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by Sam Eisenstat 1107 days ago | Benja Fallenstein and Patrick LaVictoire like this | link | parent

The argument that I had in mind was that if \(\rm{PA} \vdash \tt{A()} \ne 1\), then \(\rm{PA} \vdash \square \ulcorner \tt{A()} \ne 1 \urcorner\), so \(\rm{PA} \vdash \tt{A()} = 1\) since PA knows how the chicken rule works. This gives us \(\rm{PA} \vdash \bot\), so PA can prove that if \(\rm{PA} \vdash \tt{A()} \ne 1\), then PA is inconsistent. I’ll include this argument in my post, since you’re right that this was too big a jump.

Edit: We also need to use this argument to show that the modal UDT agent gets to the part where it iterates over utilities, rather than taking an action at the chicken rule step. I didn’t mention this explicitly, since I felt like I had seen it before often enough, but now I realize it is nontrivial enough to point out.





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