To Prove: A = (A ∩ B) ∪ (A – B)

Proof: Let X ϵ A

Now, we need to show that X ϵ (A ∩ B) ∪ (A – B)

In Case I,

X ϵ (A∩ B)

⇒ X ϵ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B)

In Case II,

X ∉A ∩ B

⇒ X ∉ B or X ∉ A

⇒ X ∉ B (X ∉ A)

⇒ X ∉ A – B ⊂ (A ∪ B) ∪ (A – B)

∴A ⊂ (A ∩ B) ∪ (A – B) (i)

It can be concluded that, A ∩ B ⊂ A and (A – B) ⊂ A

Thus, (A ∩ B) ∪ (A – B) ⊂ A (ii)

Equating (i) and (ii),

A = (A ∩ B) ∪ (A – B)

Now, we need to show, A ∪ (B – A) ⊂ A ∪ B

Let us assume that,

X ϵ A ∪ (B – A)

X ϵ A or X ϵ (B – A)

⇒ X ϵ A or (X ϵ B and X ∉A)

⇒ (X ϵ A or X ϵ B) and (X ϵ A and X ∉A)

⇒ X ϵ (B ∪A)

∴ A ∪ (B – A) ⊂ (A ∪ B) (iii)

Now, to prove: (A ∪ B) ⊂ A ∪(B – A)

Let y ϵ A∪B

yϵ A or y ϵ B

(y ϵ A or y ϵ B) and (X ϵ A and X ∉A)

⇒ y ϵ A or (y ϵ B and y ∉A)

⇒ y ϵ A ∪ (B – A)

Thus, A ∪ B ⊂ A ∪ (B – A) (iv)

∴Using (iii) and (iv), we get:

A ∪ (B – A) = A ∪ B