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Question.

What references do you recommend that dryly/precisely discuss differences between

(0) Trying to prove $\neg\mathrm{Con}(\mathrm{PA})$.

(1) Trying to exhibit an actual contradiction in $\mathrm{PA}$.

Remarks.

(0) While this is rather irrelevant for the question, I mention that I am currently neither trying to do (0), nor (1), rather would like to check a few things around this, to learn something that would take long to spell out here.

(1) Hoping for some brief relevant pointers, by people working in mathematical logic, to relevant literature, I keep this short and do not try to make precise what $\neg\mathrm{Con}(\mathrm{PA})$ and "to exhibit an actual contradiction" in (0) and (1) mean. This is an important and well-known distinction, occurring in relevant discussions of this topic.

user400188
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Peter Heinig
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  • Those are the same thing. What is different is working in a model of PA where Con(PA) is false. – Qiaochu Yuan Jul 25 '17 at 06:30
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    @QiaochuYuan They are not the same, really. In principle, there's no reason to believe that our metatheory doesn't prove $\neg$Con(PA) even if (a) our ambient metatheory is consistent and (b) PA is consistent. And even leaving aside incorrect metatheories, there's always the possibility of proving that a contradiction exists without exhibiting one. (Of course, if you prove from correct axioms that a contradiction exists, you're guaranteed to be able to find it; but there might not be a reasonable way to extract an explicit contradiction from the proof.) – Noah Schweber Jul 25 '17 at 07:09
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    Question about (0): trying to prove from what? In what meta-theory are we working? ZFC? SOA? PA+something? – Asaf Karagila Jul 25 '17 at 08:22
  • @NoahSchweber : Maybe it is almost the same. For definable numbers (are they countable or not ?) I have heard the problem is the (meta-theoretic) function sending a formula to the (real) number it defines. Here, I think we don't have such problem : we can enumerate the formulas of PA and check if they prove an inconsistency. Thus if we prove there exists an inconsistency, we can run a program and we are guaranteed (if the meta-theory where you proved it is consistent ?) to find one in a finite time. – reuns Jul 25 '17 at 08:25
  • @reuns: I think that the main issue here is "exhibit actual contradiction". What does that mean an actual contradiction? I think that most people would take it as a proof from some very weak and fairly constructive system as a meta-theory, something that intuitively "we can all agree on". My gripe with this sort of approach to this remark, is that I don't know what would such a theory be. And sure, PRA or bounded induction, both seem like good choices, but I would like to know exactly what that means. Because that would be "the meta-theory of math", to some extent. – Asaf Karagila Jul 25 '17 at 08:50
  • @AsafKaragila: to some people, "exhibit an actual contradiction" seems (I am asking this precisely because I tend to be baffled by some of the free-wheeling discussions one can easily find on this) to mean a purely syntactic, evidently-possible-in-principle, phenomenon: after fixing (0) a semantics-function $\mathrm{I}$ on the countably-infinite set of all PA-formulas, where $\mathrm{I}$ may even be allowed to be partially-defined and to sometimes take values other than $\mathsf{T}$ and $\perp$, and (1) fixing a set $\mathfrak{R}$ of ... – Peter Heinig Jul 25 '17 at 09:25
  • ... rules for tranforming an PA-formula into another PA-formula, such that, for each $R\in\mathfrak{R}$, and for any PA-$\varphi$ormula, if $\mathrm{I}(\varphi)=\mathsf{T}$, then $\mathrm{I}(R(\varphi))=\mathsf{T}$, (soundness), succeeding in finding ... – Peter Heinig Jul 25 '17 at 09:29
  • ... some PA-formulas $\varphi_0,\varphi_1,\varphi_2,\varphi_3$ such that $\mathrm{I}(\varphi_0)=\mathsf{T}$, $\mathrm{I}(\varphi_1)=\mathsf{T}$, $\varphi_i$ for both $i\in{2,3}$ is obtained (explicitly) from $\varphi_{2-i}$ by a finite set of applications of rules from $\mathfrak{R}$, and $\mathrm{I}(\varphi_2\wedge\varphi_3)$ is found to evaluate to $\perp$. I suppose your concerns are about disagreements about the function $\mathrm{I}$, right? (Definition of it, computability of it, meaning of it...) – Peter Heinig Jul 25 '17 at 09:35
  • @PeterHeinig I don't understand why you're making this so complicated. We have a perfectly good notion of what a proof from a set of axioms is; the most natural interpretation of "exhibit an actual contradiction in PA" is to give an actual formal proof, from the axioms of PA, of "$0=1$" (or similar). – Noah Schweber Jul 25 '17 at 18:13
