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$PA$ cannot prove $Prov(\phi) \to \phi$; and in particular $PA$ cannot prove "$Prov( 0 = 1 ) \to 0 = 1$".

This can be viewed as a simple consequence of Lob's theorem; but we can also interpret it as "there are (non-standard) models of $PA$ in which both $Prov(0 = 1)$ and $0 \neq 1$ are true".

Is there an intuitive way to build an example of such non-standard model of $PA$ in which $Prov( 0 = 1 )$ and $0 \neq 1$ ?

Is $\neg Con(PA)$ mandatory in such non standard model?

Vor
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2 Answers2

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Briefly:

  • "$0\not=1$" holds in every model of PA. This is because, well, PA proves "$0\not=1$." There's no way to get around this: even if PA were to also prove "$0=1$," all that would tell us is that PA has no models at all (because in every model we would have to have both $0\not=1$ and $0=1$).

  • Yes, "$Prov(0=1)$" and "$\neg Con($PA$)$" are equivalent, provably in PA (and indeed much less). Indeed, $Con($PA$)$ is often an abbreviation for "$\neg Prov(0=1)$," which makes the equivalence trivial.

  • No concrete nonstandard models of PA are really known at all. In particular, Tennenbaum's theorem prevents us from having too snappy a description of such a model. That said, the proof of the completeness theorem is rather intuitive after a while: we pass to a complete theory $T$ in a larger language which contains PA and has the "witness property" - if it proves "$\exists x\varphi(x)$," then it proves "$\varphi(t)$" for some term $t$ (generating these extra terms is why we need to expand the language) - and then the set of terms in this larger language, modulo $T$-provable equivalence, forms a model of $T$ (and hence has a reduct which is a model of PA). The reason this doesn't describe a single construction is that we have lots of freedom in building $T$, but if you blackbox that step the construction of the term model of $T$ is (in my opinion) very intuitive.

Incidentally, by the MRDP theorem we can think of a model of PA satisfying "$Prov(0=1)$" as simply a model of PA where a certain polynomial has a root which didn't before.

You may also find this question interesting.

Noah Schweber
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  • Are you using axioms for Peano arithmetic that define both 0 and 1? – Dan Christensen May 08 '18 at 04:08
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    @DanChristensen There are many completely equivalent approaches to PA. If you don't want to include $1$ from the get-go, simply replace "$0=1$" with "$\forall x[(\forall y(x\cdot y=y))\implies 0=x]$." But this is really a pedantic point irrelevant to the OP's actual question, especially since PA is frequently presented as containing an explicit symbol for $1$. – Noah Schweber May 08 '18 at 04:50
  • I guess I must have stumbled across a very atypical presentation of PA at Wikipedia, one that defines 0 but not 1. – Dan Christensen May 08 '18 at 05:04
  • @DanChristensen It's not atypical at all - it may well be in the majority for all I know - but the question is clearly not about the possible vagueness of the symbol "$1$" but rather the behavior of the internal provability predicate "$Prov$" in nonstandard models of PA. – Noah Schweber May 08 '18 at 05:16
  • I think it was your claim that $0 \neq 1$ in every model that caught my attention. It is clearly false. Maybe my answer should have been a comment on your answer? – Dan Christensen May 08 '18 at 05:47
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    @DanChristensen Neither "$Prov$" nor "$Con$" is part of the primitive language of PA, so the whole question is meaningless, right? No - we observe that those are well-known abbreviations for expressions in the language which we are choosing not to write down in detail. If I write "ZFC proves $0\not=1$," it's clear what I'm actually saying and it's clear that it's true, even though neither $0$ nor $1$ is part of the basic language of ZFC. So this is just being pedantic to the point of uselessness. – Noah Schweber May 08 '18 at 11:07
  • For that matter, "$\not=$" isn't part of the primitive language of ZFC, so "ZFC proves $0\not=1$" must be false for two reasons. Even better: I assume that, since neither "$2^{\aleph_0}$" or "$\aleph_\omega$" are part of the basic language of ZFC, the statement "ZFC proves $2^{\aleph_0}\not=\aleph_\omega$" is false? Even though this has clear meaning and is a nontrivial instance of an important set-theoretic result? And so on. – Noah Schweber May 08 '18 at 11:10
  • So many abbreviations. It's hard for the beginner to keep track. Are there others I should know about? – Dan Christensen May 08 '18 at 11:31
  • @DanChristensen Like any other field of math, logic has tons of abbreviations since it's infeasible to always write everything out in the original language. Some of these are explicitly defined (e.g. $\aleph$-notation is defined in any introductory set theory text), while others are clear from context (e.g. the symbol "$1$" when discussing PA, if we don't include it already in the language) and are freely used when no issue will result (e.g. one context where we want to be very careful with abbreviations is when we care about quantifier complexity, but (a) we don't here and (b) $1=S(0)$). – Noah Schweber May 09 '18 at 12:37
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Hint: Using the axioms at First-order theory of arithmetic. If we start with Peano's Axioms for $(N,S,0)$ with $1$ not yet defined (as at link), then from FOL we have $0=0$ and $\exists x: x=0$.

Dan Christensen
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    This has nothing to do with the question, which is about what happens in models of PA satisfying $\neg Con$(PA). The OP is aware that $0$ exists and equals itself. – Noah Schweber May 08 '18 at 04:50
  • The OP seems to be interested in the possibilities of 0=1 in Peano arithmetic. It is trivial to show, given the axioms at my link, that it is possible without generating any contradictions. – Dan Christensen May 08 '18 at 04:59
  • You just have to come up another name for the successor of 0. – Dan Christensen May 08 '18 at 05:07
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    It's very clear that that's not what they're interested in: they're asking about models of PA which think that PA is inconsistent. – Noah Schweber May 08 '18 at 05:11
  • For example,nowhere in your answer is "$Prov(0=1)$" mentioned, even though the question quite clearly asks for a model of PA in which (among other things) $Prov(0=1)$ holds. How exactly does a structure of the type you describe have this property? – Noah Schweber May 09 '18 at 12:38
  • @NoahSchweber You might consider it a comment on your answer, which the OP had already accepted when I posted. He should be made aware that, without the addition of suitable abbreviations, 0=1 is not ruled out by PA. Maybe it's nit-picking, but nit-picking is what we do here, isn't it? – Dan Christensen May 09 '18 at 13:31