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by Sam Eisenstat 877 days ago | link | parent

We can also construct an example where \({\rm PA} \vdash \phi \rightarrow \psi\) with a short proof and \(\rm PA\) also proves \(\phi \rightarrow \neg\psi\), but any such proof is much longer. We only need to put a bound on the proof length in A’s proof search. Then, the argument that \({\tt A()} = 1 \wedge {\tt U()} \ne 0\) proves its own consistency still works, and is rather short: \(O(\log n)\) as the proof length bound \(n\) increases. However, there cannot be a proof of \({\tt A()} = 1 \rightarrow {\tt U()} = 10\) within \({\tt A}\)’s proof length bound, since if it found one it would immediately take action 1. In this case \(\rm PA\) can still prove that \({\tt A()} = 1 \rightarrow {\tt U()} = 10\) simply by running the agent, but this argument shows that any such proof must be long.



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