Answer
The truth table is already given. There are two cases with a 1 bit in
the output, occurring in rows 1 and $4 .$ The corresponding subexpressions are:
Case 1: $\quad \overline{a} \cdot \overline{b}$
Case $2 : \quad a \cdot b$
The final Boolean expression is $\overline{a} \cdot \overline{b}+a \cdot b$ and the circuit diagram is
Work Step by Step
The truth table is already given. There are two cases with a 1 bit in
the output, occurring in rows 1 and $4 .$ The corresponding subexpressions are:
Case 1: $\quad \overline{a} \cdot \overline{b}$
Case $2 : \quad a \cdot b$
The final Boolean expression is $\overline{a} \cdot \overline{b}+a \cdot b$ and the circuit diagram is
