As A ⊂ B the set A is inside set B

Hence A ∪ B = B

Taking compliment

⇒ (A ∪ B)’ = B’

Using De-Morgan’s law (A ∪ B)’ = A’ ∩ B’

⇒ A’ ∩ B’ = B’ …(i)

Now we know that

B’ = (B’ – A’) + (A’ ∩ B’)

Using (i)

⇒ B’ = (B’ – A’) + B’

⇒ (B’ – A’) = B’ – B’

⇒ (B’ – A’) = 0

⇒ (B’ – A’) = {ϕ}

Hence proved