For any two sets A and B , prove that following: A ∩ ( A ∪ B ) ' = ϕ

L H S = A ∩ ( A ∪ B ) ' = A ∩ A ' ∩ B ' (Using De-morgan's law) = ( A ∩ A ' ) ∩ B ' = ϕ ∩ B ' ∵ A ∩ A ' = ϕ = ϕ = R H S ∴ L H S = R H S Hence Proved

If A , B , C are three sets such that A ⊂ B , then prove that C – B ⊂ C – A .

Given A ⊂ B To prove: C - B ⊂ C - A L e t a ∈ C - B ⇒ a ∈ C and a ∉ B ⇒ a ∈ C and a ∉ A ∵ A ⊂ B ⇒ a ∈ C - A Thus, a ∈ C - B ⇒ a ∈ C - A This is true for all x ∈ C - B Hence C - B ⊂ C - A

For any two sets A and B , prove that ( A ∪ B ) - B = A - B .

To prove ( A ∪ B ) - B = A - B L H S = ( A ∪ B ) - B = ( A ∪ B ) ∩ B ' Using the result X - Y = X ∩ Y ' = A ∩ B ' ∪ B ∩ B ' (using distributive property) = A ∩ B ' ∪ ϕ = A ∩ B ' Use the above result X - Y = X ∩ Y ' = A - B = R H S L H S = R H S Hence proved.

For any two sets A and B , prove that A - ( A ∩ B ) = A - B

L . H . S . = A - ( A ∩ B ) = A ∩ ( A ∩ B ) ' (Using the result X - Y = X ∩ Y ' ) = A ∩ A ' ∪ B ' (Using De-morgan's law) = A ∩ A ' ∪ A ∩ B ' ( ∵ ∩ is distributed over ∪ ) = ϕ ∪ A ∩ B ' = A ∩ B ' Again using the result X - Y = X ∩ Y ' = A - B = R . H . S . Hence proved.

For any two sets A and B , prove that A - ( A - B ) = A ∩ B

We need to prove A - ( A - B ) = A ∩ B L H S = A - ( A - B ) Using the Result X - Y = X ∩ Y ' ......(1) = A ∩ A - B ' = A ∩ ( A ∩ B ' ) ' using (1) = A ∩ ( A ' ∪ B ' ' ) (using De-morgan's law) = A ∩ A ' ∪ B (using law of double complementation) = A ∩ A ' ∪ A ∩ B ( ∵ ∩ is distributed over ∪ ) = ϕ ∪ ( A ∩ B ) = A ∩ B = R H S ∴ A - ( A - B ) = A ∩ B Hence proved