proposition: Lemma lc_exp_of_lc_set_exp : forall e1, lc_set_exp e1 -> lc_exp e1.
proof: Proof. pose proof lc_exp_of_lc_set_exp_mutual as H; intuition eauto. Qed.
imports: Coq.Logic.FunctionalExtensionality Coq.Program.Equality Metalib.Metatheory Metalib.LibLNgen Stlc