The circuit given below is that of:

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(C) Let the inputs be $A$ and $B$. The first gate is an $OR$ gate with one input inverted,so its output is $Y' = \bar{A} + B$.The second gate is an $A

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$NAND$ gate

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(C) Let the inputs be $A$ and $B$. The first gate is an $OR$ gate with one input inverted,so its output is $Y' = \bar{A} + B$.The second gate is an $AND$ gate with inputs $Y'$,$A$,and $B$. Its output is $Y'' = Y' \cdot A \cdot B = (\bar{A} + B) \cdot A \cdot B$.Using Boolean algebra: $Y'' = (\bar{A} \cdot A \cdot B) + (B \cdot A \cdot B) = (0 \cdot B) + (A \cdot B) = A \cdot B$.The final gate is a $NOT$ gate,so the final output is $Y = \overline{Y''} = \overline{A \cdot B}$.This is the Boolean expression for a $NAND$ gate.Truth table for the circuit:$A$$B$$\bar{A}$$Y' = \bar{A} + B$$Y'' = Y' \cdot A \cdot B$$Y = \overline{Y''}$$0$$0$$1$$1$$0$$1$$0$$1$$1$$1$$0$$1$$1$$0$$0$$0$$0$$1$$1$$1$$0$$1$$1$$0$

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