If Alice encrypts two messages $a$ and $b$, such that $x=E(a)$, $y=E(b)$. Can Alice prove (without revealing $a$, $b$ or the private key) that $a = b$?
Obviously the proof must not be too long and it should be practical to compute and verify (either interactively or non-interactively).
This is possible for the Pohlig-Hellman symmetric cipher, even if the ciphertexts are encrypted with different keys. But P-H is not public key.
If such a cryptosystem exists (and it is commutative or provides public re-encryption), then one of the limitations in Mental Poker protocols could be solved. The problem is the existence (or not) of a protocol that can provide both semantic security and abrupt drop out tolerance (without any threshold scheme). Edit: It seems that the encryption needs to be deterministic to be able to support drop-out tolerance, and I see no way to overcome this. Without determinism, I was only able to veto the cards of a single player from a new deck.
See What is the theoretical and practical status of mental poker? for a related question.