#### Answer

The truth table for this circuit would be as follows:
$\begin{array}{llll}{a} & {b} & {\text { Output }} \\ {0} & {0} & {0} \\ {0} & {1} & {0} \\ {1} & {0} & {1} \\ {1} & {1} & {0}\end{array} \quad$ (because $a$ is greater than $b )$
There is only one case where there is a 1 bit in the output. It is in the
third row and corresponds to the following Boolean expression:
$a \cdot \overline{b}$
We can skip Step 3 because with only one case, there is no combining of
Boolean expressions. Thus, the circuit diagram for this circuit is