (divTerm b nZc)).