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Recall that whether ZF is consistent is essentially a combinatorial problem about a string rewriting system: We define a set of symbols $S$ to write with, and the subset of all strings $s \in S^{\star}$ that are well-formed formula. Then we define a certain set of axioms (each is a well-formed formula), and a set of deduction rules that help us generate new sentences (this generation is concrete enough to be put into a computer, see for example metamath). A sentence that can be generated with this way is called a theorem. Then we say that ZF is inconsistent if all statements are theorems; otherwise we say it is consistent. NOTE It has nothing to do with model theory up to this point!

Recall that we still do not know whether ZF is consistent or not (cf Nelson's program). On the other hand, it was a famous theorem of Godel that ZF will never be proven to be consistent using ZF. Inspired by that PA cannot prove its own consistency either, but ZFC (stronger than PA) can prove the consistency of PA (cf Con(PA) in ZFC), I wonder whether it is possible to prove its consistency using a stronger system.

However, the answer of the question is stupidly and obviously yes. I can use ZF + Con(ZF) as the stronger axiom system, which tautologically proves the consistency of ZF! This is why I recalled that the consistency problem is essentially combinatorial in the first paragraph. Indeed, I'm not happy just with a proof using a stronger axiom system, I want to see a proof for ZF's consistency which also guarantees that if such a string rewriting algorithm is run in reality, it will never generate a statement $\phi$ and its negated statement $\neg \phi$.

Question. Is such a proof and guarantee possible?

Please note that such a guarantee falls external to mathematics. I'm really concerned with whether we will one day find a proof for some $\phi$ and $\neg \phi$ from a computer program (assuming that it's "correct").

Footnote. Proving the consistency of ZFC in some other system touched almost the same issue, but the accepted answer did not go into the direction I was asking for.

Candidate Answer. All answers so far are good, but I think @spaceisdarkgreen's addresses my concern in a direct way. It also points out that currently there isn't such answer. So I leave the decision open.

Student
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    Does the ZFC proof of consistency of PA "imply there cannot be a statement $\phi$ such that $\phi$ and $\lnot \phi$ are theorems of PA" (which is just another way of saying it implies that PA is consistent)? If so, why does the one-line proof of Con(ZF) from ZF + Con(ZF) not? – spaceisdarkgreen Oct 01 '24 at 00:17
  • @AlexKruckman I'm looking for a mathematical proof that also guarantees $\phi$ and $\neg \phi$ will never be generated from ZF using the string rewriting algorithm, if run in the reality. I have edited the question. – Student Oct 01 '24 at 00:32
  • @Student How does the proof of Fermat's last theorem guarantee that there really are not any nontrivial solutions of $x^n+y^n=z^n$ for $n>2,$ if computed in reality? Consistency means that the string rewriting algorithm can't do that... when you prove consistency you are proving that. You are asking a much more general question of how we can have a "proof we can trust to correspond to reality" about anything. (And as I mentioned before the same question applies to proofs of Con(PA) from "stronger axioms"... what makes those trustworthy?) – spaceisdarkgreen Oct 01 '24 at 00:43
  • @spaceisdarkgreen Thanks for your comment, and I'm glad you understand my point because it's indeed about having a proof that we can trust to correspond to reality. Whatever the answer is, it won't be a purely mathematical answer, but I think there should be an answer. Fermat's theorem is perhaps too daunting; but I believe our proof for the theorem that there is no integer $n$ such that $n > n+1$ does correspond to something in the "reality". On the other hand, proving Con(ZF) from the axiom Con(ZF) shows nothing about the reality. – Student Oct 01 '24 at 00:49
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    @Student I wrote an answer related to this recently that doesn’t really add that much to Giorgio’s below https://math.stackexchange.com/a/4969963/397125 – spaceisdarkgreen Oct 01 '24 at 00:55
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    @Student You have to think for yourself what you find doubtful. Others have done so and come to the conclusion that, say, the principle of mathematical induction is doubtful, or even more controversially that infinities do not exist. Obviously, these are not mainstream opinions. I think it is a personal matter. Talking to ordinary people (non-mathematicians) sometimes it feels like they even find ideas like modus tollens doubtful… – Zhen Lin Oct 01 '24 at 03:33
  • Since you seem satisfied by formal systems and syntax, perhaps I should mention that there are consistency proofs not based on the construction or existence of models but instead using well founded induction on certain countable ordinals. For some people, this is more believable than set theory. – Zhen Lin Oct 01 '24 at 03:39
  • @ZhenLin Right. This problem risked degenerating into a meaningless discussion as it's not purely about math. But I'm glad that the answer of spaceisdarkgreen, while not for ZF, turned out to be useful. It suggests I look at Gentzen's proof of Con(PA), which starts from an axiom system that should be coherent to our intuition about the string rewriting algorithm. / Please do suggest whatever related! – Student Oct 01 '24 at 03:40
  • @Student Gentzen’s proof can be carried out just as well in ZF (in fact, quite a bit more easily given ZF’s greater expressiveness). The fact that it can also be done in PRA augmented by transfinite induction, unlike the model theoretic proof, has attracted philosophical interest though. – spaceisdarkgreen Oct 01 '24 at 04:22
  • @ZhenLin Induction deniers should know that the fact that you can reach any natural number by a process of repeated succession starting at zero (the "first number") can be formally proven to be logically equivalent to the principle of mathematical induction from Peano's Axioms. – Dan Christensen Oct 01 '24 at 16:15

