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(A) The given circuit consists of a $NOT$ gate and a $NAND$ gate. The input $A$ passes through a $NOT$ gate,so the output of the $NOT$ gate is $\bar{A

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(A) The given circuit consists of a $NOT$ gate and a $NAND$ gate. The input $A$ passes through a $NOT$ gate,so the output of the $NOT$ gate is $\bar{A}$. This $\bar{A}$ and the input $B$ are then fed into a $NAND$ gate. The output $Y$ of the $NAND$ gate is given by the expression: $Y = \overline{\bar{A} \cdot B}$.Using De Morgan's theorem,$\overline{\bar{A} \cdot B} = \overline{\bar{A}} + \overline{B} = A + \overline{B}$.Now,we construct the truth table for $Y = A + \overline{B}$:| $A$ | $B$ | $\overline{B}$ | $Y = A + \overline{B}$ ||---|---|---|---|| $0$ | $0$ | $1$ | $1$ || $0$ | $1$ | $0$ | $0$ || $1$ | $0$ | $1$ | $1$ || $1$ | $1$ | $0$ | $1$ |

For the following digital logic circuit,the correct truth-table is

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