proposition: Lemma diffxB x y z : x `\` (y `\` z) = (x `\` y) `|` (x `&` z).
proof: Proof. rewrite -[y in RHS](joinIB z) diffxU joinIl diffxI -joinA joinBI join_r //. by rewrite joinBx meetKU meetA meetAC diffIK meet0x joinx0 meet_r. Qed.
imports: HB(structures) mathcomp(ssreflect ssrfun ssrbool eqtype ssrnat choice seq) mathcomp(path fintype tuple bigop finset div prime finfun) mathcomp(finset)