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Technically you have that the proof that $ZF+$ large cardinal implies the consistency of $ZF$ by showing there exists a model of $ZF$ and such a model has a truth predicate satisfying the Tarski conditions and that implies that you cannot deduce $\bot$ from the axioms of $ZF$. If you are looking for a more syntactic proof maybe something along the lines of ordinal analysis will be more satisfactory. This approach started with Gentzen, who by analyzing the structure of proofs of $PA$ deduced that there cannot be a proof of contradiction. Of course, additional assumptions are needed to be able to carry out Gentzen's proof, in particular, for the case of $PA$ one needs to assume roughly speaking that $\epsilon_0$ is well founded. This approach has been used to study the strength of many theories (see https://en.wikipedia.org/wiki/Ordinal_analysis). However, from what I am told, the theory of $ZF$ seems still quite out reach for this program.

Giorgio Genovesi
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  • Thanks for answering so quickly! It's pointed out in the comment that my original question wasn't clear enough, so I have edited it. I have to admit that this is not a math question, but more like a computational question. (By the way, what do you mean by a syntactic proof?) – Student Oct 01 '24 at 00:38
  • Essentially you are only looking at the structures of the proofs (more or less) rather than trying to construct a model of that theory in a stronger theory. For example large cardinal axioms imply there is a $\kappa$ such that $V_\kappa\vDash ZF$. This could be seen as a more semantic approach to proving the consistency of a theory. – Giorgio Genovesi Oct 01 '24 at 00:41
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Let me take a different approach than the other answers and try to convince you that model-theoretic proofs of consistency can actually be cast as proofs about combinatorial string rewriting.

If we have a string rewriting system, how do we prove some string $\varphi$ is not a theorem? We may look for an invariant: a property of strings that we can prove by induction is true of all strings produced by the system, but which we can check is not true of $\varphi$. A famous example of this is the MU puzzle by Hofstadter.

Now suppose our string rewriting system is a first-order proof system with axioms from a theory $T$. If I have access to a model $M$ of $T$ (say $\mathbb{N}$, if $T$ is PA), then I can define what it means for a sentence to be "true in $M$". This is a more complicated invariant than the one in the MU puzzle, in that it may be difficult to determine, for an arbitrary sentence, whether it is true in $M$. But it is an invariant: it is straightforward to prove that the axioms of $T$ are true in $M$ ($M$ is a model), no contradiction $\varphi\land\lnot\varphi$ is true in $M$ (by definition of truth), and all of the rewriting rules, when given inputs which are true in $M$, return outputs which are true in $M$ (soundness).

Thus, we are convinced that when we "run the rewriting system", we never reach a contradiction.

You seem concerned that model-theoretic proofs depend too much on the meta-theory. But very little about the meta-theory is used in this argument. All we need to be able to do is define the invariant and prove the soundness theorem.

