proposition: Lemma sum_nat_seq_neq0 I r (P : pred I) F : (\sum_(i <- r | P i) F i != 0)%N = has (fun i => P i && (F i != 0)%N) r.
proof: Proof. by rewrite sum_nat_seq_eq0// -has_predC; apply: eq_has => x /=; case Px: (P x). Qed.
imports: HB(structures) mathcomp(ssreflect ssrbool ssrfun eqtype ssrnat seq path) mathcomp(div fintype tuple finfun)