Show that for any sets A and B, A = (A∩B)∪(A−B) and A∪(B−A) = (A∪B)

(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