Proof using.
induction xs; intros; simpl.
- auto.
- repeat (break_match; simpl); intuition; try discriminate.
+ eauto using subseq_refl.
+ do_bool. assert (j < i) by omega.
rewrite removeAfterIndex_le with (j := i) (i := j) at 1; auto; omega.
+ do_bool. assert (i < j) by omega.
rewrite removeAfterIndex_le with (i := i) (j := j) at 2; auto; omega.
Qed.

Lemma removeAfterIndex_comm :
forall xs i j,
removeAfterIndex (removeAfterIndex xs i) j =
removeAfterIndex (removeAfterIndex xs j) i.