Proof. (1). Let a ∈ B∩Q. Since T is intra-regular, there exist s, t, x, y, z ∈
T such that a ≤ xsazaty. Thus a ≤ xsazaty ≤ xsaz(xsazaty)ty =
xs(azxsa)zatyty ∈ T T(BTTTB)T QT T T T ⊆ T T BT QT T. Therefore, B ∩
Q ⊆ (T T BT QT T].
Conversely, let a ∈ T. Then a ∈ B(a) and a ∈ Q(a), we have a ∈ B(a) ∩
Q(a). By hypothesis,