proposition: Lemma bigcup_disjoint U P F : (forall i, P i -> [disjoint U & F i]) -> [disjoint U & \bigcup_(i | P i) F i].
proof: Proof. by move/bigcup_disjointP. Qed.
imports: HB(structures) mathcomp(ssreflect ssrbool ssrfun eqtype ssrnat div seq) mathcomp(choice fintype finfun bigop)