For the logic circuit shown in the figure,the | vedclass

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(C) $1$. The first gate is a $NOR$ gate with both inputs connected to $A$. The output of this $NOR$ gate is $\overline{A + A} = \bar{A}$.$2$. The seco

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$Y = \bar{A}$

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(C) $1$. The first gate is a $NOR$ gate with both inputs connected to $A$. The output of this $NOR$ gate is $\overline{A + A} = \bar{A}$.$2$. The second gate is a $NOT$ gate. The input to this gate is $\bar{A}$,so its output is $\overline{\bar{A}} = A$.$3$. The third gate is a $NAND$ gate with both inputs connected to the output of the $NOT$ gate (which is $A$). The output of this $NAND$ gate is $\overline{A \cdot A} = \bar{A}$.$4$. Therefore,the final Boolean expression for the circuit is $Y = \bar{A}$.

For the logic circuit shown in the figure,the Boolean expression is:

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