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Jack comments on Logical Uncertainty as Probability - Less Wrong
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Comments (22)
As written axiom 1 is just false:
But given any finite set (X) of true statements in Peano arithmetic I can invent a statement in Peano arithmetic that isn't in that set but that you would instantly assign p=1.
Edit: And it's false in the other direction too! Humans aren't logically omniscient in the propositional calculus!
Anyway, I think logical uncertainty becomes demystified pretty fast when you start thinking about it in terms of computing time and computational complexity (though that isn't to say I have a solution for the problem of representing it Bayesian calculus).
It is entirely possible for P(S|X) to be < 1, although P(S|X∪{S}) = 1. There's nothing inconsistent about updating on new evidence.
So then it isn't true that P(S|X) = 1 if and only if S∈OLC(X).
If X = { "2*2=4" }, then P("2*2=4 ∨ 0=1" | X) = 1, because "2*2=4 ∨ 0=1" ∈ OLC(X).
However, P("2*3=6" | X) < 1, although P("2*3=6" | X ∪ {"2*3=6"}) = 1.