Let A = {a, b, c}, B = {b, c, d, e} and = {c, d, e, f} be subsets of U = {a, b, c, d,

Solution:

(i) A’ = {d, e, f}

$\left(A^{\prime}\right)^{\prime}=\{a, b, c\}=A$

Hence proved

(ii) $A^{U} B=\{a, b, c, d, e\}$

$\left(\mathrm{A}^{\cup} \mathrm{B}\right)^{\prime}=\{\mathrm{f}\}$

$\mathrm{A}^{\prime}=\{\mathrm{d}, \mathrm{e}, \mathrm{f}\}$

$\mathrm{B}^{\prime}=\{\mathrm{a}, \mathrm{f}\}$

$A^{\prime} \cap_{B^{\prime}}=\{f\}$

$\Rightarrow\left(A^{U} B\right)^{\prime}=\left(A^{\prime} \cap_{B^{\prime}}\right)$

Hence proved

(iii) $A^{\prime} U_{B^{\prime}}=\{a, d, e, f\}$

$A^{\cap} B=(b, c\}$

$\left(A^{\cap} B\right)^{\prime}=\{a, d, e, f\}$

$\Rightarrow\left(\mathrm{A}^{\cap} \mathrm{B}\right)^{\prime}=\mathrm{A}^{\prime} \mathrm{U}_{\mathrm{B}}^{\prime}$

Hence proved