How to prove the following xor equation?
(I use the symbol $\oplus$ here because $\wedge$ is mathematically used for conjunction ("and") instead of exclusive-or)

First we rearrange the terms,
\begin{align}
 x_1 \oplus x_2 \oplus x_3 \oplus x_4 &amp;= 0 \\
 x_1 \oplus x_3 \oplus x_2 \oplus x_4 &amp;= 0 \\
 x_1 \oplus x_3 &amp;= x_2 \oplus x_4
\end{align}
Then use $a \oplus b = a' \oplus b'$ to get (C).
$$ x_1' \oplus x_3' = x_2' \oplus x_4' $$