Answer :
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)
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