We know that (A ∩ B) ⊂ A and (A – B) ⊂ A

⇒(A ∩ B) ∩ (A – B)⊂ A….(1)

Let and x ϵ (A ∩ B) ∩ (A – B)

⇒ x ϵ (A ∩ B) and x ϵ (A–B)

⇒ x ϵ A and x ϵ B and x ϵ A and x ∉ B

⇒ x ϵ A and x ϵ A [∵ x ϵ B and x ∉ B are not possible simultaneously]

→ x ϵ A

∴ (A ∩ B) ∩ (A – B)⊂ A…(2)

From (1) and (2), we get

A = (A ∩ B) ∩ (A – B)