# Laws of Algebra of Sets

If A, B and C are three non-empty sets, then

(i) Idempotent law

1. A $\cup$ A = A
2. A $\cap$ A = A
ii) Identity law

1. A $\cup \phi$ = A
2. A $\cap$ U = A

(iii) Commutative law

1. A $\cup$ B = B $\cup$ A
2. A $\cap$ B = B $\cap$ A
(iv) Associative law
1. (A $\cup$ B) $\cup$ C = A $\cup$ (B $\cup$ C)
2. A $\cap$ (B $\cap$ C) = (A $\cap$ B) $\cap$ C
(v) Distributive law
1. (A $\cup$ B) $\cup$ C = (A $\cup$ B) $\cap$ (A $\cup$ C)
2. A $\cap$ (B $\cup$ C) = (A $\cap$ B) $\cup$ (A $\cap$ C)
(vi) De-morgan’s law
1. (A $\cup$ B)' = A' $\cap$ B'
2. (A $\cap$ B)' = A' $\cup$ B'
3. A - (B $\cup$ C) = (A - B) $\cap$ (A - C)
4. A - (B $\cap$ C) = (A - B) $\cup$ (A - C)
Sample Problem 1

If A and B are two sets, then A $\cap$ (A $\cup$B)' is equal to
(a) A
(b) B
(c) $\phi$
(d) None of these

A $\cap$ (A $\cup$B)' = A $\cap$ (A' $\cap$ B'), by de-Morgan’s law

= (A $\cap$ A') $\cap$ B', by associative law

= $\phi \cap$ B' = $\phi$

Sample Problem 2
If A and B are non-empty sets, then (A $\cap$ B) $\cup$ (A - B) is equal to
(a) B
(b) A
(c) A'
(d) B'

(A $\cap$ B) $\cup$ (A - B) = (A $\cap$ B) $\cup$ (A $\cap$ B')

because A - B = A $\cap$ B'

(A $\cap$ B) $\cup$ (A - B) = A $\cap$ (B $\cup$ B')
= A $\cap$ U
= A

Sample Problem 3
The set (A $\cup$ B $\cup$ C) $\cap$ (A $\cap$ B' $\cap$ C')' $\cap$ C' is equal to
(a) B $\cap$ C'
(b) B' $\cap$ C'
(c) A $\cap$ C
(d) None of these

(A $\cup$ B $\cup$ C) $\cap$ (A $\cap$ B' $\cap$ C')' $\cap$ C'
= (A $\cup$ B $\cup$ C) $\cap$ (A' $\cup$ B $\cup$ C) $\cap$ C'
= ($\phi$ $\cup$ B $\cup$ C) $\cap$ C'
= (B $\cup$ C) $\cap$ C'
= (B $\cap$ C') $\cup$ $\phi$ = B $\cap$ C'

Sample Problem 4

Consider the following relations
I. A - B = A - (A $\cap$ B)
II. A = (A $\cap$ B) $\cup$ (A - B)
III. A - (B $\cup$ C) = (A - B) $\cup$ (A - C)
Which of these is/are correct?
(a) I and III
(b) Only II
(c) II and III
(d) I and II

Let us consider the sets A = {1, 2, 4}, B = {2, 5, 6} and C = {1, 5, 7}

I. A - B = A - (A $\cap$ B) = {1, 4}and
A - (A $\cap$ B) = {1, 2, 4} - {2} = {1, 4}

So, A - B = A - (A $\cap$ B)

II. A = (A $\cap$ B) $\cup$ (A - B) = {2} $\cup$ {1, 4} = {1, 2, 4} = A

III. A - (B $\cup$ C) = {1, 2, 4} - {1, 2, 5, 6, 7} = {4} and
(A - B) $\cup$ (A - C) = {1, 4} $\cup$ {2, 4} = {1, 2, 4}

So, A - (B $\cup$ C) $\neq$ (A - B) $\cup$ (A - C)