proposition: Lemma subst_exp_lc_exp : forall e1 e2 x1, lc_exp e1 -> lc_exp e2 -> lc_exp (subst_exp e2 x1 e1).
proof: Proof. default_simp. Qed.
imports: Coq.Logic.FunctionalExtensionality Coq.Program.Equality Metalib.Metatheory Metalib.LibLNgen Stlc