proposition: Theorem f_equal3 : forall (A1 A2 A3 B:Type) (f:A1 -> A2 -> A3 -> B) (x1 y1:A1) (x2 y2:A2) (x3 y3:A3), x1 = y1 -> x2 = y2 -> x3 = y3 -> f x1 x2 x3 = f y1 y2 y3.
proof: Proof. destruct 1; destruct 1; destruct 1; reflexivity. Qed.
imports: Notations Ltac