proposition: Lemma fdisjointUX A B C : [disjoint A `|` B & C] = [disjoint A & C]%fset && [disjoint B & C]%fset.
proof: Proof. by rewrite -!fsetI_eq0 fsetIUl fsetU_eq0. Qed.
imports: HB(structures) mathcomp(ssreflect ssrbool ssrnat eqtype ssrfun seq) mathcomp(choice path finset finfun fintype bigop)