2|10Ax x 10,9,8,7,5,3,10,9,8,7,6BABABABABA:首先证明,BAx有,BAx对,BxAx 或故,BxAx或即ABAB因此有BAx因此有BABA:然后证明,BxAx 或有,BAx对,BxAx或故BAx即,BAx因此有BABA所以BABAUUAAUAA AAABABABAB;()()ABABABABBACBACBA)()(:左证明)()()(分配律CBCA)()(CBCA1111nniiiinniiiiAAAA ZermeloA.FraenkelJohn von Nenmann,P.Bernays,Godel;121,112.(1).nniijijkii ji j knnAAAAAAAAAAAA 1212.nnAAAUAAA150ABCNABCABCABCABACBCABCABC 祝您成功!