They can’t be both valid, because only one of the two statements is true. We assume that the predictor’s formal theory is sound (doesn’t prove false statements).
True, but still underspecified. They can’t both be valid at the same time, but the point is that a priori it’s possible that either one could be valid, and you don’t know which one without knowing additional details. If the agent one-boxes whenever the predictor predicts it one-boxes, and two-boxes whenever the predictor predicts it two-boxes, the predictor’s actual prediction is arbitrary—it will depend on the specific details of the predictor’s proof-searching algorithm. I’m guessing this means that the proof itself would have to involve those details in some way.
I think the UDT agent will still two-box in that case, for the same reason as before. To have FairBot-style cooperation, the agent must also search for one-sided proofs only.
That would depend on the details of the UDT agent, wouldn’t it? It’s hard to say without those details. If you’re assuming the proof-searching model then yes, but that’s the same old spurious proof problem isn’t it?
Why? If an agent one-boxes when both boxes are filled, that doesn’t necessarily mean the agent one-boxes unconditionally.
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.
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...
True, but still underspecified. They can’t both be valid at the same time, but the point is that a priori it’s possible that either one could be valid, and you don’t know which one without knowing additional details. If the agent one-boxes whenever the predictor predicts it one-boxes, and two-boxes whenever the predictor predicts it two-boxes, the predictor’s actual prediction is arbitrary—it will depend on the specific details of the predictor’s proof-searching algorithm. I’m guessing this means that the proof itself would have to involve those details in some way.
That would depend on the details of the UDT agent, wouldn’t it? It’s hard to say without those details. If you’re assuming the proof-searching model then yes, but that’s the same old spurious proof problem isn’t it?
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.
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...