Lemma Par_lc1 : forall G D a a' , Par G D a a' -> lc_tm a.
  intros.  induction H; auto. all: lc_solve.
Qed.

(* FIXME: find a good place for this tactic. *)
(* cannot move this to ett_ind.v because need Toplevel_lc ??? *)
Ltac lc_toplevel_inversion :=
  ma

Proof. intros G D B1 B2 c H.