Alex Kruckman
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  • I think I know what you mean! My understanding: A sound logical system (which includes the a set of axioms, a set of deduction rules, and an instruction of how it is interpreted into a certain semantics) is such that $\vdash C$ implies $\vDash C$. Hence, with $C = (\phi \wedge \neg \phi)$ being a theorem, $C$ would also be true in any model. However, in practice, this means we could prove that $C$ is never a theorem by constructing an explicit model in which $C$ is not true. Is this correct? .. But then we need to outsource our confidence into another system in which we prove that it's sound.. – Student Oct 01 '24 at 16:35
  • @Student I don't understand what you mean by "outsource". Where are we outsourcing our confidence from? – Alex Kruckman Oct 01 '24 at 17:09
  • @Student In other words: it's on you to clarify what kinds of assumptions you trust to tell the truth about reality... Have you read a proof of the soundness theorem for first-order logic? It's hard to imagine a more intuitive argument. – Alex Kruckman Oct 02 '24 at 01:15
  • I've read it and I agree it's pretty good already. But every time I think about model theory.. I hesitate a bit because I never think I know what a set really is (but a model is defined to be a set), after all, what a set really is does depend (practically slightly) on the underlying axioms. – Student Oct 02 '24 at 04:04
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    @Student (1) Amost everything in mathematics is defined to be a set. A group is a set, a topological space is a set, etc. do you hesitate a bit every time you think about group theory or topology because you don't know what a set really is? (2) It's true that changing the meta-theory can have effects on model theory. But the point I'm trying to make here is that every reasonable meta-theory that can even talk about models will prove the soundness theorem. The details about what $M$ really is are irrelevant: if we define "true in $M$", prove soundness, and prove a model $M$ exists... – Alex Kruckman Oct 02 '24 at 12:08
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    (Continued) then we can forget about what $M$ really is, and note that we have an invariant guaranteeing consistency, i.e., a property of sentences which is true of the axioms, false of all contradictions, and preserved by the rewriting rules. – Alex Kruckman Oct 02 '24 at 12:10
  • I agree! Now, to prove consistency, a way is to use model theory to produce invariant: more concretely, for each statement $\phi$, one needs to construct a model $M(\phi)$ such that $C = \phi \wedge \neg \phi$ is not true in $M(\phi)$. Is there any example that really goes like this, and are there other ways? – Student Oct 02 '24 at 14:07
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    @Student $\varphi\land \lnot\varphi$ is false in every structure $M$, no matter what $\varphi$ is. If $M$ is any model of $T$, then the set of sentences true in $M$ excludes all contradictions. This is why exhibiting a single model of PA ($\mathbb{N}$) proves consistency of PA. – Alex Kruckman Oct 02 '24 at 14:11
  • I see! And the point is that $\mathbb{N}$, while whose meaning may vary with the underlying meta theory, would 'work' given that the meta theory is reasonable enough. How about ZF? Why can't we just find a model for it to prove its consistency? Godel's theorem only says that we can't do that in ZF, but the same thing happens for PA yet we managed to come up with $\mathbb{N}$. What's the difference? – Student Oct 02 '24 at 14:29
  • @Student We can't "come up with $\mathbb N$" in PA, but we can in ZFC. We can't come up with a model of ZFC in ZFC, but mentioned already, we can in ZFC + large cardinals. The difference is only in peoples' comfort in assuming axioms stronger than PA vs assuming axioms stronger than ZFC, the latter being a theory notorious for its ability to prove virtually every accepted mathematical theorem. If that theory can't construct a model of ZFC, then why should we believe one exists? – spaceisdarkgreen Oct 03 '24 at 04:48
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    @Student (That last sentence doesn't remotely resemble my opinion, by the way. I'm just trying to explain why some might view "PA is consistent" as a "theorem" and "ZFC is consistent" as "not a theorem". Since you mentioned Nelson, I should note this isn't what he thought either: he strongly suspected even PA was inconsistent, even came up with an (erroneous) proof that it was.) – spaceisdarkgreen Oct 03 '24 at 04:52
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    @spaceisdarkgreen Thanks, this comment really threw me. I always thought that whether $\mathbb{N}$ (or other sets or math objects) exists is just a boring theoretical question to ask. But now since model theory could attempt proving inconsistency by assigning set-theoretic invariants (i.e. models), whether those "do exist" (in mind, in belief, in faith) matters for such a concrete string rewriting problem (which I believe directly corresponds to reality). This highlights the problem why this direction is 'wrong'. From now on I will doubt $\mathbb{N}$ even more. Thank you very very much! – Student Oct 03 '24 at 14:59
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The way we typically prove Con(PA) in ZFC is to observe that the set of natural numbers that we can construct in ZF are a model of PA. Note that the existence of the set of natural numbers is contingent on the axiom of infinity, and indeed, some more careful analysis shows that we cannot prove Con(PA) in ZFC - infinity.

