show that for any sets A and B A=(A intersection B)U (A-B) and A U (B-A)=(AUB)
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A U (B-A)
=A U (B n A')
=(A U B ) n (A U A')
=(A U B)n
=A U (B n A')
=(A U B ) n (A U A')
=(A U B)n
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Use relation, A – B = A ∩ B’
⇒ (A ∩ B) ∪ (A ∩ B’)
⇒A ∩ (B ∪ B’)
Use relation B ∪ B’ = U
⇒A ∩ U
⇒ A
LHS
Hence A = (A ∩ B) ∪ (A - B)
(2)A ∪ (B - A) = (A ∪ B)
LHS
A ∪ (B - A)
Use relation, B – A = B ∩ A’
⇒ A ∪ (B ∩ A’)
⇒ (A ∪ B) ∩ (A ∪ A’)
Use relation A ∪ A’ = U.
⇒ (A ∪ B) ∩ U
⇒ (A ∪ B)
RHS
Hence, A ∪ (B - A) = (A ∪ B)
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