proposition: Lemma fv_exp_subst_exp_fresh : forall e1 e2 x1, x1 `notin` fv_exp e1 -> fv_exp (subst_exp e2 x1 e1) [=] fv_exp e1.
proof: Proof. pose proof fv_exp_subst_exp_fresh_mutual as H; intuition eauto. Qed.
imports: Coq.Logic.FunctionalExtensionality Coq.Program.Equality Metalib.Metatheory Metalib.LibLNgen Stlc