proposition: Lemma sumn_map I (f : I -> nat) s : sumn [seq f i | i <- s] = \sum_(i <- s) f i.
proof: Proof. by elim: s => [|i s IHs] in f *; rewrite ?(big_nil, big_cons) //= IHs. Qed.
imports: mathcomp(ssreflect ssrbool ssrnat eqtype ssrfun seq) mathcomp(choice path finset finfun fintype bigop tuple) finmap