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

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* <math>\mathsf{PA}' \vdash X \to \Box X</math> | * <math>\mathsf{PA}' \vdash X \to \Box X</math> | ||

− | * if <math>X</math>, then <math>\mathsf{PA} \vdash X</math> | + | * if <math>X</math>, then <math>\mathsf{PA} \vdash X</math> -- i think this is the one about which Larry says "ARGH WHAT ARE YOU DOING" |

==Other questions to answer== | ==Other questions to answer== |

## Revision as of 04:28, 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

## Contents

## Notation

Precedence:

- has high precedence, so e.g. means , and means .
- is a metalogical notion and has low precedence, so means .

## Translating the puzzle using logic notation

Löb's theorem shows that if , then .

The deduction theorem says that if , then .

Applying the deduction theorem to Löb's theorem gives us .

When translating to logic notation, it becomes obvious that the application of the deduction theorem is illegitimate, because we don't actually have . This is the initial answer that Larry D'Anna gives in comments.

But now, suppose we define , and walk through the proof of Löb's theorem for this new theory . Then we would obtain the following implication: if , then . But clearly, since is one of the axioms of . Therefore by modus ponens, we have , i.e. . Now we can apply the deduction theorem to obtain . This means that our "Löb's theorem" for must be incorrect (note: the proof *is* correct for , which is why Löb's theorem is a theorem; it's just incorrect for ), 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:

## Repeating the proof of Löb's theorem for modified theory

We now repeat the proof of Löb's theorem for to see where the error is.

- by definition of
- because is one of the axioms of

So far, everything is fine. But can we assert ? For , we had the following:

- A1: For each sentence , if then

We need the following:

- A1′: For each sentence , if then

Since has more axioms than , we know that can prove everything that can. Thus, compared to A1, both the antecedent and consequent of A1′ are stronger, so A1′ does not necessarily follow from A1.

Suppose we take in A1′ to be . Then we obtain

- If then

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

- if , then -- i think this is the one about which Larry says "ARGH WHAT ARE YOU DOING"

## Other questions to answer

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