proposition: Lemma fun_ext_2 : forall (f g : forall (x1:A1) (x2:A2 x1), A3 x2), (forall x1 x2, f x1 x2 = g x1 x2) -> f = g.
proof: Proof. Proof using. repeat (intros; apply fun_ext_dep). auto. Qed.
imports: TLC(LibTactics LibAxioms)