proposition: Lemma big_hasC r (P : pred I) F : ~~ has P r -> \big[op/idx]_(i <- r | P i) F i = idx.
proof: Proof. by rewrite -big_filter has_count -size_filter -eqn0Ngt unlock => /nilP->. Qed.
imports: HB(structures) mathcomp(ssreflect ssrbool ssrfun eqtype ssrnat seq path) mathcomp(div fintype tuple finfun)