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Eliezer_Yudkowsky comments on Reflection in Probabilistic Logic - Less Wrong
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There may be an issue with the semantics here. I'm not entirely certain of the reasoning here, but here goes:
Reflection axiom: ∀a, b: (a < P(φ) < b) ⇒ P(a < P('φ') < b) = 1 (the version in Paul's paper).
Using diagonalisation, define G ⇔ -1<P('G')<1
Then P(G)<1
⇒ -1<P(G)<1 (since P > 0)
⇒ P(-1<P('G')<1)=1 (by reflection)
⇒ P(G)=1 (by definition of G).
Hence, by contradiction (and since P cannot be greater than 1), P(G)=1. Hence P('G')=1 (since the two P's are the same formal object), and hence G is false. So we have a false result that has probability 1.
But it may get worse. If we can prove P(G)=1 then (?) the logical system itself can prove P('G')=1 - and from that, can disprove G. So ¬G is a theorem of the system! And yet P(¬G)=0. So the system has theorems with probability zero...
EDIT: corrected the version of reflection to the version in Paul's paper (not the one in Eliezer's post) by removing two '.
I think this should be "Then P(G) <= 1".
No, that's correct. P(G)<1 is the premise of a proof by contradiction.
Ah. I didn't quite understand what you were trying to do there.
In general in this theory, I don't think we have (p(phi) < 1 -> p(phi) =1) -> p(phi) = 1. The theory only reflects on itself to within epsilon, so what the theory would conclude if it were certain about p(phi) being less than 1 can't be taken as the premise of a proof-by-contradiction in the same way that ~phi can be taken as a premise... ergh, I'm trying to figure out a more helpful way to state this than "I don't think you can derive that kind of probabilistic proof by contradiction from the stated properties of the theory because the system is never supposed to know the exact probabilities it assigns".
EDIT: Retracted pending further cogitation.
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).
I used the axioms as stated, and any non-intuitionist logic. If we want to capture your intuition, I think we'll have to tweak the definitions...