Proof using.
intros.
repeat (do_in_app; intuition auto 10 with *).
Qed.

Lemma In_cons_2_3_neq :
forall a x y xs ys zs,
In (A:=A) a (xs ++ x :: ys ++ y :: zs) ->
a <> x ->
a <> y ->
In a (xs ++ ys ++ zs).