# Are there different "levels" of self-referentiality in arithmetic?

Below, all sentences/formulas are first-order and in the language of arithmetic. For simplicity, we conflate numbers and numerals, and conflate sentences/formulas and their Godel numbers.

Given a formula $$\varphi(x)$$ and a sentence $$\theta$$, say that $$\theta$$ asserts its own $$\varphi$$-ness iff $$\mathsf{PA}\vdash\theta\leftrightarrow\varphi(\theta).$$ Let $$\mathsf{SR}(\theta)$$ be the set of $$\varphi$$ such that $$\theta$$ asserts its own $$\varphi$$-ness. Bounded truth predicates show that $$\mathsf{SR}(\theta)$$ is never empty. I'm curious how much $$\mathsf{SR}(\theta)$$ could actually depend on $$\theta$$; in particular, if there's a lot of potential variety here, this might give a meaningful notion of "degree of self-referentiality" of a sentence.

Here's one way to make this precise:

Are there sentences $$\theta,\theta'$$ such that $$\mathsf{SR}(\theta)\not\cong\mathsf{SR}(\theta')$$ as partial orders?

The partial ordering I have in mind is provability: $$\sigma\le\rho$$ iff $$\mathsf{PA}\vdash\forall x[\sigma(x)\rightarrow\rho(x)]$$. By considering bounded truth predicates, $$\mathsf{SR}(\theta)$$ always contains a copy of $$\mathbb{Z}$$: briefly, look at $$\tau_n^+(x)=$$ "$$x$$ is a true $$\Sigma_n$$ sentence" and $$\tau_n^-(x)=$$ "$$x$$ is not a false $$\Sigma_n$$-sentence" for $$n$$ sufficiently large. Another "canonical" element is the formula $$\varphi(x)\equiv\theta$$ (basically, "ignore input, output $$\theta$$"). Finally, $$\mathsf{SR}(\theta)$$ is always a distributive lattice. Beyond this, however, I don't see anything useful.

• Since the formulas $\varphi$ in SR($\theta$) have a free variable, when you say the order is provability, I guess you mean that PA proves $\varphi(x)\to\psi(x)$, i.e., $\forall x\ \varphi\to\psi$. Right? Dec 6, 2022 at 22:46
• @JoelDavidHamkins Yes, that's right. I've edited for clarity. Dec 6, 2022 at 22:47

If $$\varphi$$ is not required to behave the same way on Gödel codes of equivalent sentences or any such thing, then $$\mathsf{SR}(\theta)$$ is always equivalent to the preorder on all arbitrary formulas, by defining in PA a bijection $$f$$ between $$\mathbb{N} - \{\theta\}$$ and $$\mathbb{N}$$, and noting that any formula $$\varphi$$ gives rise to a formula $$\varphi' \in \mathsf{SR}(\theta)$$ via $$\varphi'(\theta) = \theta$$ and $$\varphi'(n) = \varphi(f(n))$$ for $$n \neq \theta$$.
We have that $$\varphi \leq \psi$$ iff $$\varphi' \leq \psi'$$, and that every formula $$\varphi \in \mathsf{SR}(\theta)$$ is equivalent to some $$\psi'$$ (specifically, take $$\psi(n) = \varphi(f^{-1}(n))$$). Thus, the map $$\varphi \mapsto \varphi'$$ is an equivalence from the preorder of arbitrary formulas to the preorder $$\mathsf{SR}(\theta)$$.
• Very nice. In particular, SR($\theta$) is a Boolean algebra under provable equivalence---for negation, you just negate all the other coordinates except the fixed-point coordinate, so it is still a fixed-point, but the conjunction is minimal amongst fixed-points. Dec 7, 2022 at 0:58
• Can we hope to use the same idea, but just change the formula on the coordinates of the provably equivalent to $\theta$ coordinates? PA proves this is an equivalence relation on the codes, even if it doesn't prove all the facts about that relation. Dec 7, 2022 at 1:55
• Yes, I believe we could just do that (where "provably equivalent to $\theta$ coordinates" means coordinates which satisfy a predicate in the language of PA asserting there is a proof that they are equivalent to $\theta$; thus, we would get different behavior at all coordinates $\psi$ for which PA proves $\Box (\theta \leftrightarrow \psi)$ and not just those coordinates for which PA proves $\theta \leftrightarrow \psi$). But I think that'd be enough to show a basically similar result that all $\mathsf{SR}(\theta)$ are the same, in that context, depending on how exactly we formalize things.). Dec 7, 2022 at 1:59