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My reference book is A Course on Mathematical Logic by S.M. Srivastava.

Not so long ago, MO linked me to a video of a conference by Voevodsky wherein he considered the possibility of arithmetic being inconsistent. Apparantly, there is no (known) proof of consistency for $\mathsf{PA}$ or for $\mathsf{ZF}$.

Yet, example 2.3.3 on page on page $20$ of said book claims $\mathbb{N}=\lbrace 0,1,2,\dots\rbrace$ (with the usual interpretations of $0,S,+,\times, \lt $) is a model for Peano Arithmetic, and Theorem 4.4.8 (Completeness theorem, second form) on page $63$ says "a theory $T$ is consistent iff it has a model".

Why is this not proof of the consistency of $\mathsf{PA}$? What am I misunderstanding here?

Adding to my confusion, on page $74$, Theorem 5.3.10 states that $\mathsf{PA}$ is consistent iff $\mathsf{ZF}-\mathsf{Axiom~of~Infinity}$ is consistent.

If the root of the problem is set theory, the pending question of the consistency of $\mathsf{ZF}$, how can any theory $T$ be given a model, i.e. a set with a collection of interpretations for constants, functions and relations?

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  • $\begingroup$ I can't see some of the text. $\endgroup$
    – William
    Aug 6, 2011 at 23:00
  • $\begingroup$ @William Chan is this better? $\endgroup$ Aug 6, 2011 at 23:04
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    $\begingroup$ I don't know what you mean by "there is no (known) proof of consistency for PA or for ZF." Of course neither PA nor ZF can prove their own consistency by Godel's theorem unless they are inconsistent. On the other hand, ZF can prove the consistency of PA since a model of PA can be constructed in ZF. So when you say "there is no (known) proof of consistency for PA or for ZF," what metatheory are you using? $\endgroup$ Aug 6, 2011 at 23:10
  • $\begingroup$ @Qiaochu Yuan but consistency of a theory $T$ is defined regardless of models and therefore (or so I think) of $\mathsf{ZF}$, it only requires that there be no formula $A$ written in the language of $T$ with $T\vdash A$ and $T\vdash \neg A$. $\endgroup$ Aug 6, 2011 at 23:10
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    $\begingroup$ In general, a model is used to prove that one theory is equiconsistent with another, meaning that if A is consistent, then B is. In the scenario Voevodsky is discussing, both PA and ZF would be inconsistent. The following are closely related to your question: mathoverflow.net/questions/40920/… mathoverflow.net/questions/35746/… $\endgroup$
    – user13618
    Aug 7, 2011 at 0:03

3 Answers 3

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Why is this not proof of the consistency of PA?

It is a proof of the consistency of PA. It is the same sort of proof that is used to show that the negation of the parallel postulate is consistent with the remainder of the axioms of Euclidean geometry, by constructing a model of non-Euclidean geometry.

What am I misunderstanding here?

Perhaps you heard that it is not possible to prove the consistency of PA. That is simply false; for example you can prove it directly from the axiom "PA is consistent". What is true is just that there is no proof of the consistency of PA within PA itself.

Apparantly, there is no (known) proof of consistency for PA or for ZF.

There are (at least) three proofs of the consistency of PA that are important:

  • The proof in ZFC, which proceeeds by just verifying that $\mathbb{N}$ is a model of PA.

  • The proof by Gentzen. Here consistency is proved in a much weaker theory than ZFC (but of course the theory is not included in PA).

  • The proof by Gödel using what is now called the Dialectica interpretation. This proof uses a completely different method than Gentzen's proof, but obtains the same result. It is also conducted in a weak system, but a different weak system than Gentzen's proof.

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  • $\begingroup$ I think I get it now. There is a first notion of consistency of a first order theory, which requires the theory not prove both $A$ and $\neg A$ for some formula $A$. Then there is a completely different notion, that of a model of $T$. For a model to exist, the consistency of $\mathsf{ZF}$ is needed, since a model of $T$ is a set with extra structure to reflect the language of $T$ (is that so?). We know that if a theory has a model (and thus if $\mathsf{ZF}$ itself is consistent), then it must be consistent. The completeness theorem asserts the converse. Is that about right? $\endgroup$ Aug 7, 2011 at 13:47
  • $\begingroup$ Yes, that's mostly right. If a theory has a model, it must be consistent; if $A$ and $\lnot A$ were both provable from the theory, they would both have to be true in that model, which is impossible. This doesn't require the completeness theorem, which is indeed a sort of converse. However, even if ZF was inconsistent, we might have the possibility of building a model in a weaker system than ZF, such as a type-theoretic or topos-theoretic system. The claim "if a theory has a model then it is consistent" does not require the full strength of ZF to prove. We only work in ZF(C) for convenience. $\endgroup$ Aug 7, 2011 at 14:00
  • $\begingroup$ We have the notion of a model of a theory $T$ as a set with interpretations of the various non logical symbols of the language ot $T$ (that also satisfy all axioms of $T$). Let's call a model as defined above a $\mathsf{ZF}$-model of $T$. Is there, for any other theory $S$ a notion of an $S$-model? $\endgroup$ Aug 7, 2011 at 14:13
  • $\begingroup$ You could do the same thing in type theory (e.g. in second-order arithmetic for a countable theory). In this case the model would might still be a set, but it would be a set as formalized in some theory other than ZF. There are also very different types of models, e.g. Kripke models and topoi. But the key step in proving "having a model implies consistency" is just using the set of sentences that are true in the model to obtain a consistent extension of the starting theory. The rest of the structure of the model is vital for model theory but not needed just to prove consistency of the theory. $\endgroup$ Aug 7, 2011 at 14:22
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I’m using the community wiki mode because this is not an answer. This is just a collection of links. (Some of them had already be given in the comments.) Please feel free to add others.

Vladimir Voevodsky’s talk

MO questions

Ed Nelson’s webpage

FOM

Excerpts (from the correspondence):

HF: Apparently, you think it worthwhile to spend some of your time to prove the inconsistency of PA. Un my opinion, that result would be the greatest result of all time in mathematics by orders of magnitude. And that doesn't touch on how it would compare with results in science and general human intellectual activity.

HF: It would of course be useful if you would reassess or restate your position that "the consistency of PA is a legitimate open problem in mathematics" in light of our correspondence.

VV: So far I do not plan to reassess anything in my position.

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  • $\begingroup$ Pierre-Yves Gaillard: I do think it's good to see people such as Voevodsky take an interest in foundational matters. Unfortunately, the literature is not particularly accessible, and many things are not formally published, so links like the ones you've provided should be very helpful. $\endgroup$ Aug 7, 2011 at 15:14
  • $\begingroup$ @Carl: Thanks! If you see more links to add ... $\endgroup$ Aug 7, 2011 at 15:19
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The usual proofs of the Completeness Theorem are like proofs in any other branch of mathematics. They are rigorous, but not formal proofs in ZFC.

However, the proofs can be made formal, and ZFC is strong enough to carry it out. From a proof of consistency point of view, this is not particularly helpful, since it shifts the burden of consistency to ZFC, a stronger theory.

A better candidate for a consistency proof for Peano Arithmetic is the long-ago work of Gentzen. True, it involves transfinite induction up to $\varepsilon_0$, but it has a finitary character. Unfortunately, "finitary character" is a matter of opinion, not rigorous definition.

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    $\begingroup$ Just to add slightly on the topic, there are assertions which are stronger than ZFC can prove, and when added to the axioms can prove the consistency of ZFC. However we do not know if such ZFC+P would be consistent as well. $\endgroup$
    – Asaf Karagila
    Aug 7, 2011 at 6:04

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