The logic circuit shown above is equivalent to :

Truth table of the given gate:

$\begin{cases}A & B & C \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 0\end{cases}$

Truth table of option \left(1\right)$</p><p>$\begin{cases}A & B & C \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1\end{cases}$</p><p>Truth table of option \left(2\right)$

$\begin{cases}A & B & C \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 1\end{cases}$

Truth table of option \left(3\right)$</p><p>$\begin{cases}A & B & C \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 0\end{cases}$</p><p>Truth table of option \left(4\right)$

$\begin{cases}A & B & C \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 0\end{cases}$

Since option \left(1\right)$has same truth table, hence answer is option \left(4\right)$ is correct.

Given Boolean expression can be written as

$\overset{¯}{A + \overset{¯}{B}} = C$

$∴ C = \overset{¯}{A} \cdot \overset{¯}{B} = \overset{¯}{A} \cdot B$