\documentclass[11pt]{amsart} \usepackage{geometry} \geometry{a4paper} \usepackage{graphicx} \usepackage{amssymb} \usepackage{mdframed} \usepackage{hyperref} \usepackage{lmodern} % Reproducible builds \pdfinfoomitdate=1 \pdftrailerid{} \pdfsuppressptexinfo=-1 \newmdtheoremenv{thm}{Theorem}[section] \newcommand{\prov}{\square} \newcommand{\encode}[1]{\ulcorner #1 \urcorner} \newcommand{\lob}{L\"ob's Theorem} \title{Parametric bounded version of L\"ob's theorem} \author{Patrick Stevens} \date{24th July 2016} \begin{document} \maketitle \tiny \begin{center} \url{https://www.patrickstevens.co.uk/misc/ParametricBoundedLoeb2016/ParametricBoundedLoeb2016.pdf} \end{center} \normalsize \section{Introduction} I was recently made aware of a preprint\cite{critch} of a paper which proves a bounded version of \lob. \ \begin{thm}[Parametric Bounded L\"ob] If $\prov A$ is the operator ``there exists a proof of $A$ in Peano arithmetic'' and $\prov_k A$ is the operator ``there exists a proof of $A$ in $k$ or fewer lines in Peano arithmetic'', then for every formula $p$ of one free variable in the language of PA, and every computable $f: \mathbb{N} \to \mathbb{N}$ which grows sufficiently fast, it is true that $$(\exists \hat{k})[\color{red}(\vdash [\forall k][\color{green}\prov_{f(k)} p(k) \to p(k) \color{red}])\color{black} \Rightarrow \color{blue}(\vdash [\forall k > \hat{k}][p(k)]) \color{black}]$$ \end{thm} \ (Colour is used only to emphasise logical chunks of the formula.) The paper gives plenty of motivation about why this result should be interesting and useful: section 6 of the paper, for instance, is an application to the one-shot Prisoner's Dilemma played between agents who have access to each other's source code. However, I believe that while the theorem may be true and the proof may be correct, its application may not be as straightforward as the paper suggests. \section{Background} \begin{thm}[\lob] Suppose $\prov \encode{A}$ denotes ``the formula $A$ with G\"odel number $\encode{A}$ is provable''. If $$\mathrm{PA} \vdash (\prov \encode{P} \to P)$$ then $$\mathrm{PA} \vdash P$$ \end{thm} \ \lob{} is at heart a statement about the incompatibility of the interpretation of the box as ``provable'' with the intuitively plausible deduction rule that $\prov \encode{P} \to P$. (``If we have a proof of $P$, then we can deduce $P$!'') The Critch paper has an example in Section 1.4 where $P$ is the Riemann hypothesis. \section{Problem with the paper} Suppose $\mathcal{M}$ is a model of Peano arithmetic, in which our agent is working. It is a fact of first-order logic (through the L\"owenheim-Skolem theorem) that there is no first-order way of distinguishing any particular model of PA. Therefore the model of PA could be non-standard; this is not something a first-order reasoning agent could determine. If the agent is working with a non-standard model of PA, then all the theorems of the Critch paper may well go through. However, they become substantially less useful, as follows. Let us write $M$ for the underlying class (or set) of the model $\mathcal{M}$ of PA. Then the statement $$(\exists \hat{k})[\color{red}(\vdash [\forall k][\color{green}\prov_{f(k)} p(k) \to p(k) \color{red}])\color{black} \Rightarrow \color{blue}(\vdash [\forall k > \hat{k}][p(k)]) \color{black}]$$ when relativised to the model $\mathcal{M}$ becomes $$(\exists \hat{k} \in M)[\color{red}(\vdash [\forall k \in M][\color{green}\prov^{\mathcal{M}}_{f(k)} p(k) \to p(k) \color{red}])\color{black} \Rightarrow \color{blue}(\vdash [\forall k \in M^{>\hat{k}}][p(k)]) \color{black}]$$ where $\prov^{\mathcal{M}}_{f(k)}$ is now shorthand for ``there is a proof-object $P$ in $M$ such that $P$ encodes a $M$-proof of $p(k)$ which is fewer than $f(k)$ lines long''. Notice that the quantifiers have been restricted to $M$; in particular, $\hat{k}$ might be a non-standard natural number. Likewise, the ``there is a proof'' predicate