For any three sets A , B , C prove that A - ( B ∩ C ) = ( A - B ) ∪ ( A - C )

Let x be any element of A - ( B ∩ C ) . Then, x ∈ A - ( B ∩ C ) ⇒ x ∈ A and x ∉ ( B ∩ C ) ⇒ x ∈ A and ( x ∉ B or x ∉ C ) ⇒ ( x ∈ A and x ∉ B ) or ( x ∈ A and x ∉ C ) ⇒ x ∈ ( A - B ) or x ∈ ( A - C ) ⇒ x ∈ ( A - B ) ∪ ( A - C ) ∴ A - ( B ∩ C ) ⊆ ( A - B ) ∪ ( A - C ) Similarly ( A - B ) ∪ ( A - C ) ⊆ A - ( B ∩ C ) Hence, A - ( B ∩ C ) = ( A - B ) ∪ ( A - C )

For any three sets A , B , C prove that A - ( B ∩ C ) = ( A - B ) ∪ ( A - C )

Let x be any element of A - ( B ∩ C ) . Then, x ∈ A - ( B ∩ C ) ⇒ x ∈ A and x ∉ ( B ∩ C ) ⇒ x ∈ A and ( x ∉ B or x ∉ C ) ⇒ ( x ∈ A and x ∉ B ) or ( x ∈ A and x ∉ C ) ⇒ x ∈ ( A - B ) or x ∈ ( A - C ) ⇒ x ∈ ( A - B ) ∪ ( A - C ) ∴ A - ( B ∩ C ) ⊆ ( A - B ) ∪ ( A - C ) Similarly ( A - B ) ∪ ( A - C ) ⊆ A - ( B ∩ C ) Hence, A - ( B ∩ C ) = ( A - B ) ∪ ( A - C )

If A = { 4 , 5 , 7 , 8 , 10 } , B = { 4 , 5 , 9 } and C = { 1 , 4 , 6 , 9 } , verify that ( A ∩ B ) ∩ C = A ∩ ( B ∩ C )

Given A = { 4 , 5 , 7 , 8 , 10 } , B = { 4 , 5 , 9 } and C = { 1 , 4 , 6 , 9 } ( A ∩ B ) ∩ C = A ∩ ( B ∩ C ) L H S = ( A ∩ B ) ∩ C = ( { 4 , 5 , 7 , 8 , 10 } ∩ { 4 , 5 , 9 } ) ∩ { 1 , 4 , 6 , 9 } = { 4 , 5 } ∩ { 1 , 4 , 6 , 9 } = { 4 } R H S = A ∩ ( B ∩ C ) = { 4 , 5 , 7 , 8 , 10 } ∩ ( { 4 , 5 , 9 } ∩ { 1 , 4 , 6 , 9 } ) = { 4 , 5 , 7 , 8 , 10 } ∩ { 4 , 9 } = { 4 } Hence, L H S = R H S

If A = { 4 , 5 , 7 , 8 , 10 } , B = { 4 , 5 , 9 } and C = { 1 , 4 , 6 , 9 } , verified that A ∪ ( B ∩ C ) = ( A ∪ B ) ∩ ( A ∪ C ) .

To verify, A ∪ ( B ∩ C ) = ( A ∪ B ) ∩ ( A ∪ C ) L H S = A ∪ ( B ∩ C ) = { 4 , 5 , 7 , 8 , 10 } ∪ { 4 , 5 , 9 } ∩ { 1 , 4 , 6 , 9 } = { 4 , 5 , 7 , 8 , 10 } ∪ { 4 , 9 } = { 4 , 5 , 7 , 8 , 9 , 10 } R H S = A ∪ B ∩ ( A ∪ C ) = { 4 , 5 , 7 , 8 , 10 } ∪ { 4 , 5 , 9 } ∩ { 4 , 5 , 7 , 8 , 10 } ∪ { 1 , 4 , 6 , 9 } = { 4 , 5 , 7 , 8 , 9 , 10 } ∩ { 1 , 4 , 5 , 6 , 7 , 8 , 9 , 10 } = { 4 , 5 , 7 , 8 , 9 , 10 } Hence, RHS = LHS

For any two set A and B , prove that ( A ∪ B ) ' = A ' ∩ B ' .

Let x be an arbitrary element of ( A ∪ B ) ' . Then, x ∈ ( A ∪ B ) ' ⇒ x ∉ ( A ∪ B ) ⇒ x ∉ A and x ∉ B ⇒ x ∈ A ' and x ∈ B ' ⇒ x ∈ A ' ∩ B ' ∴ ( A ∪ B ) ' ⊆ A ' ∩ B ' Again, let y be an arbitrary element of A ' ∩ B ' . Then y ∈ A ' ∩ B ' ⇒ y ∈ A ' and y ∈ B ' ⇒ y ∉ A and y ∉ B ⇒ y ∉ A ∪ B ⇒ y ∈ ( A ∪ B ) ' ∴ A ' ∩ B ' ⊆ ( A ∪ B ) ' Hence, ( A ∪ B ) ' = A ' ∩ B '