proposition: Lemma sumn_incr_first_n sh n : sumn (incr_first_n sh n) = sumn sh + n.
proof: Proof. elim: n sh => [/= | n IHn]; first by case => //=. case => [/= | s0 s /=]. - have -> : sumn (nseq n 1) = n. by elim: n {IHn} => //= n ->; rewrite add1n. by rewrite add0n add1n. - by rewrite IHn addnA addnS !addSn. Qed.
imports: HB(structures) mathcomp(all_ssreflect) mathcomp(div ssralg ssrint ssrnum binomial) tools combclass sorted ordtype