by using properties prove that

(i) a U b = a intersection b imlies a=b

(ii) for any sets a and b show that p(a intersection b) = p(a) itersection p (b)

(iii) if a,b And c be the sets such that aUb = aUc and a intersection b = a itersection c show that b = c

(iv) a intersection (aUb) = a and aU(a intersection b) = a

**(i)**

** To prove:** P (A ∪ B) = A ∩ B ⇔ A = B

**Proof: **Firstly, Let A = B Then,

A ∪ B = A and A ∩ B = A

⇒ A ∪ B = A ∩ B

Then, A = B ⇒ (A ∪ B) = (A ∩ B) ... (1)

Conversely, let A ∪ B = A ∩ B. Then we have to prove that

A = B For this, let

From (2) and (3), we get A = B

Thus, A ∪ B = A ∩ B ⇒ A = B ... (4)

From (1) and (4), we have

A ∪ B = A ∩ B ⇔ A = B

**(ii)**

In order to prove that P (A ∩ B) = P (A) ∩ P (B), it is sufficient to prove that P (A ∩ B) ⊂ P (A) ∩ P (B) and P (A) ∩ P (B) ⊂ P (A ∩ B)

First let X ∈ P (A ∩ B)

⇒ X ⊂ A ∩ B

⇒ X ⊂ A and X ⊂ B

⇒ X ∈ P (A) and X ∈ P (B)

⇒ X ∈ P (A) ∩ P (B)

∴ P (A ∩ B) ⊂ P (A) ∩ P (B) ... (1)

Now, let

*y* ∈ P (A) ∩ P (B). Then,

*y* ∈ P (A) ∩ P (B)

⇒ *y* ∈ P (A) and *y* ∈ P (B)

⇒ *y* ⊂ A and *y* ⊂ B

⇒ *y* ⊂ A ∩ B

⇒ *y* ∈ P (A ∩ B)

∴ P (A) ∩ P (B) ⊂ P (A ∩ B) ... (2)

From (1) and (2), we get

P (A ∩ B) = P (A) ∩ P (B)

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