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Truth table for the given circuit will be
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1645 Upvotes
Verified Answer
The correct answer is:
$$
\begin{array}{cc|c}
x & y & z \\
\hline 0 & 0 & 1 \\
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 1
\end{array}
$$
$$
\begin{array}{cc|c}
x & y & z \\
\hline 0 & 0 & 1 \\
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 1
\end{array}
$$
Truth table of the circuit is as follows
$$
\begin{array}{|c|c|c|c|c|c|}
\hline x & y & \bar{x} & a=x \cdot y & b=\bar{x} \cdot y & z=\overline{a \cdot b} \\
\hline 0 & 0 & 1 & 0 & 0 & 1 \\
\hline 0 & 1 & 1 & 0 & 1 & 1 \\
\hline 1 & 0 & 0 & 0 & 0 & 1 \\
\hline 1 & 1 & 0 & 1 & 0 & 1 \\
\hline
\end{array}
$$
$$
\begin{array}{|c|c|c|c|c|c|}
\hline x & y & \bar{x} & a=x \cdot y & b=\bar{x} \cdot y & z=\overline{a \cdot b} \\
\hline 0 & 0 & 1 & 0 & 0 & 1 \\
\hline 0 & 1 & 1 & 0 & 1 & 1 \\
\hline 1 & 0 & 0 & 0 & 0 & 1 \\
\hline 1 & 1 & 0 & 1 & 0 & 1 \\
\hline
\end{array}
$$
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