The circuit given below is equivalent to

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(A) The circuit consists of a $NOR$ gate,a $NAND$ gate with both inputs tied together (acting as a $NOT$ gate),and a $NOT$ gate.Let the inputs be $A$

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$NOR$ gate

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(A) The circuit consists of a $NOR$ gate,a $NAND$ gate with both inputs tied together (acting as a $NOT$ gate),and a $NOT$ gate.Let the inputs be $A$ and $B$. The output of the $NOR$ gate is $Y_1 = \overline{A+B}$.The $NAND$ gate with tied inputs acts as a $NOT$ gate,so its output is $Y_2 = \overline{Y_1} = \overline{\overline{A+B}} = A+B$.The final $NOT$ gate inverts this,so the final output $Y = \overline{Y_2} = \overline{A+B}$.This is the Boolean expression for a $NOR$ gate.| Inputs $(A, B)$ | $NOR$ Output $(Y_1)$ | $NAND$ Output $(Y_2)$ | Final Output $(Y)$ || :--- | :--- | :--- | :--- || $0, 0$ | $1$ | $0$ | $1$ || $0, 1$ | $0$ | $1$ | $0$ || $1, 0$ | $0$ | $1$ | $0$ || $1, 1$ | $0$ | $1$ | $0$ |

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