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

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(Translating the Löb's theorem back to logic)
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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
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==Translating the puzzle using logic notation==
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Löb's theorem shows that if <math>\mathsf{PA} \vdash \Box C \to C</math>, then <math>\mathsf{PA} \vdash C</math>.
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The deduction theorem says that if <math>\mathsf{PA} \cup \{H\} \vdash F</math>, then <math>\mathsf{PA} \vdash H \to F</math>.
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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 \mathsf{PA} \vdash \Box C \to C, then \mathsf{PA} \vdash C.

The deduction theorem says that if \mathsf{PA} \cup \{H\} \vdash F, then \mathsf{PA} \vdash H \to F.

Applying the deduction theorem to Löb's theorem gives us \mathsf{PA} \vdash (\Box C \to C) \to 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. \mathsf{PA} \vdash \Box L \leftrightarrow \Box(\Box L \to C)
  2. \mathsf{PA} \vdash \Box C \to C
  3. \mathsf{PA} \vdash \Box(\Box L \to C) \to (\Box \Box L \to \Box C)
  4. \mathsf{PA} \vdash \Box L \to (\Box \Box L \to \Box C)
  5. \mathsf{PA} \vdash \Box L \to \Box \Box L
  6. \mathsf{PA} \vdash \Box L \to \Box C
  7. \mathsf{PA} \vdash \Box L \to C
  8. \mathsf{PA} \vdash \Box(\Box L \to C)
  9. \mathsf{PA} \vdash \Box L
  10. \mathsf{PA} \vdash C