proposition: Lemma list_blA {A eq_A} {R : relation A} (A_bl : forall x y : A, eq_A x y = true -> R x y) {x y : list A} : list_beq eq_A x y = true -> SetoidList.eqlistA R x y.
proof: Proof. generalize dependent y; induction x as [|x xs IHxs]; simpl; intros [|??]; simpl; intro H; try discriminate; constructor; apply Bool.andb_true_iff in H; destruct H; eauto with nocore. Qed.
imports: Coq.Lists.List Coq.Lists.SetoidList Coq.Bool.Bool Coq.Arith.PeanoNat Coq.Strings.Ascii Coq.Strings.String Coq.Logic.Eqdep_dec Fiat.Common