proposition: Lemma sum_n_Sm (a : nat -> G) (n m : nat) : (n <= S m)%nat -> sum_n_m a n (S m) = plus (sum_n_m a n m) (a (S m)).
proof: Proof. intros Hnmk. rewrite (sum_n_m_Chasles _ _ m). by rewrite sum_n_n. by []. by apply Nat.le_succ_diag_r. Qed.
imports: Coq(Reals Psatz ssreflect) Rcomplements Rbar Markov Iter Lub