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(C) The logic circuit consists of an $OR$ gate,a $NAND$ gate,and a $NOR$ gate. Let the inputs be $X$ and $Y$. The output $P$ of the $OR$ gate is $P =

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$X = 1, Y = 0$

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(C) The logic circuit consists of an $OR$ gate,a $NAND$ gate,and a $NOR$ gate. Let the inputs be $X$ and $Y$. The output $P$ of the $OR$ gate is $P = \overline X + Y$. The output $Q$ of the $NAND$ gate is $Q = \overline {X \cdot \overline Y}$. The final output $R$ is the output of a $NOR$ gate with inputs $P$ and $Q$,so $R = \overline {P + Q}$.The truth table is as follows:$X$$Y$$\overline X$$\overline Y$$P = \overline X + Y$$Q = \overline {X \cdot \overline Y}$$R = \overline {P + Q}$$0$$0$$1$$1$$1$$1$$0$$0$$1$$1$$0$$1$$1$$0$$1$$0$$0$$1$$0$$0$$1$$1$$1$$0$$0$$1$$1$$0$From the truth table,it is clear that for $R = 1$,we must have $X = 1$ and $Y = 0$.

The figure shows a system of logic gates. From the study of the truth table,it can be found that to produce a high output $(1)$ at $R$,we must have

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