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paulfchristiano comments on Reflection in Probabilistic Logic - Less Wrong
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Comments (171)
I think including a condition with P in comprehension will require more subtlety---for example, such an axiom would not be consistent with most coherent distributions P . If I include the axiom P(x in S) = P(x not in x), then it better be the case that P(S in S) = 1/2.
We can still do it, but I don't see how to factor out the work into this paper. If you see a useful way to generalize this result to include axioms of L', that would be cool; I don't see one immediately, which is why it is in this form.
Sorry, we're miscommunicating somewhere. What I'm saying is that e.g. given a set
of statements, I want the axiom asserting the existence of the set %20%3E%201/2\})
, i.e. the comprehension axiom applied to the condition %20%3E%201/2)
. I don't understand how this would lead to %20=%20\mathbb{P}(x%20\notin%20x))
; could you explain? (It seems like you're talking about unrestricted comprehension of some sort; I'm just talking about allowing the condition in ordinary restricted comprehension to range over formulas in L'. Maybe the problem you have in mind only occurs in the unrestricted comprehension work which isn't in this draft?)
Consider my proposed condition that "
is consistent with 
for any coherent distribution 
". To see that this is true for ZFC in the language L', choose a standard model of ZFC in L and, for any function 
from the sentences of L' to 
, extend it to a model in L' by interpreting )
as )
; unless I'm being stupid somehow, it's clear that the extended model will satisfy ZFC-in-L' + 
.
It seems to me that the only parts of the proof that need to be re-thought are the arguments that (a)
and (b) )
are non-empty. Perhaps the easiest way to say the argument is that we extend (a) 
or (b) 
to some arbitrary complete theory 
, and set %20=%201)
if 
and 
otherwise.
I understand what you are saying. You are completely right, thanks for the observation. I don't have time to muck with the paper now, but it looks like this would work.