The truth table for the given logic circuit i | vedclass

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(A) From the logic circuit,the output $Y$ is given by the $NAND$ gate operation on the outputs of the $NOR$ gate and the $AND$ gate.Let the output of

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$A$ truth table where $Y=1$ for all inputs $(A, B)$.

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(A) From the logic circuit,the output $Y$ is given by the $NAND$ gate operation on the outputs of the $NOR$ gate and the $AND$ gate.Let the output of the $NOR$ gate be $Y_1 = \overline{A+B}$.Let the output of the $AND$ gate be $Y_2 = A \cdot B$.The final output $Y$ is the $NAND$ of $Y_1$ and $Y_2$:$Y = \overline{Y_1 \cdot Y_2} = \overline{\overline{(A+B)} \cdot (A \cdot B)}$.Using the property $\overline{X} \cdot X = 0$,we analyze the expression:$Y = \overline{(\overline{A} \cdot \overline{B}) \cdot (A \cdot B)}$$Y = \overline{(\overline{A} \cdot A) \cdot (\overline{B} \cdot B)}$Since $\overline{A} \cdot A = 0$ and $\overline{B} \cdot B = 0$,we have:$Y = \overline{0 \cdot 0} = \overline{0} = 1$.Therefore,the output $Y$ is always $1$ for all possible input combinations of $A$ and $B$.

The truth table for the given logic circuit is

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