Is the perfect secrecy of the one time pad dependent upon the xor "randomness preserving" property (xoring a uniformly random string with any string yields a uniformly random string)? I have seen different arguments in different sources.
For example, in this discussion, the first answer derives perfect secrecy without referring to the randomness preserving issue. The derivation starts with
$$Pr[C=c]=\sum{Pr[C=c|M=m']\cdot Pr[M=m']} =\sum{Pr[K=m'\oplus c]}\cdot Pr[M=m'] =\sum{2^{-n}}\cdot Pr[M=m'] =2^{-n}$$
where the second equation, which seems to be the core of the proof, is based on the fact there's a unique $K$ such that $K=m' \oplus c$. Ostensibly this would work for any other logical operator that has an inverse, not necessarily xor.
On the other hand, the other answer to that question states "The main point in any security proof for the one-time pad is that xoring a plaintext with a (uniformly) random bit string yields a (uniformly) random bit string no matter what the plaintext was.". If this is indeed the main point in the proof, then this proof would not work for other operators that have an inverse but don't preserve randomness.
Who is right?
Thanks!