proposition: Lemma fun_ext_5 : forall (f g: forall (x1:A1) (x2:A2 x1) (x3:A3 x2) (x4:A4 x3) (x5:A5 x4), A6 x5), (forall x1 x2 x3 x4 x5, f x1 x2 x3 x4 x5 = g x1 x2 x3 x4 x5) -> f = g.
proof: Proof. Proof using. repeat (intros; apply fun_ext_dep). auto. Qed.
imports: TLC(LibTactics LibAxioms)