Laws of sets | Introduction to SETs, Part IV

SOME USEFUL AND IMPORTANT LAWS
Continuation of previous posts

1. Introduction to SETs, Part1 
2. Introduction to SETs, Part II
3. Introduction to SETs, Part III

Topics covered in this post are

1. Idempotent Laws 
2. Identity Laws 
3. Commutative Laws 
4. Associative Laws 
5. Distributive Laws 
6. De-Morgan’s Laws 
7. Verify De-Morgan’s Laws
8. Verify Distributive Laws

Some commonly used laws of sets are,

1. Idempotent Laws

For any set A
(i) A ∪ A = A
(ii) A ∩ A = A

2. Identity Laws

For any subset A of the universal set U
A ∪ Ø = A, A ∩ U = A, where Ø is empty set

3. Commutative Laws

For any two sets A and B
(i) A ∪ B = B ∪ A
(ii) A ∩ B = B ∩ A

4. Associative Laws

If A, B, C are any three sets then
(i) (A ∪ B) ∪ C = A ∪ (B ∪ C)
(ii) (A ∩ B) ∩ C = A ∩ (B ∩ C)

5. Distributive Laws

If A, B, C are any three sets then
(i) A ∪ (B ∩ C) = (A ∪ B)∩ (A ∪ C)
(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

6. De-Morgan’s Laws

For any two sets A and B
(i) (A ∪ B)’ = A’ ∩ B’
(ii) (A ∩ B)’ = A’ ∪ B’

Example: Now let’s verify (a) De-Morgan’s Laws and (b) Distributive Laws with the sets, A = {1, 3, 5}, B = {3, 5, 7, 9}, C = {2, 6, 8, 9} are subsets of universal set U = {1, 2, 3, 5, 6, 7, 8, 9}.

(a) De-Morgan’s laws state that
(i) (A ∪ B)’ = A’ ∩ B’
(ii) (A ∩ B)’ = A’ ∪ B’

To verify (i)

A ∪ B = {1, 3, 5, 7, 9}
\ (A ∪ B)’ = {2, 6, 8}

A’ = {2, 6, 7, 8, 9}
B’ = {1, 2, 6, 8}
\ A’ ∩ B’ = {2, 6, 8}

Hence verified that, (A ∪ B)’ = A’ ∩ B’

To verify (ii)

A ∩ B = {3, 5}
\ (A ∩ B)’ = {1, 2, 6, 7, 8, 9} and
A’ ∪ B’ = {1, 2, 6, 7, 8, 9}

Hence verified that, (A ∩ B)’ = A’ ∪ B’

(b) Distributive laws state that
(i) A ∪ (B ∩ C) = (A ∪ B)∩ (A ∪ C)
(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

To verify (i)

B ∩ C = {9}
\ A ∪ (B ∩ C) = {1, 3, 5, 9}

(A ∪ B) = {1, 3, 5, 7, 9}
(A ∪ C) = {1, 2, 3, 5, 6, 8, 9}
\ (A ∪ B)∩ (A ∪ C) = {1, 3, 5, 9}

Hence verified that, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

To verify (ii)

B ∪ C = {2, 3, 5, 6, 7, 8, 9}
\ A ∩ (B ∪ C) = {3, 5}

A ∩ B = {3, 5}
A ∩ C = {}
\ (A ∩ B)∪ (A ∩ C) = {3, 5}

Hence verified that, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)