proposition: Lemma disjointFl A B x : [disjoint A & B] -> x \in B -> x \in A = false.
proof: Proof. rewrite disjoint_sym; exact: disjointFr. Qed.
imports: HB(structures) mathcomp(ssreflect ssrfun ssrbool eqtype ssrnat seq choice) mathcomp(path div)