Proof using.
intros a l1 l2 H.
apply (permutation_cons_ex_aux a (a :: l1) l2 H nil l1); simpl in |- *; auto.
Qed.

(* A permutation can be simply inverted if the two list starts with a cons *)
Theorem permutation_inv :
forall (a : A) (l1 l2 : list A),
permutation (a :: l1) (a :: l2) -> permutation l1 l2.