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

This page gives spoilers for the puzzle!

This pages states the puzzle and gives the solution.

Notation

Precedence:

  • \Box has high precedence, so e.g. \Box C \to C means (\Box C) \to C, and \Box \Box L \to \Box C means (\Box \Box L) \to (\Box C).
  • \vdash is a metalogical notion and has low precedence, so \mathsf{PA} \vdash \Box C \to C means \mathsf{PA} \vdash (\Box C \to C).

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 (note: the proof is correct for \mathsf{PA}, which is why Löb's theorem is a theorem; it's just incorrect for \mathsf{PA}'), 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. \mathsf{PA} \vdash \Box L \leftrightarrow \Box(\Box L \to C) by construction of L
  2. \mathsf{PA} \vdash \Box C \to C by hypothesis
  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

We now repeat the proof of Löb's theorem for \mathsf{PA}' := \mathsf{PA} \cup \{\Box C \to C\} to see where the error is.

  1. \mathsf{PA}' \vdash \Box L \leftrightarrow \Box(\Box L \to C) by construction of L
  2. \mathsf{PA}' \vdash \Box C \to C because \Box C \to C is one of the axioms of \mathsf{PA}'
  3. \mathsf{PA}' \vdash \Box(\Box L \to C) \to (\Box \Box L \to \Box C) instance of A3
  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

So far, everything is fine. But can we assert \mathsf{PA}' \vdash \Box(\Box L \to C)? For \mathsf{PA}, we had the following:

A1: For each sentence X, if \mathsf{PA}\vdash X then \mathsf{PA} \vdash \Box X

We need the following:

A1′: For each sentence X, if \mathsf{PA}'\vdash X then \mathsf{PA}' \vdash \Box X

Since \mathsf{PA}' has more axioms than \mathsf{PA}, we know that \mathsf{PA}' can prove everything that \mathsf{PA} can. Thus, compared to A1, both the antecedent and consequent of A1′ are weaker, so A1′ does not necessarily follow from A1.

Suppose we take X in A1′ to be \Box C \to C. Then we obtain

If \mathsf{PA}'\vdash \Box C \to C then \mathsf{PA}' \vdash \Box (\Box C \to C)

Other ways we might try to get step 8 to work:

  • \mathsf{PA}' \vdash X \to \Box X
  • if X, then \mathsf{PA} \vdash X -- i think this is the one about which Larry says "ARGH WHAT ARE YOU DOING". Or maybe he means "if \Delta \vdash X, then \Delta \vdash \Box X" for arbitrary \Delta.
    • Now one might wonder, why is \Delta=\mathsf{PA} so special? I think this happens because \Box is the provability predicate of \mathsf{PA} and not some other theory. In particular, if \Delta \supseteq \mathsf{PA}, then if \Delta \vdash X we only know the weaker statement that \Delta proves X, not that \mathsf{PA} proves it. So we shouldn't be able to prove that \Delta can prove \Box X. What if \Delta \subseteq \mathsf{PA}? Then I think we are in a better position. e.g. if \Delta = \mathsf{Q}, then I think \mathsf Q can still represent the provability predicate of \mathsf{PA} (?) so then we have \mathsf Q \vdash \Box X. (this part might be wrong)

But can we prove that we can never get step 8 for \mathsf{PA}'? Well, this is exactly what the puzzle statement does -- it assumes the proof goes through, and derives an absurdity! (? actually, i'm not sure this is actually a contradiction. If we assume PA is consistent, then it seems to be a contradiction. I think this is the point nate soares and alex flint are trying to make in the comments. but we can re-word the puzzle to say one should find the first error.)

Other questions to answer

  • does \Box in \mathsf{PA}' still mean provability about \mathsf{PA}, or about itself?
  • are we removing a layer or adding one?
  • can't \mathsf{PA}' prove everything \mathsf{PA} can? yes, but this doesn't mean we can substitute implications automatically because the consequent also gets stronger