User:IssaRice/Computability and logic/Eliezer Yudkowsky's Löb's theorem puzzle

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original link: https://web.archive.org/web/20160319050228/http://lesswrong.com/lw/t6/the_cartoon_guide_to_l%C3%B6bs_theorem/

current LW link: https://www.lesswrong.com/posts/ALCnqX6Xx8bpFMZq3/the-cartoon-guide-to-loeb-s-theorem

Translating the puzzle using logic notation

Löb's theorem shows that if PACC, then PAC.

The deduction theorem says that if PA{H}F, then PAHF.

Applying the deduction theorem to Löb's theorem gives us PA(CC)C.

When translating to logic notation, it becomes obvious that the application of the deduction theorem is illegitimate, because we don't actually have PA{CC}C. This is the initial answer that Larry D'Anna gives in comments.

But now, suppose we define PA':=PA{CC}, and walk through the proof of Löb's theorem for this new theory PA'. Then we would obtain the following implication: if PA'CC, then PA'C. But clearly, PA'CC since CC is one of the axioms of PA'. Therefore by modus ponens, we have PA'C, i.e. PA{CC}C. Now we can apply the deduction theorem to obtain PA(CC)C. This means that our "Löb's theorem" for PA' must be incorrect, and somewhere in the ten-step proof is an error.

Translating the Löb's theorem back to logic

http://yudkowsky.net/assets/44/LobsTheorem.pdf

Since the solution to the puzzle refers back to the proof of Löb's theorem, we first translate the proof from the cartoon version back to logic:

  1. PAL(LC)
  2. PACC
  3. PA(LC)(LC)
  4. PAL(LC)
  5. PALL
  6. PALC
  7. PALC
  8. PA(LC)
  9. PAL
  10. PAC

Repeating the proof of Löb's theorem for modified theory

  1. PA'L(LC)
  2. PA'CC
  3. PA'(LC)(LC)
  4. PA'L(LC)
  5. PA'LL
  6. PA'LC
  7. PA'LC
  8. PA'(LC)
  9. PA'L
  10. PA'C