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Manfred comments on What should a Bayesian do given probability of proving X vs. of disproving X? - Less Wrong Discussion
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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.
They're not necessarily exhaustive. Part of my question was how your answer changes in the cases where you know those possibilities are exhaustive (because the question is decidable).
If they are mutually exclusive but not exhaustive, and you have no other information besides P(X|proven)=1, P(X|disproven)=0, (usually not true, but useful for toy cases), then we have P(X)=P(proven)+1/2*(1-P(disproven) -P(proven)).
This is only equivalent to P(X) = P(proven) / (P(proven) + P(disproven)) if P(proven)+P(disproven)=1.
Basically what's going on is that if you can neither prove nor disprove X, and you don't have any information about what to think when there's neither proof nor disproof, your estimate just goes to 1/2.
Appendix: The equations we want in this case come from P(X)= P(X|A)*P(A)+P(X|B)*P(B)+P(X|C)*P(C), for A, B, C mutually exclusive and exhaustive. So for example if "proven" and "disproven" are mutually exclusive but not exhaustive, then we can add some extra category "neither." We take as starting information that P(X|proven)=1 and P(X|disproven)=0, and in the lack of other information P(X|neither)=1/2. So P(X)=P(X|proven)*P(proven)+P(X|disproven)*P(disproven)+P(X|neither)*P(neither).