(i) A ∩ (A’ ∪ B) = A ∩ B

Consider LHS A ∩ (A’ ∪ B)

On expanding

(A ∩ A’) ∪ (A ∩ B)

As we know, (A ∩ A’) =ϕ

⇒ ϕ ∪ (A∩ B)

⇒ (A ∩ B)

∴ LHS = RHS

Thus proved.

(ii) A – (A – B) = A ∩ B

For the any sets A and B we have De-Morgan’s law

(A ∪ B)’ = A’ ∩ B’, (A ∩ B) ‘ = A’ ∪ B’

Considering LHS

= A – (A–B)

= A ∩ (A–B)’

= A ∩ (A∩B’)’

= A ∩ (A’ ∪ B’)’) (Here, (B’)’ = B)

= A ∩ (A’ ∪ B)

= (A ∩ A’) ∪ (A ∩ B)

= ϕ ∪ (A ∩ B) (Here, A ∩ A’ = ϕ)

= (A ∩ B) (Here, ϕ ∪ x = x, for any set)

= RHS

∴ LHS=RHS

Thus proved.

(iii) A ∩ (A ∪ B’) = ϕ

Consider LHS A ∩ (A ∪ B’)

= A ∩ (A ∪ B’)

= A ∩ (A’∩ B’) (By De–Morgan’s law)

= (A ∩ A’) ∩ B’ (here, A ∩ A’ = ϕ)

= ϕ ∩ B’

= ϕ (since, ϕ ∩ B’ = ϕ)

= RHS

∴ LHS=RHS

Thus proved.

(iv) A – B = A Δ (A ∩ B)

Consider RHS A Δ (A ∩ B)

A Δ (A ∩ B) (here, E Δ F = (E–F) ∪ (F–E))

= (A – (A ∩ B)) ∪ (A ∩ B –A) (here, E – F = E ∩ F’)

= (A ∩ (A ∩ B)’) ∪ (A ∩ B ∩ A’)

= (A ∩ (A’ ∪ B’)) ∪ (A ∩ A’ ∩ B) (by using De-Morgan’s law and associative law)

= (A ∩ A’) ∪ (A ∩ B’) ∪ (ϕ ∩ B) (by using distributive law)

= ϕ ∪ (A ∩ B’) ∪ ϕ

= A ∩ B’ (here, A ∩ B’ = A–B)

= A – B

= LHS

∴ LHS=RHS

Thus proved.