  • @Noah: I suspect that I'm the source of this overcomplexity (or maybe I was just the one to bring it to the surface?), by asking what does it mean to "actually exhibit". I'm guessing that you're reading this as "write down on a piece of paper a proof of 0=1 from PA"? – Asaf Karagila Jul 25 '17 at 19:15
  • @Noah: I had in mind a proof in PA rather than in a metatheory for 0). Is there still an important difference there? – Qiaochu Yuan Jul 25 '17 at 20:51
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    @QiaochuYuan Yes: in theory, PA could prove $\neg$Con(PA) and still be consistent! – Noah Schweber Jul 25 '17 at 21:52
  • @Noah: this confuses me. I had the impression that if PA is consistent then Con(PA) would be true in the standard model $\mathbb{N}$, since otherwise there would be a standard integer encoding a proof of a contradiction in PA. So if PA is consistent but proves ¬Con(PA) does that imply that the standard model doesn't exist / isn't a model of PA? – Qiaochu Yuan Jul 25 '17 at 22:00
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    @QiaochuYuan Yes, this is operating under the assumption that we are not committed to the claim "PA is true of the standard natural numbers" - either because we doubt their existence, or we doubt that they satisfy PA. But note that we are already at least this skeptical as soon as we doubt the consistency of PA in the first place: no inconsistent theory can have a model, so if we're committed to PA being true in the standard naturals that already amounts to being committed to its consistency. – Noah Schweber Jul 25 '17 at 22:20
  • @AsafKaragila Yes,I'm reading it that way. – Noah Schweber Jul 25 '17 at 22:20
  • @Qiaochu: If we work with ZFC, or some other sufficiently strong meta-theory, then you're correct. If we work with a very weak meta-theory which does not prove the consistency of PA, then this is no longer necessarily the case (also the notion of "standard model" becomes a bit more shaky, I guess). – Asaf Karagila Jul 25 '17 at 22:22
  • @Noah: right, yes, of course. Thanks! – Qiaochu Yuan Jul 25 '17 at 22:47
  • Many thanks to each of the three of you for this illuminating discussing. At least for for my own education, I take the liberty of giving a rendition of/comment on the comment containing 'incorrect metatheories', which I think clarifies an often-glossed-over important aspect: (...) – Peter Heinig Jul 28 '17 at 05:30
  • @Noah: The commentwith 'incorrect metatheories' however can be seen to contain a triple-negative and some (slight) redundancy (which can be helpful of course), hence not that easy to understand. More seriously, the comment contains (what I perceive to be) sort-of-an-small-error, in that it uses 'incorrect metatheory' for 'consistent metatheory which, in a world in which PA is syntactically/actually consistent, but comes to the weird conclusion $\neg\mathrm{Con}(\mathrm{PA})$.' – Peter Heinig Jul 28 '17 at 05:32
  • Then the comment with "incorrect metatheories" later accidentally uses the phrase 'correct axiom' for 'consistent axiom', although the latter use of 'correct' was just an oversight, not meant to have the interesting more complex meaning of 'incorrect'. Side-question: do you agree that 'correct axioms' in your comment should be 'consistent axioms', in a usual sense? Moreover, the comment with 'incorrect metatheories' contains two reasonably distinct aspects. – Peter Heinig Jul 28 '17 at 05:34
  • So here goes: 'The phrases 'prove $\neg\mathrm{Con}(\mathrm{PA})$' and 'to exhibit an actual contradiction in PA' are not the same. There are at least two aspects in which they differ. Aspect 0. Assuming that PA is consistent, no general reason is known which would rule out that there may be a reasonable, but unusual, metatheory $M$, such that (0) $M$ is consistent and (1) there is a proof in $M$ of $\neg \mathrm{Con}(\mathrm{PA})$. Briefly, reasonable-seeming metatheories may come to weird conclusions, even in a world in which PA is syntactically/actually/physically consistent. – Peter Heinig Jul 28 '17 at 05:37
  • I'll put my rendition of Aspect 1 on hold until the question around the interesting term "incorrect metatheories" seems to be clarified. – Peter Heinig Jul 28 '17 at 05:39
  • @Noah Schweber: re Aspect 0, would you agree with the definition (PA actually consistent)$:=$(PA syntactically consistent)$:=$(there does not exist a physical derivation of 0=1 from PA), the latter existence-statement construed with a Tarskian-definition of truth, i.e., it is true if and only if its interpretation in the world is true, and then, with the definition (incorrect metatheory)$:=$(metatheory $M$ which in a world with "PA actually consistent" allows to physically derive, within $M$, the statement $\neg\mathrm{Con}(\mathrm{PA}$) ? In that sense, such an $M$ would be "incorrect". – Peter Heinig Jul 28 '17 at 06:49
