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
Work Step by Step
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
