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

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

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. ${\mathsf {PA}}'\vdash \Box (\Box L\to C)\to (\Box \Box L\to \Box C)$
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 stronger, 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"

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