by Sam Eisenstat 9 days ago | Jessica Taylor likes this | link | parent Two minor comments. First, the bitstrings that you use do not all correspond to worlds, since, for example, $$\rm{Con}(\rm{PA}+\rm{Con}(\rm{PA}))$$ implies $$\rm{Con}(\rm{PA})$$, as $$\rm{PA}$$ is a subtheory of $$\rm{PA} + \rm{Con}(\rm{PA})$$. This can be fixed by instead using a tree of sentences that all diagonalize against themselves. Tsvi and I used a construction in this spirit in A limit-computable, self-reflective distribution, for example. Second, I believe that weakening #2 in this post also cannot be satisfied by any constant distribution. To sketch my reasoning, a trader can try to buy a sequence of sentences $$\phi_1, \phi_1 \wedge \phi_2, \dots$$, spending $$\2^{-n}$$ on the $$n$$th sentence $$\phi_1 \wedge \dots \wedge \phi_n$$. It should choose the sequence of sentences so that $$\phi_1 \wedge \dots \wedge \phi_n$$ has probability at most $$2^{-n}$$, and then it will make an infinite amount of money if the sentences are simultaneously true. The way to do this is to choose each $$\phi_n$$ from a list of all sentences. If at any point you notice that $$\phi_1 \wedge \dots \wedge \phi_n$$ has too high a probability, then pick a new sentence for $$\phi_n$$. We can sell all the conjunctions $$\phi_1 \wedge \dots \wedge \phi_k$$ for $$k \ge n$$ and get back the original amount payed by hypothesis. Then, if we can keep using sharper continuous tests of the probabilities of the sentences $$\phi_1 \wedge \dots \wedge \phi_n$$ over time, we will settle down to a sequence with the desired property. In order to turn this sketch into a proof, we need to be more careful about how these things are to be made continuous, but it seems routine.

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