PROOF. Necessity. Let p =p5p, where 5 ESx. Let o = 6p, then o2 = andp =po-. It is known in [1] that L(of) and L(p) are complete lattices in which joins are unions and, moreover, L(of) is completely distributive. If A CX, then it is easy to show that {t(A) =4t(4(A)) and +(A) =x(+1(A)) where c(A)EL(o-), ql(A)EL(p) and X(A)EL(a). Define the mapping 0 of L(of) onto L(p) as follows: if q(A)EL(o-), then 6(c (A)) =+/(A). Clearly, 0 preserves set-inclusion order and is one-toone. Hence, L(of) is completely isomorphic with L(p). This proves that L(p) is completely distributive. Sufficiency. Let L(p) be a completely distributive complete lattice. Define the binary relation 5 as follows: (x, y) E, iff p(x, y)pCp. Obviously, papCp. For each zEX, define K, = {'( {v }): zEzI({v }) }. For any yCX, let K= {K,:zECip ({y})} and S(K) denote the set of mappings s of 41({y}) into L(p) such that for every zEt'({y}), s(z)EK,. Then V{AK,: zEt({y}) } =A { Vs(1'({y})): sES(K) }. Since lattice joins are unions, we have Vs(qJ({y}))Dqt({y}), for each sES(K), and hence A { Vs(V( {y } )): sES(K) } t1( {y } ). Therefore, U {AK,: z&P({y})} = V{AKz: zQI({y}) }y I ( {y})