proposition: Lemma string_of_term_sub : forall s l, string_of_term (sub s l) = string_of_term l ++ string_of_term (s (var l)).
proof: Proof. intro s. apply term_ind_forall. refl. intros f t IH. rewrite sub_fun1. simpl. rewrite !(Vcast_cons (is_unary_sig f)). simpl. rewrite IH. refl. Qed.
imports: CoLoR(LogicUtil RelUtil SN VecUtil ListUtil NatUtil)