Figure (a) shows A as the NOR gate’s two inputs and $Y$ as the output. As a result, the circuit’s output is
$\overline{\mathrm{A}+\mathrm{A}}$
So, the output will be $Y=\overline{\mathrm{A}}$
Therefore, the gate is a NOT gate.
The truth table is given as
(b) A and B are the inputs, while $Y$ is the output of the given circuit in figure (b). We can deduce that the outputs of the first two NOR gates are $\overline{A}$ and $\overline{B}$ using the result obtained in solution (a). This is the NOR gate’s input. As a result, the circuit’s output is $\mathbf{Y}=\overline{\overline{\mathbf{A}}+\overline{\mathbf{B}}}=\mathbf{A} \cdot \mathbf{B} \cdot$. As a result, this serves as an AND gate. The following is a truth table.
