Assume an ElGamal Cryptosystem. Assume a set of three players, $P_1$, $P_2$ and $P_3$. The private key $x$ is shared among the players. The player $P_1$ has a piece of the private key $x_1$, $P_2$ has the piece $x_2$ and $P_3$ has $x_3$.
Assume the following messages, $m_1$ and $m_2$.
Let $E(m_1)$ and $E(m_2)$ be two ElGamal ciphertexts.
The elements $r_1$ and $r_2$ are random numbers, and $y = g ^ x$, where $g$ is a generator of the group.
Encrypted message m1: $E(m_1) = (g ^ {r_{1}}, m_1 \cdot y ^ {r_{1}}) = (a_1, b_1)$
Encrypted message m2: $E (m_2) = (g ^ {r_{2}}, m_2 \cdot y ^ {r_{2}}) = (a_2, b_2)$
The question is: how can the holders (players) of the private key execute the oblivious test of plaintext equality proposed by Schnorr and Jakobsson? Specifically, how can the three players can determine if $m_1 = m_2$ without revealing the messages?
PS: The protocol oblivious test of plaintext equality proposed by Schnorr and Jakobsson is in the article entitled "Efficient Oblivious Proofs of Correct Exponentiation".