If A = {1, 3, 5, 7, 9, 11, 13, 15, 17} B = {2, 4, ... , 18} and N the set of natural numbers is the universal set, then A′ ∪ (A ∪ B) ∩ B′ is

Given: A = {1, 3, 5, 7, 9, 11, 13, 15, 17} B = {2, 4, ... , 18}

To find: A′ ∪ (A ∪ B) ∩ B′

A′ ∪ (A ∪ B) ∩ B′

= A′ ∪ [(B’ ∩ A) ∪ (B’ ∩ B)]

{∵ Distributive property of set:

(A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C)}

= A′ ∪ [(A ∩ B’) ∪ Φ]

{∵ (B’ ∩ B) = Φ}

= A′ ∪ (A ∩ B’)

= (A’ ∪ A) ∩ (A’ ∪ B’)

{∵ Distributive property of set:

(A ∪ B) ∩ (A ∪ C) = A ∪ (B ∩ C)}

= Φ ∩ (A’ ∪ B’)

{∵ (A’ ∩ A) = Φ}

= (A’ ∪ B’)

= (A ∩ B)’

{∵ (A’ ∪ B’) = (A ∩ B)’}

A′ ∪ (A ∪ B) ∩ B′ = (A ∩ B)’

A contains all odd numbers and B contains all even numbers

Hence, A ∩ B = Φ

⇒ A′ ∪ (A ∪ B) ∩ B′ = {Φ}’

⇒ A′ ∪ (A ∪ B) ∩ B′ = N