補充一(錯誤)
heron0520 寫到:Let A and B be two subsets of E. It there exists a subset C of E such that...(略)
It there exists...似乎該寫作If there exists...
補充二(其他方法)
設P,Q是2個任一集合,E是宇集合,則
P=P∩E=P∩(Q∪Q')=(P∩Q)∪(P∩Q')
P=P∪∅=P∪(Q∩Q')=(P∪Q)∩(P∪Q')
證明:
A=A
由A配出A∪C
A=(A∪C)∩(A∪C')
由A配出A∩C
A=(A∪C)∩(((A∩C)∪(A∩C'))∪C'),使用結合律
A=(A∪C)∩((A∩C)∪((A∩C')∪C'))
由C'配出A∩C'
A=(A∪C)∩((A∩C)∪((A∩C')∪((C'∩A)∪(C'∩A')))),使用結合律
A=(A∪C)∩((A∩C)∪(((A∩C')∪(C'∩A))∪(C'∩A'))),使用交換律
A=(A∪C)∩((A∩C)∪(((C'∩A)∪(C'∩A))∪(C'∩A')))
A=(A∪C)∩((A∩C)∪((C'∩A)∪(C'∩A')))
A=(A∪C)∩((A∩C)∪C')
同理
B=(B∪C)∩((B∩C)∪C')
∵B∩C=A∩C,B∪C=A∪C
∴(A∪C)∩((A∩C)∪C')=(B∪C)∩((B∩C)∪C')
∴A=B