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Let $\mathbb T$ be a probability distribution over complete and consistent theories of first order arithmetic that contain $PA$. Additionally, we will require that for any sentence $\phi$ in the language of arithmetic, $P(\mathbb T \vdash \phi)$ is defined (that is, $\{T : T \in dom(\mathbb T), T \vdash \phi\}$ is a measurable set).

Is there any such $\mathbb T$ such that $p_{\mathbb T}(\phi) = P(\mathbb T \vdash \phi)$ is a computable function?

Christopher King
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  • Slightly related – Christopher King Feb 19 '19 at 13:44
  • "Given an oracle for $p_\mathbb{T}$" Given that $p_\mathbb{T}$ is a map from sentences to reals, we have to be careful with what this means. In particular, your claim that $p_\mathbb{T}\ge_T\mathbb{T}_{a,s}$ is not true for all such notions. – Noah Schweber Feb 19 '19 at 14:27
  • @NoahSchweber oh yeah, I forgot the subtleties of real numbers and computation. I deleted that paragraph. – Christopher King Feb 19 '19 at 14:35
  • Outsider question: is the set of theories of first order arithmetic countable? If not, how do you define a sigma algebra on it? – Jack M Feb 19 '19 at 14:43
  • It is easier to replace it by $P($ ZFC proves $\mathbb{T} $ is consistent $\land \mathbb{T} \vdash \phi)$ because you can then choose a distribution on an enumerable set of integers $S$, interpret the $s_j \in S$ as theories $T_j$, enumerate the theorems of ZFC trying to prove $T_j$ is consistent and proves $\phi$, return $\sum_{j=1}^J P(s_j) 1_{ZFC \vdash \text{ in J steps} (T_j \text{ is consistent} \land T_j \vdash \phi)}$ and ask if a bound for the remainder $\sum_{j > J} P(s_j) ...$ is computable – reuns Feb 19 '19 at 15:07
  • @JackM The real numbers aren't countable, but we can define a $\sigma$-algebra on them ... – Noah Schweber Feb 19 '19 at 15:10
  • @NoahSchweber Yes, but the point is, you wouldn't ask a question about probability distributions on the reals without specifying the sigma algebra (well, you would, but only because there happens to be one very commonly used sigma algebra for $\mathbb R$). – Jack M Feb 19 '19 at 15:12
  • @JackM There also is for the set of all theories - via Godel numbering, the set of all sets of first-order sentences (in a fixed countable language) is in bijection with Cantor space $2^\omega$, which comes with a natural $\sigma$-algebra since it comes with a natural topology. (And this topological approach theories is standard in logic - e.g. its why the "compactness theorem" is called that.) – Noah Schweber Feb 19 '19 at 15:19
  • @JackM yes, the cantor space one. – Christopher King Feb 19 '19 at 22:15
  • @JackM although, I would be fine with the one generated by the sets in the first paragraph. – Christopher King Feb 19 '19 at 22:35

1 Answers1

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Unless I'm missing something, there is not. The key observations are $(i)$ positive probability implies consistency and $(ii)$ we can't split our space into two measure-zero sets.

First, we need to settle on how we're representing such a $p_\mathbb{T}=p$ so that we can talk about its (non)computability. I'm going to do the following: we conflate $p$ with the set $$\{\langle \varphi,q\rangle\in Sent\times\mathbb{Q}_{>0}: p(\varphi)\ge q\}.$$

(But any reasonable representation should give the same result.)

Suppose I have such a $p$ as an oracle; I'll show how to use $p$ to compute a (consistent) completion of PA.

Fix an enumeration $(a_i)_{i\in\omega}$ of the positive rationals. We say that a sentence $\varphi$ is $p$-consistent iff there is some $i$ such that $\langle \varphi, a_i\rangle\in p$ and no $j<i$ has $\langle \neg\varphi,a_j\rangle\in p$. Note that we can have both $\varphi$ and $\neg\varphi$ be $p$-consistent; we could modify the definition to rule this out, but there's no need to.

The key points now are:

  • In order for $p$ to indeed come from a probability distribution, for every sentence $\varphi$ we must have at least one of $\varphi$ and $\neg\varphi$ be $p$-consistent. Otherwise $\mathbb{T}$ would assign probability $0$ to both $\{T: \varphi\in T\}$ and $\{T: \neg\varphi\in T\}$, hence to the whole space.

  • Given the previous bulletpoint, the set of $p$-consistent sentences is computable from $p$; to tell whether $\varphi$ is $p$-consistent, we just search through the positive rationals until you find the first positive rational witnessing $p$-consistency of at least one of $\varphi$ or $\neg\varphi$.

  • Finally, if $\varphi$ is $p$-consistent, then PA $\cup\{\varphi\}$ is consistent. (Otherwise $\{T: \varphi\in T\}$ isn't even a subset of our probability space!)

So now we just use the usual greedy algorithm:

  • Enumerate the sentences as $(\varphi_k)_{k\in\omega}$.

  • Define the sequence $(\psi_k)_{k\in\omega}$ of sentences recursively by setting $\psi_k=\varphi_k$ if $\varphi_k\wedge(\bigwedge_{l<k}\psi_l)$ is $p$-consistent, and $\psi_k=\neg\varphi_k$ otherwise.

  • The theory PA $\cup\{\psi_k:k\in\omega\}$ is then a consistent completion of PA which is $p$-computable.

In particular, any such $p$ is of PA degree - and conversely, given a PA degree $\bf d$ we can construct such a $p$ (just focus on some fixed completion of PA computed by $\bf d$). So in fact we've (unsurprisingly) characterized the degrees of such $p$s as precisely the PA degrees.

Noah Schweber
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  • So basically we start with PA, and add a statement or negation to PA based on which one has positive probability (together with the previous axioms), right? And do that for each statement in sequence. – Christopher King Feb 22 '19 at 20:43
  • @PyRulez Yup. The only issue is to make sure we know what to do in case both have positive probability - and that's easy, we just go with the "first" one. – Noah Schweber Feb 22 '19 at 21:54
  • I guess you'd technically get a lot of redundant axioms too (i.e. equivalent in PA), right? – Christopher King Feb 22 '19 at 21:58
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    @PyRulez Sure, but PA itself is extremely redundant (the induction scheme isn't minimal in any sense). – Noah Schweber Feb 22 '19 at 22:00