proposition: Lemma subst_exp_intro_rec : forall e1 x1 e2 n1, x1 `notin` fv_exp e1 -> open_exp_wrt_exp_rec n1 e2 e1 = subst_exp e2 x1 (open_exp_wrt_exp_rec n1 (var_f x1) e1).
proof: Proof. pose proof subst_exp_intro_rec_mutual as H; intuition eauto. Qed.
imports: Coq.Logic.FunctionalExtensionality Coq.Program.Equality Metalib.Metatheory Metalib.LibLNgen Stlc