For set A and B,
[B' ⋃ (B' – A')]'
= (B')' ⋂ (B' – A')' [By De-Morgan's Law]
= (B')' ⋂ (B' ⋂ (A')')'
= (B')' ⋂ (B' ⋂ A)'
= (B')' ⋂ (B ⋃ A') [By De-Morgan's Law]
= B ⋂ (B ⋃ A')
= B ⋂ (B ⋃ A')
= (B ⋂ B) ⋃ (B ⋂ A')
= B ⋃ (B ⋂ A')
[Since B ⋂ A' ⊆ B]
= B
i.e [B' ⋃ (B' – A')]' = B