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(A) From the figure,the output $y_1$ is the output of a $NAND$ gate with inputs $(A \cdot B)$ and $B$. Thus,$y_1 = \overline{(A \cdot B) \cdot B}$.Giv

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$1$,$0$

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(A) From the figure,the output $y_1$ is the output of a $NAND$ gate with inputs $(A \cdot B)$ and $B$. Thus,$y_1 = \overline{(A \cdot B) \cdot B}$.Given $A=0$ and $B=1$,$A \cdot B = 0 \cdot 1 = 0$.So,$y_1 = \overline{0 \cdot 1} = \overline{0} = 1$.The output $y_2$ is the output of a $NOR$ gate with inputs $(B+C)$ and $B$. Thus,$y_2 = \overline{(B+C) + B}$.Given $B=1$ and $C=1$,$B+C = 1+1 = 1$.So,$y_2 = \overline{1 + 1} = \overline{1} = 0$.Therefore,the values of $y_1$ and $y_2$ are $1$ and $0$ respectively.

Four logic gates are connected as shown in the figure. If the inputs are $A=0$,$B=1$ and $C=1$,then the values of $y_1$ and $y_2$ respectively are

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