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(C) Let the inputs be $A$,$B$,and $C$. The circuit consists of $NAND$ gates and $OR$ gates.$1$. The top $NAND$ gate has input $A$ connected to both te

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$\bar{A}+\bar{B}+\bar{C}$

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(C) Let the inputs be $A$,$B$,and $C$. The circuit consists of $NAND$ gates and $OR$ gates.$1$. The top $NAND$ gate has input $A$ connected to both terminals,so its output is $\overline{A \cdot A} = \bar{A}$.$2$. The middle two $NAND$ gates have inputs $(A, B)$ and $(B, C)$ respectively,giving outputs $\overline{A \cdot B}$ and $\overline{B \cdot C}$.$3$. These two outputs are fed into an $OR$ gate,resulting in $\overline{A \cdot B} + \overline{B \cdot C}$.$4$. The bottom $NAND$ gate has input $C$ connected to both terminals,so its output is $\overline{C \cdot C} = \bar{C}$.$5$. Finally,all these signals are combined by an $OR$ gate to give the output $Y$:$Y = \bar{A} + (\overline{A \cdot B} + \overline{B \cdot C}) + \bar{C}$Using De Morgan's theorem,$\overline{A \cdot B} = \bar{A} + \bar{B}$ and $\overline{B \cdot C} = \bar{B} + \bar{C}$.Substituting these:$Y = \bar{A} + (\bar{A} + \bar{B}) + (\bar{B} + \bar{C}) + \bar{C}$Using the idempotent law $\bar{A} + \bar{A} = \bar{A}$,we get:$Y = \bar{A} + \bar{B} + \bar{C}$

For the logic gates shown below,the correct output is

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