The truth table for the given circuit is:

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(C) The circuit consists of two $AND$ gates, one $NOT$ gate, and one $NAND$ gate.Let the inputs be $x$ and $y$.The upper $AND$ gate receives $x$ and $

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$x$$y$$z$$0$$0$$1$$0$$1$$1$$1$$0$$1$$1$$1$$1$

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(C) The circuit consists of two $AND$ gates, one $NOT$ gate, and one $NAND$ gate.Let the inputs be $x$ and $y$.The upper $AND$ gate receives $x$ and $y$, so its output is $a = x \cdot y$.The lower $AND$ gate receives $\bar{x}$ (from the $NOT$ gate) and $y$, so its output is $b = \bar{x} \cdot y$.The final $NAND$ gate receives $a$ and $b$, so its output is $z = \overline{a \cdot b} = \overline{(x \cdot y) \cdot (\bar{x} \cdot y)}$.Using Boolean algebra: $z = \overline{(x \cdot \bar{x}) \cdot (y \cdot y)} = \overline{0 \cdot y} = \overline{0} = 1$.Since the output $z$ is $1$ for all combinations of inputs $(x, y)$, the correct truth table is option $C$.

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