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

From Machinelearning

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 PA⊢◻C→C, then PA⊢C.

The deduction theorem says that if PA∪{H}⊢F, then PA⊢H→F.

Applying the deduction theorem to Löb's theorem gives us PA⊢(◻C→C)→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∪{◻C→C}⊢C. This is the initial answer that Larry D'Anna gives in comments.

But now, suppose we define PA':=PA∪{◻C→C}, and walk through the proof of Löb's theorem for this new theory PA'. Then we would obtain the following implication: if PA'⊢◻C→C, then PA'⊢C. But clearly, PA'⊢◻C→C since ◻C→C is one of the axioms of PA'. Therefore by modus ponens, we have PA'⊢C, i.e. PA∪{◻C→C}⊢C. Now we can apply the deduction theorem to obtain PA⊢(◻C→C)→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. PA⊢◻L↔◻(◻L→C)
  2. PA⊢◻C→C
  3. PA⊢◻(◻L→C)→(◻◻L→◻C)
  4. PA⊢◻L→(◻◻L→◻C)
  5. PA⊢◻L→◻◻L
  6. PA⊢◻L→◻C
  7. PA⊢◻L→C
  8. PA⊢◻(◻L→C)
  9. PA⊢◻L
  10. PA⊢C