proposition: Lemma subsetD A B C : (A \subset B :\: C) = (A \subset B) && [disjoint A & C].
proof: Proof. by rewrite setDE subsetI -disjoints_subset. Qed.
imports: HB(structures) mathcomp(ssreflect ssrbool ssrfun eqtype ssrnat div seq) mathcomp(choice fintype finfun bigop)