Chapter B, Problem 1E

Explanation of Solution

$\text{A}$	$B$	$\overline{\text{A}}$	$\overline{B}$	$\overline{\text{A}\text{+}\text{B}}$	$\overline{\text{A}}·\overline{\text{B}}$	$\overline{\text{A}·\text{B}}$	$\overline{\text{A}}\text{+}\overline{\text{B}}$
0	0	1	1	1	1	1	1
0	1	1	0	0	0	1	1
1	0	0	1	0	0	1	1
1	1	0	0	0	0	0	0

We have, A = 0 and B = 0

Then, $\overline{\text{A}}\text{=}\text{1}$ and $\overline{\text{B}}\text{=}\text{1}$

$\begin{array}{c}\Rightarrow \overline{\text{A+}\text{B}}\text{=}\overline{\text{A}}·\overline{\text{B}}\\ \Rightarrow \overline{\text{0}\text{+}\text{0}}\text{=}\overline{\text{0}}·\overline{\text{0}}\\ \Rightarrow \overline{\text{0}}\text{=}\text{1}·\text{1}\\ \Rightarrow \text{1}\text{=}\text{1}\\ \therefore \text{LHS}\text{=}\text{RHS}\end{array}$

$\begin{array}{c}\Rightarrow \overline{\text{A+}\text{B}}\text{=}\overline{\text{A}}·\overline{\text{B}}\\ \Rightarrow \overline{\text{0}\text{+}1}\text{=}\overline{\text{0}}·\overline{1}\\ \Rightarrow \overline{1}\text{=}\text{1}·0\\ \Rightarrow 0\text{=}0\\ \therefore \text{LHS}\text{=}\text{RHS}\end{array}$

$\begin{array}{c}\Rightarrow \overline{\text{A+}\text{B}}\text{=}\overline{\text{A}}·\overline{\text{B}}\\ \Rightarrow \overline{1\text{+}\text{0}}\text{=}\overline{1}·\overline{\text{0}}\\ \Rightarrow \overline{1}\text{=}0·\text{1}\\ \Rightarrow 0\text{=}0\\ \therefore \text{LHS}\text{=}\text{RHS}\end{array}$

$\begin{array}{c}\Rightarrow \overline{\text{A+}\text{B}}\text{=}\overline{\text{A}}·\overline{\text{B}}\\ \Rightarrow \overline{1\text{+}1}\text{=}\overline{1}·\overline{1}\\ \Rightarrow \overline{1}\text{=}0·0\\ \Rightarrow 0\text{=}0\\ \therefore \text{LHS}\text{=}\text{RHS}\end{array}$

$\begin{array}{c}\Rightarrow \overline{\text{A}·\text{B}}\text{=}\overline{\text{A}}+\overline{\text{B}}\\ \Rightarrow \overline{0·0}\text{=}\overline{0}+\overline{0}\\ \Rightarrow \overline{0}\text{=}1+1\\ \Rightarrow 1\text{=}1\\ \therefore \text{LHS}\text{=}\text{RHS}\end{array}$

$\begin{array}{c}\Rightarrow \overline{\text{A}·\text{B}}\text{=}\overline{\text{A}}+\overline{\text{B}}\\ \Rightarrow \overline{0·1}\text{=}\overline{0}+\overline{1}\\ \Rightarrow \overline{0}\text{=}1+0\\ \Rightarrow 1\text{=}1\\ \therefore \text{LHS}\text{=}\text{RHS}\end{array}$

$\begin{array}{c}\Rightarrow \overline{\text{A}·\text{B}}\text{=}\overline{\text{A}}+\overline{\text{B}}\\ \Rightarrow \overline{1·0}\text{=}\overline{1}+\overline{0}\\ \Rightarrow \overline{0}\text{=}0+1\\ \Rightarrow 1\text{=}1\\ \therefore \text{LHS}\text{=}\text{RHS}\end{array}$

$\begin{array}{c}\Rightarrow \overline{\text{A}·\text{B}}\text{=}\overline{\text{A}}+\overline{\text{B}}\\ \Rightarrow \overline{1·1}\text{=}\overline{1}+\overline{1}\\ \Rightarrow \overline{1}\text{=}0+0\\ \Rightarrow 0\text{=}0\\ \therefore \text{LHS}\text{=}\text{RHS}\end{array}$