Prove that (A ∪ B)’=A’ ∩B’

Proof: Assume a to be an arbitrary element that belongs to (A U B)’

⇒ a ∈ (A U B)’

⇒ a ∉ (A U B) [Because an element belonging to the complement of a set cannot belong to the set]

⇒ a ∉ A an a ∉ B

⇒ a ∈ A’ and a ∈ B’ [Using complement of a set definition]

⇒ a ∈ A’ ∩ B’

⇒ (A U B)’ ⊆ A’ ∩ B’ — (1)

Next, let us assume b to be an arbitrary element in A’ ∩ B’

⇒ b ∈ A’ ∩ B’

⇒ b ∈ A’ and b ∈ B’

⇒ b ∉ A and b ∉ B [Because an element belonging to the complement of set cannot belong to the set]

⇒ b ∉ A U B

⇒ b ∈ (A U B)’

⇒ A’ ∩ B’ ⊆ (A U B)’ — (2)

From (1) and (2), we get (A U B)’ = A’ ∩ B’. Hence, we can say that A Union B Complement is equal to the intersection of the complements of the two sets A and B.