by Sam Eisenstat 1010 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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