proposition: Theorem appendfDisjoint : forall L1 L2 : list A, Disjoint L1 L2 -> forall L3 : list A, Disjoint L1 L3 -> Disjoint L2 L3 -> Disjoint (appendf L1 L2) L3.
proof: Proof. intros L1 L2 HL1L2 L3 H' H'0. apply appendDisjoint with (L1 := L1) (L2 := L2); auto. Qed.
imports: Coq(Arith) Coq(Lexicographic_Exponentiation)