For any two sets, prove that :
(i) A∪(A∩B)=A
(ii) A∩(A∪B)=A
(i) A∪(A∩B)=(A∪A)∩(A∪B)
[ ∴ union ∪ is distributive overintersection ∩]
= A∩(A∪B) [∴A∪A=A]
= A [∴A⊂(A∪B), as union of two sets is bigger than each of the individual sets]
Hence , A∪(A∩B)=A Proved.
(ii) A∩(A∪B)=(A∩A)∪(A∩B)
[∵∩ distributes over ∪]
= A∪(A∩B)[∵A∩A=A]
= A [using (i)]