To show:

A = (A ∩ B) ∪ (A – B)

Let x ∈ A We have to show that

x ∈ (A ∩ B) ∪ (A – B)

Case I x ∈ A ∩ B

Then,

x ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B)

Case II

x ∉ A ∩ B ⇒ x ∉ A or x ∉ B

∴ x ∉ B [x ∉ A]

∴ x ∉ A – B ⊂ (A ∪ B) ∪ (A – B)

∴ A ⊂ (A ∩ B) ∪ (A – B) … (1)

It is clear that A ∩ B ⊂ A and (A – B) ⊂ A

∴ (A ∩ B) ∪ (A – B) ⊂ A … (2)

From (1) and (2), we obtain

A = (A ∩ B) ∪ (A – B) To prove:

A ∪ (B – A) ⊂ A ∪ B Let x ∈ A ∪ (B – A) 8

⇒ x ∈ A or x ∈ (B – A)

⇒ x ∈ A or (x ∈ B and x ∉ A)

⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∉ A)

⇒ x ∈ (A ∪ B)

∴ A ∪ (B – A) ⊂ (A ∪ B) … (3)

Next,

we show that

(A ∪ B) ⊂ A ∪ (B – A).

Let

y ∈ A ∪ B

⇒ y ∈ A or y ∈ B

⇒ (y ∈ A or y ∈ B) and (y ∈ A or y ∉ A)

⇒ y ∈ A or (y ∈ B and y ∉ A)

⇒ y ∈ A ∪ (B – A)

∴ A ∪ B ⊂ A ∪ (B – A) … (4)

Hence, from (3) and (4), we obtain

A ∪ (B – A) = A ∪B.