proposition: Lemma fdisjointU1X x A B : [disjoint x |` A & B]%fset = (x \notin B) && [disjoint A & B]%fset.
proof: Proof. by rewrite fdisjointUX fdisjoint1X. Qed.
imports: HB(structures) mathcomp(ssreflect ssrbool ssrnat eqtype ssrfun seq) mathcomp(choice path finset finfun fintype bigop)