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(B) Let the output of the first $NAND$ gate be $P = \overline{A \cdot B}$.The inputs to the upper $NAND$ gate are $A$ and $P$, so its output is $Q = \

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$(0, 0, 1)$

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(B) Let the output of the first $NAND$ gate be $P = \overline{A \cdot B}$.The inputs to the upper $NAND$ gate are $A$ and $P$, so its output is $Q = \overline{A \cdot P}$.The inputs to the lower $NAND$ gate are $P$ and $B$, so its output is $R = \overline{P \cdot B}$.The final output $Y$ is the output of a $NAND$ gate with inputs $Q$ and $R$, so $Y = \overline{Q \cdot R}$.Using De Morgan's Law: $Y = \overline{Q} + \overline{R} = \overline{\overline{A \cdot P}} + \overline{\overline{P \cdot B}} = (A \cdot P) + (P \cdot B) = P \cdot (A + B)$.Substituting $P = \overline{A \cdot B} = \overline{A} + \overline{B}$, we get $Y = (\overline{A} + \overline{B}) \cdot (A + B) = \overline{A} \cdot B + A \cdot \overline{B}$.This is the expression for an $XOR$ gate.For $A = 0, B = 0$: $Y = 0 \oplus 0 = 0$.For $A = 1, B = 1$: $Y = 1 \oplus 1 = 0$.For $A = 0, B = 1$: $Y = 0 \oplus 1 = 1$.Thus, the outputs are $(0, 0, 1)$.

Select the outputs $Y$ of the combination of gates shown below for inputs $A = 0, B = 0$; $A = 1, B = 1$ and $A = 0, B = 1$ respectively.

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