proposition: Lemma args_eq_4 : forall (f:A1->A2->A3->A4->B) x1 y1 x2 y2 x3 y3 x4 y4, x1 = y1 -> x2 = y2 -> x3 = y3 -> x4 = y4 -> f x1 x2 x3 x4 = f y1 y2 y3 y4.
proof: Proof. Proof using. intros. subst~. Qed.
imports: TLC(LibTactics LibAxioms)