proposition: Lemma eqr_sum_div I r P (f : I -> F) c a : c != 0 -> \big[+%R/0]_(x <- r | P x) (f x / c) == a = (\big[+%R/0]_(x <- r | P x) f x == a * c).
proof: Proof. by move=> ?; rewrite -mulr_suml -(divr1 a) eqr_div ?oner_eq0// mulr1 divr1. Qed.
imports: HB(structures) mathcomp(ssreflect ssrfun ssrbool eqtype ssrnat div seq) mathcomp(choice fintype finfun bigop prime binomial)