proposition: Lemma prodf_seq_neq0 I r (P : pred I) (F : I -> R) : (\prod_(i <- r | P i) F i != 0) = all (fun i => P i ==> (F i != 0)) r.
proof: Proof. rewrite prodf_seq_eq0 -all_predC; apply: eq_all => i /=. by rewrite implybE negb_and. Qed.
imports: HB(structures) mathcomp(ssreflect ssrfun ssrbool eqtype ssrnat div seq) mathcomp(choice fintype finfun bigop prime binomial)