What should a Bayesian do given probability of proving X vs. of disproving X?

Is it reasonable to assign P(X) = P(will_be_proven(X)) / (P(will_be_proven(X)) + P(will_be_disproven(X))) ?

No I don't think so, consider the following example:

I flip a coin. If it comes up heads I take two green marbles, else I take one red and one green marble. Then I offer to let you see a random marble and I destroy the other one without showing you.

Then, suppose you wish to test whether my coin came up tails. if the marble is red, you have proven the coin came up tails and the chance of tails being disproven is zero, so your expression is 1, but it should be 0.5.

If so, consider X = "free will exists". One could argue that the term "free will" is defined such that it is impossible to detect it, or to prove that it exists. But if one could prove that the many worlds interpretation of quantum mechanics is correct, that would constitute a disproof of X. Then P(will_be_proven(X)) / (P(will_be_proven(X)) + P(will_be_disproven(X))) = 0.

This has to be one of the worst examples I've ever seen that wasn't actively trying to push an agenda.

Not a great example. "But if one could prove that the many worlds interpretation of quantum mechanics is correct, that would constitute a disproof of X." -- where X is free will? Probably not by many definitions of it.

In general, I don't see how this question is different from "what a Bayesian should do" with any other probabilistic statement.

Is it reasonable to assign P(X) = P(will_be_proven(X)) / (P(will_be_proven(X)) + P(will_be_disproven(X))) ?

It's also possible that X will never be either proven or disproven.

For a proposed proof, now that proofs can be checked by machines, it does make sense to expect the proof to be validated or invalidated.

But if one could prove that the many worlds interpretation of quantum mechanics is correct, that would constitute a disproof of X.

I'm not sure I follow. But then, I don't know of any definition of free will that isn't self-contradictory, so it's not surprising that I don't understand what would convince those who seriously consider free will.

In general, the assumption you're asking about is whether will_be_proven(X) and will_be_disproven(X) are mutually exclusive and exhaustive. If the objects you're calling "proof" and "disproof" can ever coexist, the mutual exclusivity part breaks. If there's any option other than a proof existing or a disproof existing, then the exhaustiveness part breaks down.

These assumptions are omnipresent because they're incredibly useful. But they're really just quantitative common sense. If using them feels like a "trick," that's evidence there's something fishy.

Is it reasonable to assign P(X) = P(will_be_proven(X)) / (P(will_be_proven(X)) + P(will_be_disproven(X)))

No. For example, there exist statements that are unprovable but true. To assign a numerator of 0 (or epsilon) to all unprovable statements is wrong.

Commenters minunderstand your problem and your argument for its solution. I take your problem to be "What could the probability of a mathematical proposition be besides its comparative likelihood of proof or disproof?

Perhaps the answer is that there are reasons besides proof to believe even a mathematical proposition. Empirical reasons, that is.

Doing means acting. There's nothing to be done about the 4-color problem as it's mostly about information. To do you have to use an utility function that ways possible outcomes and then you can decide to do A or do B.

Is it reasonable to assign P(X) = P(will be proven(X)) / (P(will be proven(X)) + P(will be disproven(X))) ?

No. Consider Goldbach's conjecture: any even number above four is the sum of two odd primes. If it's false, it can be disproved via counterexample. If it's true, the proof must be much more sophisticated. At this point, we might be able to say that if it were feasible to disprove via counterexample that would have already happened, but that certainly wasn't always the case.

Edit: Fixed Goldbach's conjecture.