# How to prove the following xor equation?

In a test paper, a question is given as :

Let, x1^x2^x3^x4 =0 where x1, x2, x3, x4 are Boolean variables,and ^ is the XOR operator.

Which one of the following must always be TRUE?
(A) x1x2x3x4 = 0
(B) x1x3+x2 = 0
(C)  x1'^x3' = x2'^x4'
(D) x1+x2+x3+x4 = 0

+ is OR and concatenation is AND operator.


Correct answer is C, which I guessed because when all booloean variables are 1, then, neither A,B, or D will hold,But what is the correct way to get to option C?

(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 &= 0 \\ x_1 \oplus x_3 \oplus x_2 \oplus x_4 &= 0 \\ x_1 \oplus x_3 &= x_2 \oplus x_4 \end{align} Then use $a \oplus b = a' \oplus b'$ to get (C). $$x_1' \oplus x_3' = x_2' \oplus x_4'$$