Therefore, B=AUB-A intersection (cub) = {0 0,2,4,6,8,9, 10}

A crosses B={3}, so the * * * of AB is only 3. Besides 3, A also has B. Because CuB crosses A={9}, A has 9 and B does not. If there is any one of 157 in A, such as 5, then there is no one in B and there is 5 in CuB. The intersection of CuB and a is not only 9, so a = {3,9}.