LHS = A ∩ (A B)’

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

⇒ LHS = A ∩ (A’ ∩ B’)

⇒ LHS = (A ∩ A’) ∩ (A ∩ B’)

We know that A ∩ A’ = φ

⇒ LHS = φ ∩ (A ∩ B’)

We know that intersection of null set with any set is null set only

Let (A ∩ B’) be any set X hence

⇒ LHS = φ ∩ X

⇒ LHS = φ

⇒ LHS = RHS

Hence proved

Prove that A ∩ (A B)’ = φ

Prove that A ∩ (A B)’ = φ