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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
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  • @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.). – Andrew Mar 03 '16 at 07:39

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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: $$ \overline{x_1 \bigoplus x_2} \bigoplus x_3 = \bar{x_1}\bar{x_2}\bar{x_3} + x_1x_2\bar{x_3} + \bar{x_1}x_2x_3 + x_1\bar{x_2}x_3$$

megamind
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