If you’re assuming the proof-searching model then yes, but that’s the same old spurious proof problem isn’t it?
Yeah.
If an agent one-boxes when both boxes are filled then the predictor will predict that the agent one-boxes when both boxes are filled, and so both boxes will be filed, and so the agent one-boxes unconditionally.
I’m confused, which implementation of the predictor are you assuming in that sentence? I don’t think that every predictor will be able to figure out every true statement about the agent...
Yeah, a predictor won’t necessarily figure out every true statement, but my original point was about the transparent Newcomb’s, in which case if Omega finds that the agent one-boxes whenever Omega fills both boxes, then Omega will fill both and the agent will one-box.
In the case of the proof-searching model things aren’t quite as nice, but you can still get pretty far. If the predictor has a sufficiently high proof limit and knows that it fills both boxes whenever it finds a proof of one-boxing, then a short enough proof that the agent one-boxes whenever both boxes are filled should be sufficient (I’m quite sure a Löbian argument holds here).
So yes, from the predictor’s point of view, a proof that the agent one-boxes whenever both boxes are filled should be sufficient. Now you just need the agent to not be stupid...
Yeah.
I’m confused, which implementation of the predictor are you assuming in that sentence? I don’t think that every predictor will be able to figure out every true statement about the agent...
Yeah, a predictor won’t necessarily figure out every true statement, but my original point was about the transparent Newcomb’s, in which case if Omega finds that the agent one-boxes whenever Omega fills both boxes, then Omega will fill both and the agent will one-box.
In the case of the proof-searching model things aren’t quite as nice, but you can still get pretty far. If the predictor has a sufficiently high proof limit and knows that it fills both boxes whenever it finds a proof of one-boxing, then a short enough proof that the agent one-boxes whenever both boxes are filled should be sufficient (I’m quite sure a Löbian argument holds here).
So yes, from the predictor’s point of view, a proof that the agent one-boxes whenever both boxes are filled should be sufficient. Now you just need the agent to not be stupid...