The figure shows a system of logic gates. Fro | vedclass

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(C) The circuit consists of an $OR$ gate,a $NAND$ gate,and two $NOT$ gates. Let the inputs be $X$ and $Y$. The output of the top $OR$ gate is $P = \ba

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$X = 1, Y = 0$

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(C) The circuit consists of an $OR$ gate,a $NAND$ gate,and two $NOT$ gates. Let the inputs be $X$ and $Y$. The output of the top $OR$ gate is $P = \bar{X} + Y$. The output of the bottom $NAND$ gate is $Q = \overline{X \cdot \bar{Y}} = \bar{X} + Y$. The final output $R$ is the output of a $NOR$ gate with inputs $P$ and $Q$,so $R = \overline{P + Q}$.$X$$Y$$\bar{X}$$\bar{Y}$$P = \bar{X} + Y$$Q = \bar{X} + Y$$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 $R = 1$ only when $X = 1$ 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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