Proof using.
intros A1 B1 f l1 l2 l3 H; generalize l3; elim H; clear H l1 l2 l3.
intros l3; case l3; simpl in |- *; auto.
intros H; exists (nil (A:=A1)); auto.
intros; discriminate.
intros a0 l1 l2 H H0 l3; case l3; simpl in |- *; auto.
intros; discriminate.
intros a1 l H1; case (H0 l); auto.
injection H1; auto.
intros l5 (H2, H3); exists (a1 :: l5); split; simpl in |- *; auto.
apply f_equal2 with (f := cons (A:=B1)); auto; injection H1; auto.
intros a0 b l l3; case l3.
intros; discriminate.
intros a1 l0; case l0; simpl in |- *.
intros; discriminate.
intros a2 l1 H; exists (a2 :: a1 :: l1); split; simpl in |- *; auto.
repeat apply f_equal2 with (f := cons (A:=B1)); injection H; auto.
intros l1 l2 l3 H H0 H1 H2 l0 H3.
case H0 with (1 := H3); auto.
intros l4 (HH1, HH2).
case H2 with (1 := HH2); auto.
intros l5 (HH3, HH4); exists l5; split; auto.
apply permutation_trans with (1 := HH3); auto.
Qed.

(* Permutation of a map can be inverted *)
Theorem permutation_map_ex :
forall (A B : Type) (f : A -> B) l1 l2,
permutation (map f l1) l2 ->
exists l3 : _, permutation l3 l1 /\ l2 = map f l3.