proposition: Theorem fappendfIncl1 : forall (L1 L2 : list A) (HL1L2 : Disjoint L1 L2), incl L1 (fappendf L1 L2).
proof: Proof. intros L1 L2 HL1L2. apply appendDisjIncl1 with (L2 := map f L2); auto. apply Disjointf; auto. Qed.
imports: Coq(Arith) Coq(Lexicographic_Exponentiation)