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

Lemma before_2_3_reduce :
forall xs ys zs x y a b,
before(A:=A) a b (xs ++ x :: ys ++ y :: zs) ->
a <> x ->
a <> y ->
before a b (xs ++ ys ++ zs).