EXOR gate gives output as 1 when we have odd number of 1’s in the input.
(EXNOR gate gives output as 1 when we have even number of 0’s in the input.)
Now, given x1⊕x2⊕x3⊕x4=0 which means number of 1’s in the input is not odd (but even).
Possible input values of x1,x2,x3,x4 are: Zero 1’s, two 1’s or all four 1’s.
Now let’s check the options:
A) x1x2x3x4=0. False when we have all 1’s in the input.
B) x1x3+x2=0. False when two 1’s occur for the inputs x1,x3.
C) x¯1⊕x¯3=x¯2⊕x¯4. This is same as x1⊕x3=x2⊕x4
Now, since the number of 1’s in the input is even,
if number of 1’s in x1,x3 is even then number of 1’s in x2,x4 is also even and if number of 1’s in x1,x3 is odd then number of 1’s in x2,x4 is also odd. Hence this is always true.
D) x1+x2+x3+x4=0. False when we have zero 1’s in the input.