proposition: Lemma prod_nat_seq_eq0 I r (P : pred I) F : (\prod_(i <- r | P i) F i == 0)%N = has (fun i => P i && (F i == 0%N)) r.
proof: Proof. by rewrite (big_morph _ (id1 := false) muln_eq0)// big_has_cond. Qed.
imports: HB(structures) mathcomp(ssreflect ssrbool ssrfun eqtype ssrnat seq path) mathcomp(div fintype tuple finfun)