proposition: Lemma diffxI x y z : x `\` (y `&` z) = (x `\` y) `|` (x `\` z).
proof: Proof. apply/eqP; rewrite eq_diff leUx !leBx //= joinIl joinA joinCA !diffKU. rewrite joinCA -joinA [_ `|` x]joinC ![x `|` _]join_l //. by rewrite -joinIl leUr /= meetUl {1}[_ `&` z]meetC ?meetBI joinx0. Qed.
imports: HB(structures) mathcomp(ssreflect ssrfun ssrbool eqtype ssrnat choice seq) mathcomp(path fintype tuple bigop finset div prime finfun) mathcomp(finset)