Zero-Knowledge proof of inequality I think it's neat how a person can prove that two discrete logarithms are equal without actually revealing the value.

So, conversely, is there a way for me to prove that two discrete logarithms are not equal without revealing either logarithm's value?  Assume that I know (by construction or by oracle) the discrete logarithm value for any pair of elements (from the 4 given elements) and I thereby know that all are unique.
A zero-knowledge algorithm for proving inequality of two discrete logarithms was given e.g. by Camenisch and Shoup in Section 6 of Practical Verifiable Encryption and Decryption of Discrete Logarithms (CRYPTO'03).

Quoting  from the paper:


  Let $G=\langle g \rangle$ be a group of prime order $q$. The prover
  and verifier have common inputs $g, h, y, z \in G$ where $g, h$ are
  generators of $G$ and $log_g y \neq log_h z$. The prover has the
  additional input $x = log_g y$. The prover and the verifier then
  engage in the following protocol.
  
  
  The prover randomly chooses $r \in Z_q$, computes the auxiliary commitment $C = (h ^x/z)^r$, and sends $C$ to the verifier.
  The prover executes the protocol denoted $PK\left\{(\alpha, \beta) : C = h^\alpha (\frac{1}{z})^\beta \wedge 1 = g^\alpha (\frac{1}{y})^\beta \right\}$ with the verifier. (Note that this is a
  proof of equality, so we know how to do this).
  The verifier accepts if it accepts in Step 2, and if $C \neq 1$; otherwise, the verifier rejects.
  
  
  This constitutes an honest-verifier proof system for proving that
  $log_g y \neq hog_h z$.