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Comments on When is proof of provability a proof?

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When is proof of provability a proof?

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I am writing some notes on mathematical logic and I hit upon something I never really understood. I am pretty sure my question is elementary; nevertheless, I would appreciate help clearing my confusion.

We start with a bare-bones metatheory, enough to manipulate and reason about finite strings (I read that a weak fragment of arithmetic is sufficient but I don't know the details). In this setting we can study the syntax of first-order logic (FOL) and, in particular, formulate the ZFC axioms.

Using the language of ZFC, we can again develop the syntax of FOL (now everything is a set underneath rather than primitive symbols) and formulate the ZFC axioms. So ZFC can reason about ZFC. Thus far everything is clear to me.


Questions

Let $\varphi$ be a sentence in the language of ZFC. Suppose that

$$\mathrm{ZFC} \vdash (\mathrm{ZFC} \vdash \varphi).$$

In other words, ZFC proves that ZFC proves that $\varphi$ is true. Does it follow that $\mathrm{ZFC} \vdash \varphi$? I suspect the answer is yes, so how can I convince myself of that fact? Do I need to specify my metatheory more precisely? Do I need to assume ZFC is consistent?

More generally, if $T$ is a first-order theory and $\psi$ a sentence in the language of $T$, it makes sense to ask whether

$$\mathrm{ZFC} \vdash (T \vdash \psi) \quad\text{ implies }\quad T \vdash \psi.$$

Is this stronger statement true? What conditions does ZFC satisfy so that this holds?


Motivation

Suppose one wishes to study a first-order theory $T$. The metatheory is too weak to use standard mathematical techniques. However inside ZFC we have semantics (i.e. models which are sets) and we can prove Gödel's completeness and compactness theorems, and thus all of model theory applies to $T$. But I am left wondering whether results about $T$ derived using these methods translate directly to an honest proof in the weak metatheory.

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I suspect the answer is yes, so how can I convince myself of that fact? Do I need to specify my metatheory more precisely? Do I need to assume ZFC is consistent?

You're on the right track. The correct assumption to make here is known as $\Sigma_1^0$-soundness. A $\Sigma_1^0$-sound theory is only able to prove true $\Sigma_1^0$-statements—statements of the form $\exists n^{\in \mathbb{N}} (P(n))$ where $P(n)$ is an arithmetical statement with bounded quantifiers (or, equivalently, a statement checkable in bounded time by a Turing machine).

As Gödel showed, provability is a $\Sigma_1^0$-statement[1]. Conversely, all $\Sigma_1^0$-statements can be thought of as provability statements [2]. Hence $\Sigma_1^0$-soundness is exactly equivalent to proof of provability implying actual provability!


Two tangential notes on this question:

  1. Any arithmetic that is powerful enough to talk about addition and multiplication is incomplete. This was proven with the MRDP theorem that showed that universal computation (and hence self-referential statements, true-but-unprovable statements, etc.) can be implemented using Diophantine equations. However, the MRDP theorem itself I believe can be proven in PRA—a fragment of arithmetic in which only primitive-recursive functions can be defined.

  2. Nonstandard models of arithmetic provide a nice illustrative counterexample to $\Sigma_1^0$-soundness. Remember that Gödel showed that no theory can prove its own consistency; this means that every theory is consistent with a statement affirming its own inconsistency. So take, say, $\text{PA} + ¬\text{Con(PA)}$ (which must be consistent). This theory proves that $\exists n (n \text{ is the length of a proof of $\bot$ in PA})$. But since this $\Sigma_1^0$-statement is clearly false and yet provable in $\text{PA} + ¬\text{Con(PA)}$, we see that $\text{PA} + ¬\text{Con(PA)}$ is consistent but not $\Sigma_1^0$-sound. In other words there is a "nonstandard number" $n$—the length of the shortest proof of $\bot$ in PA—which $\text{PA} + ¬\text{Con(PA)}$ proves to exist but also to be greater than any "standard" (actually existing) number!



  1. In fact this, rather than his famous diagonalization argument, was the main difficulty in Gödel's proof of his incompleteness theorem! ↩︎

  2. (e.g. in a "theory" $T$ in which $T \vdash S$ simply means that there exists a "proof" $n \in \mathbb{N}$ such that $\phi_S(n)$ is true) ↩︎

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Small clarification needed (8 comments)
Small clarification needed
Hernán Ibarra Mejia‭ wrote 5 months ago · edited 5 months ago

Thanks, this seems to be exactly what I'm looking for. I am convinced that provability is in $\Sigma^0_1$ ("there exists a finite string of symbols such that they form a proof"). But something is still unclear to me. Do we know whether ZFC is $\Sigma^0_1$-sound?

