Probably Correct

Alright, Canada day seems like a good time to post the solution for the labeled masses problem. Working on it yourself you probably found it is best to assume that the coins are labeled correctly and attempt to prove that rather than something else. Any method that would actually prove the coins are labeled correctly would also show that they are labeled incorrectly if they are not (if it didn't do that, one would hardly accept that it proves they are labeled correctly). Actually, this gives us something of a rewording of the problem, which I will later transform into the followup problem. Assume the coins are unlabeled, but you know which coin is which, you must prove to an audience (of mathematical puzzle solvers) that you know which coin is which using only 2 measurings.

Anyway, moving on to the solution (to the problem as initial worded). Its tricky to do anything involving weighing 1 coin against 2 right away, since if you weigh 1+2 against 3, even if it balances you cannot be sure that it was not the 1+3 against 4 or the 2+4 against the 6 or something. You could also try having 4 coins on one side, but that cannot work since on the second measurement you must split up those 4 coins or you cannot tell them apart. Perhaps something involving 2 vs. 2 works, but I didn't find anything. The thing that does work is to weigh the 1+2+3 against the 6. If it is imbalanced, we are done, if it is balanced then you have identified the 6 (it is the only one that can have as much mass as 3 others combined) and you have identified a group that is the {1,2,3}, since they are the only 3 light enough that a single coin could balance them.

So, in one measurement we have identified the {1,2,3}, the {4,5} and the {6}. The next measurement is to do the 6 and the 1 against the 5 and the 3.

If they are labeled correctly, the 5+3 should be heavier than the 6+1 (so, if that doesn't happen we are done). Since the 6 is identified, and the 1 must weight at least 1, the 6+1 must weigh at least 7, but since the 5+3 came out heavier, the 5 could not be any lighter and neither could the 3, so the 5 and 3 are correct. By a similar logic, the 5+3 cannot together weight more than 8, so the 1 cannot afford to be any heavier if the 6+1 is to be less than 8, thus the 1 must be the 1. That leaves the 2 and 4 both uniquely identified.

This completes the solution. I cannot prove that there aren't any other solutions, but I greatly doubt their existence.

Next, the follow-up:

A mathemagician has 8 coins, with masses 1 through 8, he knows which coin is which. He intends to prove to an audience that he knows the correct mass of at least one of the coins using a single measurement of a balance scale. What measurement can he perform to prove the mass of any one of the coins?

You may assume that the audience knows that the coins have masses 1 through 8, but they do not know which one is which. After one measurement, the mathemagician must be able to select a single coin and say "I have now proven this one is the x" for some value of x. You may also assume the audience is a bunch of mathematicians, they will not be tricked by some sort of lies.