  • @NoahSchweber: and, do you agree that, with the above, "incorrect metatheory" is (0) mathematically defined, (1) can nevertheless never be decided for a metatheory at hand because one of the hypotheses of the definitions, namely PA being physically-consistent, is known not to be provable by a finite formal proof. So, at the risk of sounding too physical, in that sense, knowing a meta-theory to be "incorrect" would required some-sort-of omniscience about the physical universe? – Peter Heinig Jul 28 '17 at 06:53
  • @PeterHeinig If you believe that statements like "is consistent" have meaning, then PA is either consistent or it isn't; and if it's consistent, then metatheories proving that it isn't are, indeed, incorrect. I don't really know what you mean by "physically derive" - the possible finiteness of the universe, the second law of thermodynamics, quantum mechanics, and other aspects of reality make me seriously doubt that physical reality is good for anything other than small finite numbers, and a proof of $0=1$ might be very long. I think omniscience re: the physical universe is far too little. – Noah Schweber Aug 03 '17 at 21:19
  • @NoahSchweber: Your last comment makes me curious; if the physical reality does not have an embedding of the natural numbers, then what would a "proof of 0=1 might be very long" even mean? Either there is a proof (a string of symbols stored in some physical medium) or there is no such proof. I don't see how one can claim to refer to abstract proof if PA itself has no physical interpretation. – user21820 Aug 05 '17 at 14:17
  • @user21820 I don't consider mathematical objects to be physical objects - the statement "Either there is a proof (a string of symbols stored in some physical medium) or there is no such proof" is not one that I agree with, since I don't think a proof is a string of symbols stored in some physical medium any more than I think that the size of the universe puts a bound on what natural numbers can exist. There are of course other (e.g. ultrafinitist) takes on mathematical existence, but this is I think the standard one - mathematicians tend to view mathematical objects as "ideal" entities. – Noah Schweber Aug 05 '17 at 14:34
  • @NoahSchweber: That's what I don't understand about your comment. We invented PA only because (physical) counting numbers seem to obey those axioms. Without the physical basis for PA, there just isn't any reason to assume that PA has a model. Clearly the mathematical objects that you call proofs rely on the meta-system's assumption that such a model exists or equivalent (if you axiomatize finite strings rather than natural numbers). Note that even basic classical logic relies on the unlimited ability to concatenate any two strings. – user21820 Aug 05 '17 at 14:41
  • @user21820 I don't really think that comments are a good place to have a serious discussion about mathematical philosophy. I'll just point out that statements like "Note that even basic classical logic relies on the unlimited ability to concatenate any two strings" point to the insufficiency of the physical universe as a basis for mathematics - if the universe is finite,then there is a limit on our ability to concatenate finite strings (if we ground mathematical existence in physical reality)! (This debate is fairly standard fare in math. phil. and is treated extensively in theliterature.) – Noah Schweber Aug 05 '17 at 14:46
  • @NoahSchweber: To be clear, I can readily understand the phrase "ideal entities" in the sense of merely "notions" with properties described via an axiomatization, and hence these notions may be utterly meaningless (or with no instantiation) like the notion of a finite field with 6 elements. That is fine. But if you think that mathematics is meaningful (and let's just focus on PA first), I can't see your reason if not for physical reality. And you are welcome to the logic chat-room to continue this discussion. – user21820 Aug 05 '17 at 14:52
  • @user21820 "But if you think that mathematics is meaningful (and let's just focus on PA first), I can't see your reason if not for physical reality. And you are welcome to the logic chat-room to continue this discussion." That's a good conversation to have, but a long one, and I don't have the time or energy right now. But you could ask this as a question here (e.g. "how do "aphysicalist" mathematicians think about mathematical existence vs. physical reality, and how can mathematics acquire meaning if it tries to detach itself from physical reality?") and I think you'd get useful responses. – Noah Schweber Aug 05 '17 at 14:54

