proposition: Lemma pow_pow_in_inv: forall c1 c2 n1 n2 w, In w (App (Pow (Char c1) n1) (Pow (Char c2) n2)) -> w = Util.pow1 c1 n1 ++ Util.pow1 c2 n2.
proof: Proof. intros. apply app_in_inv in H. destruct H as (w1, (w2, (?, (Ha, Hb)))). subst. apply pow_char_in_inv in Ha. apply pow_char_in_inv in Hb. subst. reflexivity. Qed.
imports: Coq.Lists.List