XOR will never go out of bounds because it combines bits and doesn’t create new bits where no bits were set before.
The result 5 is correct. Look at the binary representation of your value and the XOR result
10 00001010
20 00010100
30 00011110
5 00000101
20 00010100
10 00001010
30 00011110
--------------
00000101 => 5
An easy help for calculating a result of many XORed values is: The result will have a bit set where an odd number of bits are combined, no bit set for even number of bits.
If it is not possible that this can happen then is there a proof for this?
XOR is equivalent to addition without carry on the individual bits. When you add bits without carry, no overflow can happen and so the int value can’t go out of bounds.