Proof with congruence || auto.
induction e; intros u k Fr; simpl in *; f_equal...
Case "exp_bvar".
destruct (k === n)... simpl. destruct (x == x)...
Case "exp_fvar".
destruct (a == x)... contradict Fr; fsetdec.
Qed.

Lemma subst_ee_intro : forall x e u,
x `notin` fv_ee e ->
open_ee e u = subst_ee x u (open_ee e x).