(i) A=(A∩B)∪(A−B)
Consider RHS=(A∩B)∪(A−B)
=(A∩B)∪(A∩B′) (by def of difference of sets, A−B=A∩B′)
=A∩(B∪B′) (by distributive )
=A∩U (∵A∪A′=U)
=A
=LHS
Hence, A=(A∩B)∪(A−B)
(ii) A∪(B−A)=A∪B
Consider, A∪(B−A)
=A∪(B∩A′) (by def of difference of sets, A−B=A∩B′)
=(A∪B)∩(A∪A′) (by distributive property)
=(A∪B)∩U (∵A∪A′=U)
=A∪B