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Benja comments on Reflection in Probabilistic Logic - Less Wrong
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Stuart's proof by contradiction goes through, as far as I can see. (The speculation in the last paragraph doesn't, as Paul notes, and I don't know what "Hence P('G')=1" is supposed to mean, but the proof by contradiction part does work.)
ETA: I now think that "Hence P('G') = 1" is supposed to mean that this statement is implied by the first-order theory (T + the reflection axioms).