proposition: Lemma big_hasC : forall r (P : pred I) F, ~~ has P r -> \big[op/idx]_(i <- r | P i) F i === idx.
proof: Proof. move=> r P F; rewrite <- big_filter; rewrite has_count count_filter. case: filter => // _; exact: big_nil. Qed.
imports: ssreflect ssrbool ssrfun eqtype ssrnat seq div choice fintype finfun paths Setoid Morphisms OperationClasses bigops