proposition: Theorem f_equal5 : forall (A1 A2 A3 A4 A5 B:Type) (f:A1 -> A2 -> A3 -> A4 -> A5 -> B) (x1 y1:A1) (x2 y2:A2) (x3 y3:A3) (x4 y4:A4) (x5 y5:A5), x1 = y1 -> x2 = y2 -> x3 = y3 -> x4 = y4 -> x5 = y5 -> f x1 x2 x3 x4 x5 = f y1 y2 y3 y4 y5.
proof: Proof. admit. Defined.