proposition: Lemma cons_eq_cons_inv : forall x1 x2 l1 l2, x1 :: l1 = x2 :: l2 -> x1 = x2 /\ l1 = l2.
proof: Proof. Proof using. introv H. inverts* H. Qed.
imports: Coq.Classes.Morphisms TLC(LibTactics LibLogic LibReflect LibOperation LibProd LibOption LibNat LibInt LibWf LibMonoid LibRelation)