Yeah, the problem i have with that though is that I'm left asking: why did I change my probability in that? Is it because i updated on something else? Was I certain of that something else? If not, then why did I change my probability of that something else, and on we go down the rabbit hole of an infinite regress.
Wait, actually, I'd like to come back to this. What programming language are we using? If it's one where either grue is primitive, or one where there are primitives that make grue easier to write than green, then true seems simpler than green. How do we pick which language we use?
Here's my problem. I thought we were looking for a way to categorize meaningful statements. I thought we had agreed that a meaningful statement must be interpretable as or consistent with at least one DAG. But now it seems that there are ways the world can be which can not be interpreted even one DAG because they require a directed cycle. SO have we now decided that a meaningful sentence must be interpretable as a directed, cyclic or acyclic, graph?
In general, if I say all and only statements that satisfy P are meaningful, then any statement that doesn't satisfy P must be meaningless, and all meaningless statements should be unobservable, and therefor a statement like "all and only statements that satisfy P are meaningful" should be unfalsifiable.
What is Markov relative?
Does EY give his own answer to this elsewhere?
Wait... this will seems stupid, but can't I just say: "there does not exist x where sx = 0"
Here's a new strategy.
Use guess culture as a default. Use guess tricks to figure out whether other communicator speaks Ask. Use Ask tricks to figure out whether communicator speaks Tell.
Let's forget about the oracle. What about the program that outputs X only if 1 + 1 = 2, and else prints 0? Let's call it A(1,1). The formalism requires that P(X|A(1,1)) = 1, and it requires that P(A(1,1)) = 2 ^-K(A(1,1,)), but does it need to know that "1 + 1 = 2" is somehow proven by A(1,1) printing X?
In either case, you've shown me something that I explicitly doubted before: one can prove any provable theorem if they have access to a Solomonoff agent's distribution, and they know how to make a program that prints X iff theorem S is provable. All they have to do is check the probability the agent assigns to X conditional on that program.
Awesome. I'm pretty sure you're right; that's the most convincing counterexample I've come across.
I have a weak doubt, but I think you can get rid of it:
let's name the program FTL()
I'm just not sure this means that the theorem itself is assigned a probability. Yes, I have an oracle, but it doesn't assign a probability to a program halting; it tells me whether it halts or not. What the Solomoff formalism requires is that "if (halts(FTL()) == true) then P(X|FTL()) = 1" and "if (halts(FTL()) == false) then P(X|FTL()) = 0" and "P(FTL()) = 2^-K(FTL())". Where in all this is the probability of Fermat's last theorem? Having an oracle may imply knowing whether or not FTL is a theorem, but it does not imply that we must assign that theorem a probability of 1. (Or maybe, it does and I'm not seeing it.)
Edit: Come to think of it... I'm not sure there's a relevant difference between knowing whether a program that outputs True iff theorem S is provable will end up halting, and assigning probability 1 to theorem S. It does seem that I must assign 1 to statements of the form "A or ~ A" or else it won't work; whereas if the theorem S is is not in the domain of our probability function, nothing seems to go wrong.
In either case, this probably isn't the standard reason for believing in, or thinking about logical omniscience because the concept of logical omniscience is probably older than Solomonoff induction. (I am of course only realizing that in hindsight; now that I've seen a powerful counter example to my argument.)
Upvoted for cracking me up.