# 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 – (AB) = A(A B) (since A – B = AB)

= 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?

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