## Invitation to Computer Science 8th Edition

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
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