# Laws of Algebra of Sets

# Laws of Algebra of Sets

Here we will learn about some of the laws of algebra of sets.

**1. Commutative Laws:**

For any two finite sets A and B;

(i) A U B = B U A

(ii) A ∩ B = B ∩ A

**2. Associative Laws:**

For any three finite sets A, B and C;

(i) (A U B) U C = A U (B U C)

(ii) (A ∩ B) ∩ C = A ∩ (B ∩ C)

Thus, union and intersection are associative.

**3. Idempotent Laws:**

For any finite set A;

(i) A U A = A

(ii) A ∩ A = A

**4. Distributive Laws:**

For any three finite sets A, B and C;

(i) A U (B ∩ C) = (A U B) ∩ (A U C)

(ii) A ∩ (B U C) = (A ∩ B) U (A ∩ C)

Thus, union and intersection are distributive over intersection and union respectively.

**5. De Morgan’s Laws:**

For any two finite sets A and B;

(i) A – (B U C) = (A – B) ∩ (A – C)

(ii) A - (B ∩ C) = (A – B) U (A – C)

De Morgan’s Laws can also we written as:

(i) (A U B)’ = A' ∩ B'

(ii) (A ∩ B)’ = A' U B'

**More laws of algebra of sets:**

**6.** For any two finite sets A and B;

(i) A – B = A ∩ B'

(ii) B – A = B ∩ A'

(iii) A – B = A ⇔ A ∩ B = ∅

(iv) (A – B) U B = A U B

(v) (A – B) ∩ B = ∅

(vi) A ⊆ B ⇔ B' ⊆ A'

(vii) (A – B) U (B – A) = (A U B) – (A ∩ B)

**7.** For any three finite sets A, B and C;

(i) A – (B ∩ C) = (A – B) U (A – C)

(ii) A – (B U C) = (A – B) ∩ (A – C)

(iii) A ∩ (B - C) = (A ∩ B) - (A ∩ C)

(iv) A ∩ (B △ C) = (A ∩ B) △ (A ∩ C)

**Example** Use the properties of sets to prove that for all the sets A and B

A – (A ∩ B) = A – B

**Solution** We have

A – (A ∩ B) = A ∩ (A ∩ B)′ (since A – B = A ∩ B′)

= A ∩ (A′ ∪ B′) [by De Morgan’s law)

= (A ∩ A′) ∪ (A ∩ B′) [by distributive law]

=φ ∪ (A ∩ B′)

= A ∩ B′ = A – B** **

**Example**

** **For all sets A, B and C

Is (A – B) ∩ (C – B) = (A ∩ C) – B?

Justify your answer.

**Solution **Yes

Let x ∈ (A – B) ∩ (C – B)

⇒ x ∈ A – B and x ∈ C – B

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

⇒ (x ∈ A and x ∈ C) and x ∉ B

⇒ (x ∈ A ∩ C) and x ∉ B

⇒ x ∈ (A ∩ C) – B

So (A – B) ∩ (C – B) ⊂ (A ∩ C) – B ... (1)

Now, conversely

Let y ∈ (A ∩ C) – B

⇒ y ∈ (A ∩ C) and y ∉ B

⇒ (y ∈ A and y ∈ C) and (y ∉ B)

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

⇒ y ∈ (A – B) and y ∈ (C – B)

⇒ y ∈ (A – B) ∩ (C – B)

So, (A ∩ C) – B ⊂ (A – B) ∩ (C – B)…. (2)

From equation (1) and (2) we get

(A – B) ∩ (C – B) = (A ∩ C) – B