The Boolean function for the output Z of the circuit shown in the figure is most nearly:
 

A.  $$XY$$
B.  $$barXbarY$$
C.  $$XoplusY$$
D.  $$bar(XoplusY)$$

This circuit consists of 4 NAND gates. A NAND gate conforms to the following truth table:

Based on the truth table, the NAND gate's Boolean expression is: $\overline{XY}$. Using this expression, we can walk through and write down the expressions at the output of each gate until we get one for Z:

Now that we have an expression for Z, we are interested in simplification. We will need to use DeMorgan's law for NAND which states:

$\overline{AB} =\overline{A}+\overline{B}$

We can therefore split up our expression like this:

$$\begin{gathered} Z = \overline{\overline{X\overline{XY}}\overline{Y\overline{XY}}} \\ Z = \overline{\overline{X\overline{XY}}}+\overline{\overline{Y\overline{XY}}} \end{gathered}$$

Two top bars indicate a double negative, which would cancel each other out into a positive. We can eliminate the two bars over each expression:

$Z = X\overline{XY}+Y\overline{XY}$

Now, we apply DeMorgan's once more on both parts of our equation:

$Z = X(\overline{X}+\overline{Y})+Y(\overline{X}+\overline{Y})$

Next, we distribute:

$Z = X\overline{X}+X\overline{Y}+Y\overline{X}+\overline{Y}Y$

Both $X\overline{X}$ and $\overline{Y}Y$ terms will become 0, leaving us with our final expression:

$Z = X\overline{Y}+Y\overline{X}$

This expression is representative of the XOR logic, which can be expressed through this symbol: $\oplus$. Therefore, another way to write our expression is:

$Z = X\oplus Y$

This corresponds to answer C.