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

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current LW link: https://www.lesswrong.com/posts/ALCnqX6Xx8bpFMZq3/the-cartoon-guide-to-loeb-s-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 <math>\mathsf{PA} \vdash \Box C \to C</math>, then <math>\mathsf{PA} \vdash C</math>.
The deduction theorem says that if <math>\mathsf{PA} \cup \{H\} \vdash F</math>, then <math>\mathsf{PA} \vdash H \to F</math>.
Applying the deduction theorem to Löb's theorem gives us <math>\mathsf{PA} \vdash (\Box C \to C) \to C</math>.


==Translating the Löb's theorem back to logic==
==Translating the Löb's theorem back to logic==

Revision as of 03:19, 10 February 2019

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.

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