If it helps, I'm taking the position of a ZFC-skeptic. For such a skeptic, the fact that ZFC proves that "there exists a derivation from a set of axioms $T$ of the sentence $\varphi$" might not convince them that there is actually a derivation of $\varphi$. After all, such a skeptic might say, ZFC is non-constructive so we have no witnesses to the $\Sigma^0_1$-sentence. Also ZFC might be inconsistent.

However, if we could convince the skeptic that ZFC is $\Sigma^0_1$-sound (better yet, that all arithmetical statements proved in ZFC are "true") then the skeptic might reluctantly agree that ZFC is useful in studying arithmetic (and hence first-order logic).

clemens‭ wrote 5 months ago · edited 5 months ago

Hernán Ibarra Mejia‭ No, certainly we don't. I looked at such things a while back, and a very open-minded skeptic might acquiesce to certain much weaker set theories (like Kripke–Platek set theory) but not to anything like $ZFC$.

That being said, what do you need ZFC for if you're just studying first-order logic?

clemens‭ wrote 5 months ago

Hernán Ibarra Mejia‭ Now, your intuition here is not incorrect that people can sometimes agree to the usefulness of theories on account of proofs of $\Sigma_1^0-soundness$. E.g., some people were convinced of the validity of Peano Arithmetic because it is $\Pi_1^0$-conservative (I think more, but I forget) over its intuitionistic variant, Heyting Arithmetic; i.e., any $\Pi_1^0$-statement provable in Peano Arithmetic is provable in Heyting Arithmetic. So such arguments do work sometimes; they just wouldn't work for ZFC.

clemens‭ Wow, I had no idea that ZFC is not known to be arithmetically sound; I find this very disturbing. While one can argue empirically for the consistency of ZFC, I have not seen any convincing reasons as to why ZFC should be arithmetically sound. I found some interesting literature in this direction, I will probably self-answer later explaining what I found.

That being said, what do you need ZFC for if you're just studying first-order logic?

Well there are countless examples of model theorists, who usually work in ZFC, deducing purely syntactic statements, including $\Sigma_0^1$-statements. Sometimes these methods are the only way we know how to prove even simple things like, e.g., constructive logic does not prove the law of excluded middle. But how can we trust these methods if we don't trust the arithmetical soundness of ZFC?

More generally, how can we trust, e.g., the proof of Fermat's last theorem which uses inaccesible cardinals?

clemens‭ Anyways, these philosophical questions are only rethorical; you've already clarified the situation for me. I will mark this answer as accepted.

Derek Elkins‭ wrote 5 months ago

Hernán Ibarra Mejia‭ You may find WHAT DOES IT TAKE TO PROVE FERMAT’S LAST THEOREM? PDF even if you were only using FLT as an example. This is not a technical paper but more a discussion around the disconnect between the strong assumptions Wiles' proof nominally requires and the (much) weaker assumptions that it probably actually requires.

clemens‭ wrote 5 months ago · edited 5 months ago

Hernán Ibarra Mejia‭ Yes, I too find it unsettling, especially given how mathematicians' intuitions on what constitutes a set have been proven wrong in the past (e.g. with Frege). As you probably know ZFC is based on the use of self-referential definitions: a set can be defined in terms of the properties of all sets. Hence it is called an "impredicative" theory (though the terms "predicative" and "impredicative" are not really formally defined). The theory $ATR_0$ (arithmetical transfinite recursion) is the "least powerful" impredicative theory: as its name suggests, it defines functions using general transfinite recursion and thus is able to define all ordinals definable using the $2$-element Veblen function (IIRC).

Personally I'm comfortable with some types of impredicativity (I find the existence of the Feferman-Schütte ordinal $\Gamma_0$ fairly intuitive, for instance) and not so much with others. It is an enjoyable philosophical investigation.

clemens‭ wrote 5 months ago

Hernán Ibarra Mejia‭ That being said, if ZFC is consistent (not necessarily sound), and it proves some other theory consistent, then the other theory must actually be consistent. Otherwise ZFC would prove a statement of the form $\forall x (P(x))$, whereas there would actually exist an $x$ such that $¬P(x)$, which would yield a finite proof of a contradiction. So I believe it's almost always only the consistency of (extended variants of) ZFC that matters for the model-theoretic proofs you're talking about.