# Negation of XOR

I feel pretty confident with expanding an XOR, but when it is negated, it throws me for a loop a bit.

The problem I am trying to prove: $$\overline{x_1 \bigoplus x_2} \bigoplus x_3 = \bar{x_1}\bar{x_2}\bar{x_3} + x_1x_2\bar{x_3} + \bar{x_1}x_2x_3 + x_1\bar{x_2}x_3$$

How I tried to prove it: $$\overline{x_1 \bigoplus x_2} x_3 + x_1\bar{x_2}x_3 + \bar{x_1}x_2\bar{x_3}$$ $$\bar{x_1}x_2x_3 + x_1x_2\bar{x_3}+x_1\bar{x_2}x_3+\bar{x_1}x_2\bar{x_3}$$

However I don't believe this to be correct, and I definitely feel I made a mistake somewhere.

• @Tanner Then you have an expression containing only con- and dis-junction (AND and OR). Next you might convert it to a normal form, using other algebraic rules (distributivity, commutativity, etc.). Mar 3, 2016 at 7:39

Your 1st step is correct. But expansion of $\overline{x_1 \bigoplus x_2}$ is not correct.
$\overline{x_1 \bigoplus x_2}$ is equivalent to $x1$ XNOR $x2$.
Now according to De Morgan's Laws: $$\overline{x_1 + x_2} = \overline{x_1}.\overline{x_2}$$ $$\overline{x_1.x_2} = \overline{x_1}+\overline{x_2}$$
Hence, $$\overline{x_1 \bigoplus x_2} = \overline{x_1.\overline{x_2}+\overline{x_1}.x_2}$$ $$=\overline{x_1.\overline{x_2}}.\overline{\overline{x_1}.x_2}=(\overline{x_1}+x_2).(x_1 + \overline{x_2}) = \overline{x_1}.\overline{x_2}+x_1.x_2$$
Using this expansion would result in: $$\overline{x_1 \bigoplus x_2} \bigoplus x_3 = \bar{x_1}\bar{x_2}\bar{x_3} + x_1x_2\bar{x_3} + \bar{x_1}x_2x_3 + x_1\bar{x_2}x_3$$