proposition: Lemma disjoints1 A x : [disjoint [set x] & A] = (x \notin A).
proof: Proof. by rewrite (@eq_disjoint1 _ x) // => y; rewrite !inE. Qed.
imports: HB(structures) mathcomp(ssreflect ssrbool ssrfun eqtype ssrnat div seq) mathcomp(choice fintype finfun bigop)