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by Alex Appel 114 days ago | link | parent

Intermediate update:

The handwavy argument about how you’d get propositional inconsistency in the limit of imposing the constraint of “the string cannot contain \(a\wedge b\wedge c...\to\neg\phi\) and \(a\) and \(b\) and… and \(\phi\)

is less clear than I thought. The problem is that, while the prior may learn that that constraint applies as it updates on more sentences, that particular constraint can get you into situations where adding either \(\phi\) or \(\neg\phi\) leads to a violation of the constraint.

So, running the prior far enough forward leads to the probability distribution being nearly certain that, while that particular constraint applied in the past, it will stop applying at some point in the future by vetoing both possible extensions of a string of sentences, and then less-constrained conditions will apply from that point forward.

On one hand, if you don’t have the computational resources to enforce full propositional consistency, it’s expected that most of the worlds you generate will be propositionally inconsistent, and midway through generating them you’ll realize that some of them are indeed propositionally inconsistent.

On the other hand, we want to be able to believe that constraints capable of painting themselves into a corner will apply to reality forevermore.

I’ll think about this a bit more. One possible line of attack is having \(\mathbb{P}(\phi)\) and \(\mathbb{P}(\neg\phi)\) not add up to one, because it’s possible that the sentence generating process will just stop cold before one of the two shows up, and renormalizing them to 1. But I’d have to check if it’s still possible to \(\varepsilon\)-approximate the distribution if we introduce this renormalization, and to be honest, I wouldn’t be surprised if there was a more elegant way around this.

EDIT: yes it’s still possible to \(\varepsilon\)-approximate the distribution in known time if you have \(\mathbb{P}(\phi)\) refer to \(\frac{probability to encounter \phi first}{1-probability to halt first}\), although the bounds are really loose. This is because if most of the execution paths involve halting before the sentence is sampled, \(\varepsilon\)-error in the probability of sampling \(\phi\) first will get blown up by the small denominator.

Will type up the proof later, but it basically proceeds by looking at the probability mass associated with “sample the trivial constraint that accepts everything, and sample it again on each successive round”, because this slice of probability mass has a near-guarantee of hitting \(\phi\), and then showing that even this tiny slice has substantially more probability mass than the cumulative probability of ever sampling a really rare sentence or not hitting any of \(\phi\), \(\neg\phi\), or the string terminating.





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