proposition: Lemma prodf_seq_eq0 I r (P : pred I) (F : I -> R) : (\prod_(i <- r | P i) F i == 0) = has (fun i => P i && (F i == 0)) r.
proof: Proof. by rewrite (big_morph _ mulf_eq0 (oner_eq0 _)) big_has_cond. Qed.
imports: HB(structures) mathcomp(ssreflect ssrfun ssrbool eqtype ssrnat div seq) mathcomp(choice fintype finfun bigop prime binomial)