(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 |xC}

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

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

= { x |xC} ∪ { 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 |xC}

= { a, c, e, g, x}

= {x, a, c, e, g }

= { x |xC} ∪ { 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 C ∩ B ….(i)

Now let x be an element of C ∩ B

Then, xC ∩ B

x C and x B

x B and x C [by definition of intersection]

x B ∩ C

C ∩ B 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 C ∩ A ….(i)

Now let x be an element of C ∩ A

Then, xC ∩ A

x C and x A

x A and x C [by definition of intersection]

x A ∩ C

C ∩ A 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 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) (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 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) (A ∩ B) ∩ C …..(ii)

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

[every set is a subset of itself]

Hence, proved.