An infinite set transcends the finite sets (or finite numbers), which are perfectly content to be on their own. Of course once we add one, we get a whole bunch of infinite sets -- way too many by some accounts -- that constitute the set theoretical universe described by ZFC.

That leads to the idea of proving Con(ZFC) by assuming some "stronger axiom of infinity" and iterating. This is the large cardinal hierarchy, discussed in Giorgio's answer to this question and Asaf's answer you linked. For instance you can assume an inaccessible cardinal $\kappa$ exists, and $V_\kappa$ is a model of ZFC, and voila, ZFC is consistent. Just like we talk about the natural numbers in PA, then we "encapsulate" them in ZFC to get a proof of consistency, we talk about a universe of sets in ZFC and then encapsulate it in ZFC + inaccessible to get consistency.

This is all very interesting, but seems to be going in the opposite direction one might hope to be going in if we want to really be convinced that there's no contradiction in these theories. We're just assuming a richer and richer ontology and thus seemingly putting ourselves at higher risk of inconsistency in order to prove consistency. At the same time, the fact that we can do this in a sensible way and there's a whole tower of stronger theories that nobody has found an inconsistency in might be encouraging (though there has been one case where a very strong large cardinal property proposed and taken seriously but was found relatively quickly to be inconsitent).

Of course, going purely in the "right direction" is impossible due to the incompleteness theorem. But, per Giorgio's comments on Gentzen's proof (see also my answer here), that doesn't mean there's absolutely no hope of a proof that would convince a set theory skeptic.

spaceisdarkgreen
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  • Thank you. I think you're the only one who understand my concern so far. Instead of a formal proof for Con(ZF), I want a formal proof from an axiom system that matches 'our intuition' about the rewriting algorithm. And any other proofs, say by the 'cheating' method from assuming Con(ZF) as an axiom, or by some other sophisticated methods (e.g. the strong axiom of infinity you mentioned), are not satisfactory. But now I understand that even the 'wrong direction' is difficult enough, I don't think I should pursue the 'right direction', and therefore should leave this thread open. – Student Oct 01 '24 at 02:42
  • I'm happy to step back and look at PA instead of ZF. You mentioned that to prove Con(PA) from ZFC, it's essentially about checking that $\mathbb{N}$ is a model. However, I wish to see a direct proof, not a model-theoretic proof because model-theoretic 'truth' depends on the metatheory (https://mathoverflow.net/questions/464234/truth-in-a-different-universe-of-sets) which makes the problem even more complicated. Q1 Is there a direct proof? - And I'm also interested in Gentzen's proof of Con(PA). Do you have a reference for that proof, which hopefully stresses on the 'right direction'? – Student Oct 01 '24 at 02:47
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    @Student Well... the existence of a syntactic proof depends on the metatheory in the exact same sense (in fact JDH outlines how these flow from one another... the existence in of a proof is an arithmetical statement which is a statement about what is "true in $\mathbb N$"). – spaceisdarkgreen Oct 01 '24 at 03:08
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    @Student I think Gentzen's proof (or things in the same vague proof-theoretical bucket like Godel's Dialectica proof) would qualify as "direct proofs" in the sense you mean. As for references, unfortunately, expositing proof theory at an introductory level seems to be somewhat of an open research problem. There is an older book by Takeuti that has a proof. There is also a new introductory book by Mancosu Galvan and Zach that has a very long and detailed exposition of the proof. You can also see Blass's answer here for the flavor. – spaceisdarkgreen Oct 01 '24 at 03:12
  • Galvan-Zach's book is indeed very nice (and long). Blass's answer is also very flavorful! I'm curious why Gentzen's need to assume transfinite induction to $\epsilon_0$, which serves as an upper bound ordinal for the 'cut size's for all proofs for $0=1$ (in Blass's term). After all, the size of all proofs should be less than countably infinite.. so why do we need more than $\omega$ cuts? – Student Oct 01 '24 at 04:41
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    $\epsilon_0$ is also countable. Just because your proof is finite doesn't mean that induction up to $\omega$ will do what you need! – Zhen Lin Oct 01 '24 at 05:43