(i) (A ∪ B)’ = A’ ∩ B’

Now, firstly let us consider the LHS

A ∪ B = {x: x ∈ A or x ∈ B}

= {2, 3, 5, 7, 9}

(A∪B)’ means Complement of (A∪B) with respect to universal set U.

Therefore, (A∪B)’ = U – (A∪B)’

U – (A∪B)’ is defined as {x ∈ U: x ∉ (A∪B)’}

U = {2, 3, 5, 7, 9}

(A∪B)’ = {2, 3, 5, 7, 9}

U – (A∪B)’ = ϕ

Then, RHS

A’ means Complement of A with respect to universal set U.

Therefore, A’ = U – A

(U – A) is defined as {x ∈ U: x ∉ A}

U = {2, 3, 5, 7, 9}

A = {3, 7}

A’ = U – A = {2, 5, 9}

B’ means Complement of B with respect to universal set U.

Therefore, B’ = U – B

(U – B) is defined as {x ∈ U: x ∉ B}

U = {2, 3, 5, 7, 9}

B = {2, 5, 7, 9}

B’ = U – B = {3}

A’ ∩ B’ = {x: x ∈ A’ and x ∈ C’}.

= ϕ

∴ LHS = RHS

Thus verified.

(ii) (A ∩ B)’ = A’ ∪ B’

Now, firstly let us consider the LHS

(A ∩ B)’

(A ∩ B) = {x: x ∈ A and x ∈ B}.

= {7}

(A∩B)’ means Complement of (A ∩ B) with respect to universal set U.

Therefore, (A∩B)’ = U – (A ∩ B)

U – (A ∩ B) is defined as {x ∈ U: x ∉ (A ∩ B)’}

U = {2, 3, 5, 7, 9}

(A ∩ B) = {7}

U – (A ∩ B) = {2, 3, 5, 9}

(A ∩ B)’ = {2, 3, 5, 9}

Then, RHS

A’ means Complement of A with respect to universal set U.

Therefore, A’ = U – A

(U – A) is defined as {x ∈ U: x ∉ A}

U = {2, 3, 5, 7, 9}

A = {3, 7}

A’ = U – A = {2, 5, 9}

B’ means Complement of B with respect to universal set U.

Therefore, B’ = U – B

(U – B) is defined as {x ∈ U: x ∉ B}

U = {2, 3, 5, 7, 9}

B = {2, 5, 7, 9}

B’ = U – B = {3}

A’ ∪ B’ = {x: x ∈ A or x ∈ B}

= {2, 3, 5, 9}

∴ LHS = RHS

Thus verified.