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All of Quarkster's Comments + Replies

Where Recursive Justification Hits Bottom
Quarkster20

You got my meaning. I have a bad habit of under-explaining things.

As far as the second part goes, I'm wary of the math. While I would imagine that your argument would tend to work out much of the time, it certainly isn't a proof, and Bayes' Theorem doesn't deal with the respective complexity of the canonical events A and B except to say that they are each more probable individually than separately. Issues of what is meant by the complexity of the events also arise. I suspect that if your assertion was easy to prove, then it would have been proven by ... (read more)

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Where Recursive Justification Hits Bottom
Quarkster40

No, because you can't say anything about the relationship of P(A) in comparison to P(C|D)

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1Psy-Kosh
Not sure why you were silently voted down into negatives here, but if I understand your meaning correctly, then you're basically saying this: P(A)*P(B|A) vs P(C) aren't automatically comparable because C, well, isn't A? I'd then say "if C and A are in "similar terms"/level of complexity... ie, if the principle of indifference or whatever would lead you to assign equivalent probabilities to P(C) and P(A) (suppose, say, C = ~A and C and both have similar complexity), then you could apply it. (or did I miss your meaning?)
Belief in Belief
Quarkster40

"Those who find this confusing may find it helpful to study mathematical logic, which trains one to make very sharp distinctions between the proposition P, a proof of P, and a proof that P is provable"

This is a bit of a side question, but wouldn't a proof that P is provable be a proof of P? In fact, it sounds like a particularly elegant form of proof.

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0LizzardWizzard
Here's an example of what Doug Hofstadter writes in I Am A Strange Loop. Kurt Goedel discovered that Principia Mathematica by Bertrand Russell does provide reference to itself. So Russell in his book yields the propositions and their proofs, and then Goedel assigns specific numbers to proofs and therefore proves that there is a proof that they are in fact, provable
0Anomylous
Outside of mathematics, a statement that is provable is also disprovable. Then it's called a hypothesis. I'm reminded of the joke where an engineer, a physicist, and a mathematician are going to a job interview. The interviewer has rigged a fire to start in the wastepaper basket, to see how they react in a crisis situation. The engineer sees the fire, sees the water cooler, grabs the water cooler and dumps it on the fire. The physicist sees the fire, sees the water cooler, grabs pencil and paper, calculates the exact amount of water needed to extinguish the fire, then pours that amount of water into the basket, exactly extinguishing the fire. The mathematician sees the fire, sees the water cooler, and says, "Ah! A solution exists!".
6Eliezer Yudkowsky
If you trust base system B, then a proof that P is provable in B is good as gold to you. But it is not a proof in B. http://lesswrong.com/lw/t6/the_cartoon_guide_to_l%C3%B6bs_theorem/
3Psy-Kosh
Hrm... if the system is isn't necessarily trustworthy, then that the system proves that it can prove P doesn't mean that it's actually true that it can prove P, I guess. EDIT: actually, having it as an explicit axiom "If this proves P, then P" runs you into trouble in any system that has something like Lob's theorem. ("if some specific subset of the rest of this system, (ie, other than this axiom) proves P, then P" can potentially be okay, though)
2RobinZ
Seconded - this is an interesting question. (And I suspect that there are some interesting cases in which a proof that P is provable does not constitute a proof, but this is mainly because I've seen mathematicians break similarly intuitive propositions before.)