proposition: Lemma tuto_disjoint_sym : forall A B, [disjoint A & B] = [disjoint B & A].
proof: Proof. by move=> A B; congr eqn; apply: eq_card => x; apply: andbC. Qed.
imports: HB(structures) mathcomp(ssreflect ssrfun ssrbool eqtype ssrnat seq path) mathcomp(choice fintype tuple finset)