proposition: Lemma fdisjointX1 x A : [disjoint A & [fset x]] = (x \notin A).
proof: Proof. by rewrite fdisjoint_sym fdisjoint1X. Qed.
imports: HB(structures) mathcomp(ssreflect ssrbool ssrnat eqtype ssrfun seq) mathcomp(choice path finset finfun fintype bigop)