5

Lob's theorem states that:

Let $\textbf{Prov}(n^A)$ be the arithmetic statement such that $PA\vdash$ $A$ iff $\textbf{Prov}(n^A)$, where $PA$ is peano arithmetic, and $n^A$ is the godel number of $A$. Then

$$PA \vdash\textbf{Prov}(n^A)\rightarrow A\quad\quad\text{ implies} \quad \quad PA\vdash A$$

Where the same applies to any system that is at least as powerful as peano arithmetic.

My question is: Is there an intuitive explanation of why this is true? I've read an $n$ step proof in modal logic, but my intuition is not improved by it.

user56834
  • 12,323

1 Answers1

11

It's just a reformulation of Gödel's theorem : if you have some intuition for Gödel's theorem then you can transfer it to Löb's theorem.

Indeed if PA proves $Prov(A)\to A$ then it means that the provability of $A$ is enough to conclude the truth of $A$ : but we know that some models may believe that some things are provable while they're not : this is Gödel's theorem with $A= "0=1"$.

So if $Prov(A)\to A$ is provable it means that any model that thinks it has a proof of $A$ is not wrong about it : this can only happen if $A$ is true.

To see more precisely why it's just Gödel's theorem : assume you have such an $A$, and assume $T=$PA + $\neg A$ is consistent. In particular $T$ is a consistent recursively axiomatizable extension of PA, so it doesn't prove its own consistency (Gödel); in particular there is no $\varphi$ such that $T\vdash \neg Prov_T(\varphi)$.

However, $PA\vdash Prov_{PA}(A)\to A$ so $T\vdash \neg Prov_{PA}(A)$ hence $T\vdash \neg Prov_T(0=1)$ and so $T\vdash Con(T)$ : a contradiction.

Conversely, Löb's theorem can be used to prove Gödel's theorem for PA: if PA is consistent, then PA$\nvdash 0=1$ and so by Löb PA$\nvdash Prov_{PA}(0=1) \to (0=1)$ hence PA $\nvdash \neg Prov_{PA}(0=1)$: PA doesn't prove its own consistency (this of course works for any nice $T$ extending PA)

Addendum : I've reduced "intuition for Löb's theorem" to "intuition for Gödel's theorem" but haven't given the latter. For this one it's "simply" that a theory satisfying the hypotheses can actually talk about itself and its proofs and so can in effect reproduce the liar's paradox : there is a sentence that represents "I am not provable"

Maxime Ramzi
  • 45,086
  • Thanks for the answer. There's three things that make the interpretation of the answer a bit hard for me: Firstly, there's one part where a appears, but I don't know what type of statements those can be. Secondly, you might be overusing "$A$" a bit. You say "such an $A$", but you also seem to have a particular $A$ in mind, or "types" of $A$ (e.g. inconsistency encodings). Thirdly, can you specify what you mean by a "model thinking of having a proof". I'm not sure if you just mean if the statement is true in the model, or whether you switch proof framework away from the syntactic theory. – Nikolaj-K Aug 12 '22 at 22:16
  • 1
    @Nikolaj-K : $\varphi$ can be anything; and $A$ is the same as in the OP, that is, an $A$ such that $Prov_T(A) \to A$ is provable in $T$. Finally, "a model thinking it has a proof [of $B$]" means that $Prov_T(B)$ is satisfied in the model – Maxime Ramzi Aug 13 '22 at 15:31
  • Thanks for the clarifications. – Nikolaj-K Aug 13 '22 at 19:17