proposition: Lemma big_mkcondr I r (P Q : pred I) F : \big[*%M/1]_(i <- r | P i && Q i) F i = \big[*%M/1]_(i <- r | P i) (if Q i then F i else 1).
proof: Proof. by rewrite -big_filter_cond big_mkcond big_filter. Qed.
imports: HB(structures) mathcomp(ssreflect ssrbool ssrfun eqtype ssrnat seq path) mathcomp(div fintype tuple finfun)