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

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

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

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 definition 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 $X$, if $\mathsf{PA}\vdash X$ then $\mathsf{PA} \vdash \Box X$

We need the following:

A1′: For each $X$, 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.