In the logic circuit diagram,when all the fou | vedclass

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(B) The circuit consists of an $OR$ gate,a $NAND$ gate,a $NOT$ gate,and a $NOR$ gate.$Y_1$ is the output of the $OR$ gate with inputs $A$ and $B$,so $

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$(0, 0, 1)$

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(B) The circuit consists of an $OR$ gate,a $NAND$ gate,a $NOT$ gate,and a $NOR$ gate.$Y_1$ is the output of the $OR$ gate with inputs $A$ and $B$,so $Y_1 = A + B$.$Y_2$ is the output of the $NOT$ gate connected to the output of the $NAND$ gate with inputs $C$ and $D$. The $NAND$ output is $\overline{C \cdot D}$,so $Y_2 = \overline{(\overline{C \cdot D})} = C \cdot D$.$Y_3$ is the output of the $NOR$ gate with inputs $Y_1$ and $Y_2$,so $Y_3 = \overline{Y_1 + Y_2}$.When $A = 0, B = 0, C = 0, D = 0$:$Y_1 = 0 + 0 = 0$.$Y_2 = 0 \cdot 0 = 0$.$Y_3 = \overline{0 + 0} = \overline{0} = 1$.Thus,the outputs $(Y_1, Y_2, Y_3)$ are $(0, 0, 1)$.

In the logic circuit diagram,when all the four inputs $A, B, C, D$ are $1$,the outputs $Y_1, Y_2, Y_3$ are respectively $(1, 1, 0)$. When all the inputs $A, B, C, D$ are changed to $0$,the outputs $Y_1, Y_2, Y_3$ respectively change to

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