Proof using.
induction 1; intros; simpl in *; intuition.
- auto using op_equivalent_refl.
- eauto.
Qed.

Lemma op_equivalent_all_Is_I_middle :
forall xs ys k,
(forall x, In x xs -> exists k, x = I k) ->
op_equivalent (xs ++ I k :: ys) (I k :: xs ++ ys).