proposition: Lemma subsets_disjoint A B : (A \subset B) = [disjoint A & ~: B].
proof: Proof. by rewrite subset_disjoint; apply: eq_disjoint_r => x; rewrite !inE. Qed.
imports: HB(structures) mathcomp(ssreflect ssrbool ssrfun eqtype ssrnat div seq) mathcomp(choice fintype finfun bigop)