Proof using.
induction xs; intros; simpl in *; intuition.
Qed.

Lemma before_2_3_insert :
forall xs ys zs x y a b,
before(A:=A) a b (xs ++ ys ++ zs) ->
b <> x ->
b <> y ->
before a b (xs ++ x :: ys ++ y :: zs).