2 Answers2

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The phrases you've quoted are informal, and as such can be interpreted in different ways. I think the most natural interpretations, however, are:

  • To "exhibit an actual contradiction in PA" means to find a formal proof - e.g. in the sense of sequent calculus - of "$0=1$" from the axioms of PA. Note that a formal proof is computer-verifiable, so this really is a totally airtight notion: if you claim to have an actual contradiction from PA, that's an objective statement and can be objectively verified.

  • To "prove that PA is inconsistent" means to prove the statement "PA is inconsistent" (appropriately coded) in some theory, possibly other than PA, which we are confident is $\Sigma^0_1$-correct. This is a somewhat informal notion, on two counts: "appropriately coded" and "we are confident" are each informal properties, and each is reasonably subtle. That said, once we fix a representation $\varphi$ of "PA is inconsistent" in some fixed language (e.g. the language of arithmetic, or the language of set theory) and some theory $T$ in the appropriate language, this can become totally objective if we interpret "prove" as "produce a formal proof from the axioms of $T$).

These are obviously in principle different tasks; however, most logicians agree that in practice they amount to the same thing. Namely, if some theory $T$ which is stronger than PA proves that PA is inconsistent, I think this would cause most logicians to doubt the $\Sigma^0_1$-correctness of $T$, rather than the consistency of PA; and most logicians are confident in the appropriateness of the representation of "PA is inconsistent" in the language of PA (although this has been studied) - so the non-objective parts of the second task are removed (work in PA itself, and use the usual coding). Meanwhile, if PA proved "PA is inconsistent," it is likely that this proof would easily yield an actual contradiction, even though a priori PA could be consistent even if it proved its own inconsistency.

I don't know a specific source which discusses this, but when in doubt: define your terms and phrases precisely, and then you don't need another source to clarify what you mean. And that said, I believe the above captures accurately how these phrases are used in logic the vast majority of the time.

Noah Schweber
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  • In your comment to the OP, containing "they are not the same", do you mean "incorrect metatheory"$=$"non-$\Sigma_1^0$-correct theory"? @Noah Schweber – Peter Heinig Jul 31 '17 at 11:00
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If you already know the precise versions of what your (0) and (1) are usually taken to mean, then the answer should have been clear. $ \def\pa{\text{PA}} \def\con{\text{Con}} $

As usual assume PA is consistent. Consider $T = \pa + \neg \con(\pa)$. Then $T \nvdash \bot$ but $T \vdash \neg \con(T)$. So clearly an explicit proof of contradiction over PA may not exist even if PA proves $\neg \con(\pa)$. Of course, in the latter case we clearly should start doubting the meaningfulness of PA, since it would be $Σ_1$-unsound, which is really really bad.

user21820
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  • I think I know what $\Sigma_1$-unsound means, but could you please briefly comment on the relation of the three technical terms (0) "incorrect metatheory", (1) " $\Sigma_1^0$-correct" and (2) "$\Sigma_1$-unsound", the first two having been used by another commenter, the latter having been used by you? @user21820 – Peter Heinig Jul 31 '17 at 18:51
  • @PeterHeinig: Based on the content of the other answer to which you are referring, his use of (1) means exactly the same as "Σ1-sound". Most texts use (2) as the technical term for it, which is probably why you said you think you know what it means. To eliminate remaining doubt, (working in the meta-system which already has a structure satisfying PA) we call a formal system S that interprets arithmetic Σ1-sound iff S proves only (the translations of) true arithmetical sentences. Clearly, (in the meta-system MS) PA is Σ1-sound. If PA proves ¬Con(PA) then we get a contradiction (in MS). Bad! – user21820 Aug 01 '17 at 15:19
  • @PeterHeinig: No reasonable meta-system can do without basic string manipulation plus induction, which is essentially equivalent to PA. So it better be the case that PA does not prove ¬Con(PA). For the same reason, MS had better not prove ¬Con(PA). It's not much better if MS proves ¬Con(MS); although this doesn't give an actual contradiction in MS, it does imply that MS is Σ1-unsound, and very weak meta-systems can prove the Σ1-unsoundness of such an MS. – user21820 Aug 01 '17 at 17:19
  • @PeterHeinig: As for "incorrect meta-theory", you'd have to ask the user of such a term precisely what is meant, but it certainly includes at least meta-systems that prove their own arithmetical unsoundness, since that would be 'self-defeating'. – user21820 Aug 01 '17 at 17:22