is now ``there is an object which $M$ unpacks into a proof''; but such objects may be non-standard naturals themselves, and unpack into non-standard proofs (which $\mathcal{M}$ still believes are proofs, because it doesn't know the difference between ``standard'' and ``non-standard''). \subsection{Aside: non-standard proof objects} What is a non-standard proof object? Let's imagine we have some specific statements $a_i$ for each natural $i$ such that $a_i \to a_{i+1}$ for each $i$, and such that $a_0$ is an axiom of PA. I'm using $a_i$ only for shorthand; the reader should imagine I had some specific statements and specific proofs of $a_i \to a_{i+1}$. Consider the following proof of $a_2$: \begin{enumerate} \item $a_0$ (axiom) \item $a_1$ (by writing out the proof of $a_0 \to a_1$ above this line) \item $a_2$ (by writing out the proof of $a_1 \to a_2$ above this line) \end{enumerate} If we take a simple G\"odel numbering scheme, namely ``take the number to be an ASCII string in base $256$'', it's easy to see that this proof has a G\"odel number. After all, we're imagining that I have specific proofs of $a_i \to a_{i+1}$, so I could just write them in. Then you're reading this document which was originally encoded as ASCII, so the G\"odel numbering scheme must have worked. Similarly, there is a G\"odel number corresponding to the following: \begin{enumerate} \item $a_0$ (axiom) \item $a_1$ (by writing out the proof of $a_0 \to a_1$ above this line) \item \dots \item $a_k$ (by writing out the proof of $a_{k-1} \to a_k$ above this line) \end{enumerate} Now, suppose we're working in a non-standard model, and fix non-standard $K$. Then there is a (probably non-standard) natural $L$ corresponding to the following proof: \begin{enumerate} \item $a_0$ (axiom) \item $a_1$ (by writing out the proof of $a_0 \to a_1$ above this line) \item \dots \item $a_K$ (by writing out the proof of $a_{K-1} \to a_K$ above this line) \end{enumerate} Now, this is not a ``proof'' in our intuitive sense of the word, because from our perspective it's infinitely long. However, the model still thinks this is a proof, and that it's coded by the (non-standard) natural $L$. \subsection{Implication for PBL} So the model $\mathcal{M}$ believes there is a natural $\hat{k}$ such that \dots But if that natural is non-standard (and remember that this is not something the model can determine without breaking into second-order logic!) then PBL doesn't really help us. It simply tells us that all sufficiently-large non-standard naturals have a certain property; but that doesn't necessarily mean any standard naturals have that property. And the application to the Prisoners' Dilemma in Critch's paper requires a standard finite $\hat{k}$. If we, constructing the agent Fairbot, could somehow guarantee that it would be working within the standard model of PA, then all would be well. However, we can't do that within first-order logic. It could be the case that when constructing Fairbot, the only sufficiently-large naturals turn out to be non-standard. When we eventually come to run $\mathrm{Fairbot}_k(\mathrm{Fairbot}_k)$, it could therefore be that it will take nonstandardly-many proof steps to discover the ``(coooperate, cooperate)'' outcome. In practice, therefore, the agents would not find that outcome: we can only run them for standardly-many steps, and all non-standard naturals look infinite to us. \section{Acknowledgements} My thanks are due to Mi\"etek Bak (who persuaded me that there might be a problem with the article) and to John Aspden (who very capably forced Mi\"etek to clarify his objection until I finally understood it). As ever, any mistakes in this article are due only to me. \begin{thebibliography}{9} \bibitem{critch} Andrew Critch, \emph{Parametric Bounded L\"ob's Theorem and Robust Cooperation of Bounded Agents}, \url{http://arxiv.org/abs/1602.04184v4} \end{thebibliography} \end{document}