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

From Machinelearning
Jump to: navigation, search
(Translating the puzzle using logic notation)
Line 31: Line 31:
# <math>\mathsf{PA} \vdash \Box L</math>
# <math>\mathsf{PA} \vdash \Box L</math>
# <math>\mathsf{PA} \vdash C</math>
# <math>\mathsf{PA} \vdash C</math>
==Repeating the proof of Löb's theorem for modified theory==
# <math>\mathsf{PA}' \vdash \Box L \leftrightarrow \Box(\Box L \to C)</math>
# <math>\mathsf{PA}' \vdash \Box C \to C</math>
# <math>\mathsf{PA}' \vdash \Box(\Box L \to C) \to (\Box \Box L \to \Box C)</math>
# <math>\mathsf{PA}' \vdash \Box L \to (\Box \Box L \to \Box C)</math>
# <math>\mathsf{PA}' \vdash \Box L \to \Box \Box L</math>
# <math>\mathsf{PA}' \vdash \Box L \to \Box C</math>
# <math>\mathsf{PA}' \vdash \Box L \to C</math>
# <math>\mathsf{PA}' \vdash \Box(\Box L \to C)</math>
# <math>\mathsf{PA}' \vdash \Box L</math>
# <math>\mathsf{PA}' \vdash C</math>

Revision as of 03:33, 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.

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

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

Translating the Löb's theorem back to logic


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

Repeating the proof of Löb's theorem for modified theory

  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