(i) LHS = A ∪ B 

= {a, b, c, d, e}∪ {a, c, e, g} 

= { a, b, c, d, e, g} 

= {a, c, e, g}∪{a, b, c, d, e} 

= B ∪ A 

= RHS 

Hence proved. 

(ii) To prove: A ∪ C = C ∪ A 

Since the element of set C is not provided, 

let x be any element of C. 

LHS = A ∪ C 

= {a, b, c, d, e} ∪ { x |x∈C} 

= { a, b, c, d, e, x} 

= {x, a, b, c, d, e }

= { x |x∈C} ∪ { a, b, c, d, e} 

= C ∪ A 

= RHS

Hence proved. 

(iii) To prove: B ∪ C = C ∪ B 

Since the element of set C is not provided, 

let x be any element of C. 

LHS = B ∪ C 

= { a, c, e, g } ∪ { x |x∈C} 

= { a, c, e, g, x} 

= {x, a, c, e, g } 

= {x |x∈C} ∪ { a, c, e, g }

= C∪B 

= RHS 

Hence proved. 

(iv) LHS = A ∩ B 

= {a, b, c, d, e}∪ {a, c, e, g} 

= {a, c, e}

RHS = B ∩ A 

= {a, c, e, g}∩{a, b, c, d, e} 

= {a, c, e} 

∴ A ∩ B = B ∩ A 

(v) Let x be an element of B ∩ C 

x∈B ∩ C 

x∈B and x∈C

x∈C and x∈B [by definition of intersection] 

x∈C∩B 

B∩C \(\subset\)C ∩ B ….(i) 

Now let x be an element of C∩B 

Then, x∈C∩B 

x∈C and x∈B

x∈B and x∈C [by definition of intersection]

x∈B∩C

C∩B \(\subset\)B ∩ C ….(ii) 

From (i) and (ii) we have, 

B∩C = C∩B [ every set is a subset of itself] 

Hence proved. 

(vi) Let x be an element of A ∩ C 

x∈A∩C 

x∈A and x∈C

x∈C and x∈A [by definition of intersection]

x∈C∩A 

A∩C \(\subset\)C∩A ….(i) 

Now let x be an element of C ∩ A

 Then, x∈C∩A

x∈C and x∈A

x∈A and x∈C [by definition of intersection] 

x∈A ∩ C

C ∩ A\(\subset\) A ∩ C ….(ii) 

From (i) and (ii) we have, 

A ∩ C = C ∩ A [ every set is a subset of itself] 

Hence proved. 

(vii) Let x be any element of (A ∪ B) ∪ C 

x∈(A ∪ B) or x∈C

x∈A or x∈B or x∈C 

x∈A or x∈(B ∪ C) 

x∈A ∪ (B ∪ C) 

(A ∪ B) ∪ C\(\subset\) A ∪ (B ∪ C) …..(i)

Now, let x be an element of A ∪ (B ∪ C) 

Then, x∈ A or (B ∪ C)

x∈A or x∈B or x∈C

x∈(A ∪ B) or x∈C

x∈(A ∪ B) ∪ C 

A ∪ (B ∪ C) \(\subset\)(A ∪ B) ∪ C …..(ii) 

From i and ii, (A ∪ B) ∪ C = A ∪ (B ∪ C)

[ every set is a subset of itself]

Hence , proved.

(viii) Let x be any element of (A ∩ B) ∩ C 

x∈(A ∩ B) and x∈C

x∈A and x∈B and x∈C

x∈A and x (B ∩ C)

x∈A ∩ (B ∩ C) 

(A ∩ B) ∩ C \(\subset\)A ∩ (B ∩ C) …..(i) 

Now, let x be an element of A ∩ (B ∩ C) 

Then, x∈A and (B ∩ C)

x∈A and x∈B and x∈C

x∈(A ∩ B) and x∈C

x∈(A ∩ B) ∩ C

A ∩ (B ∩ C) \(\subset\)(A ∩ B) ∩ C …..(ii) 

From i and ii, (A ∩ B) ∩ C = A ∩ (B ∩ C)

[every set is a subset of itself] 

Hence, proved.