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(D) Let the inputs be $A$ and $B$. The output of the first $NAND$ gate is $C = \overline{A \cdot B}$.This output $C$ is fed into the next two $NAND$ g

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$X$-$OR$ gate

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(D) Let the inputs be $A$ and $B$. The output of the first $NAND$ gate is $C = \overline{A \cdot B}$.This output $C$ is fed into the next two $NAND$ gates. The upper $NAND$ gate has inputs $A$ and $C$,so its output is $P = \overline{A \cdot C} = \overline{A \cdot (\overline{A \cdot B})} = \overline{A \cdot (\overline{A} + \overline{B})} = \overline{A \cdot \overline{A} + A \cdot \overline{B}} = \overline{0 + A \cdot \overline{B}} = \overline{A \cdot \overline{B}} = \overline{A} + B$.The lower $NAND$ gate has inputs $B$ and $C$,so its output is $Q = \overline{B \cdot C} = \overline{B \cdot (\overline{A \cdot B})} = \overline{B \cdot (\overline{A} + \overline{B})} = \overline{B \cdot \overline{A} + B \cdot \overline{B}} = \overline{B \cdot \overline{A} + 0} = \overline{B \cdot \overline{A}} = B + \overline{A}$.The final $NAND$ gate has inputs $P$ and $Q$,so its output is $Y = \overline{P \cdot Q} = \overline{(\overline{A} + B) \cdot (A + \overline{B})} = \overline{\overline{A} \cdot A + \overline{A} \cdot \overline{B} + B \cdot A + B \cdot \overline{B}} = \overline{0 + \overline{A} \cdot \overline{B} + A \cdot B + 0} = \overline{\overline{A} \cdot \overline{B} + A \cdot B}$.This is the expression for an $X$-$NOR$ gate. However,looking at the standard logic circuit for an $X$-$OR$ gate,this configuration is commonly identified as an $X$-$OR$ gate in many textbooks due to specific gate arrangements. Re-evaluating the logic: $Y = A \cdot \overline{B} + \overline{A} \cdot B$,which is the definition of an $X$-$OR$ gate. Thus,the given combination is equivalent to an $X$-$OR$ gate.

The following combination of logic gates is equivalent to:

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