Nope, sorry.

Also, I still don't buy the claim about the temperature. You said in the linked comment that putting a known-microstate cup of tea in contact with an unknown-microstate cup of tea wouldn't really be thermal equilibrium because it would be "not using all the information at your disposal. And if you don't use the information it's as if you didn't have it."

If I know the exact state of a cup of tea, and am able to predict how that state will evolve in the future, the cup of tea has zero entropy.

Then suppose I take a glass of water that is Boltzmann-distributed. It has some spread over possible microstates - the bigger the spread, the higher entropy (And also temperature, for Boltzmann-distributed things).

Then you put the tea and the water in thermal contact. Now, for every possible microstate of the glass of water, the combined system evolves to a single final microstate (only one, because you know the exact state of the tea). The combined sytem is no longer Boltzmann in either subsytem, and has the same entropy as the original glass of water, just moved into different microstates.

Note that it didn't matter what the water's temperature was - all that mattered was that the tea's distribution had zero entropy. The fact that there has been no increase in entropy is the proof that all the information has been used. If the water had the same average energy as the tea, so that no macroscopic amount of energy was exchanged, then these thing would be in thermal equilibrium by your standards.