apply eqA_trans with (y := multA (divA A1 a nZa) a); auto.
apply divA_multA_comp_r with (1 := cs); auto.
apply eqA_sym; apply divA_is_multA with (1 := cs); auto.
Qed.
Hint Resolve divA_A1.

Theorem divA_nZ :
forall a b : A,
~ eqA b A0 -> forall nZa : ~ eqA a A0, ~ eqA (divA b a nZa) A0.