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

Lemma remove_all_cons :
forall ys a l,
(remove_all ys (a :: l) = remove_all ys l /\
In a ys) \/
(remove_all ys (a :: l) = a :: (remove_all ys l) /\
~In a ys).