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It is well known that there is a fomula which means the consistency of PA in PA,that is Con(PA).

Then, Is Con(PA) true in standard model? or not?

Moreover,how to prove it?

I'm sorry. I edit this question as follows.

I can't understand well the relation of between [...is true in a relevant meta-theory] and [...is true in the standard model].

One requires a meta-theory to treat PA as object-theory.

Here, I adopt ZFC as meta-theory.

In ZFC, It is provable that PA has a model,and then Con(PA) is true.

On the other hand, it is said that Con(PA) is true in the standard model iff

Con(PA) is evaluated true at usual interpretation.

Above two notions slightly different (at least on definitions).

I can't understand well the relation of two notions - still after reading

Reese's answer.

maimai
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  • Possible duplicate of Why bother with Mathematics, if Gödel's Incompleteness Theorem is true? – Frits Veerman Jun 01 '17 at 15:41
  • I'm sorry about the lack of context. I've edited the question. But, your linked question isn't close to mine (at least in my thought) – maimai Jun 02 '17 at 11:49
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    What do you understand by "the standard model"? Since your metatheory is ZFC, it seems you mean the structure $(\omega,+,\times,S,<,0) $ defined within the universe of sets (or some variant of this, depending on what specific signature you use to state the axioms of PA). Here, $+,\times $ are ordinal addition and ordinal multiplication, 0 is $\emptyset $, $S (\alpha)=\alpha\cup{\alpha } $, and $a <b $ iff $a\in b $. If this is indeed the structure you mean, you can easily prove that the second-order formulation of PA holds for it arguing about ordinals in ZFC. – Andrés E. Caicedo Jun 02 '17 at 12:58
  • Once you have this, the rest is the usual sketch: The statement that PA is consistent holds, since you exhibited a model. Its formalization is a statement about numbers, and the argument in the given answer can be quickly formalized in ZFC. – Andrés E. Caicedo Jun 02 '17 at 13:29
  • Thank you for your comment! Hmm, so [Con(PA) is true in ZFC] and [Con(PA) is true in the standard model(also in ZFC)] are equivalent? – maimai Jun 02 '17 at 14:00
  • A posteriori.${} $ – Andrés E. Caicedo Jun 02 '17 at 14:36
  • I've understand it at last! I'm sorry about very very basic misunderstanding and bothering you. My misunderstanding was as follows. – maimai Jun 03 '17 at 14:13
  • One of most basic factors of Godel's incompleteness theorem is, many properties of a logical system(here PA) are formalized in that system. Particularly, provability is formalizable, and in Sigma_1. So, Con(PA) is in Pi_1, actually "Sigma_n is true" is formalizable for all n. However, for all formulas P, "P is true" is not definable (i.e. Tarski's theorem) . – maimai Jun 03 '17 at 14:13
  • So, one can take a proposition Con(PA)(in PA) such that Con(PA) is evaluated true at the standard model(model theoretically defined notion) iff "PA is consistent"(not in PA! but in meta-theory, naively meta-theoretically defined notion) is true. Then, provided PA is consistent, "Con(PA) is true in the standard model" holds obviously! – maimai Jun 03 '17 at 14:13
  • I've not understand well such a very very basic notion. I'm so sorry! – maimai Jun 03 '17 at 14:13

1 Answers1

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Yes, provided $PA$ is consistent. $Con(PA)$ is just the statement "there is no proof of $1 = 0$ from $PA$". If it were not true, then there would be such a proof, and that proof would be coded by some standard natural number $n$. But then we could decode it back into the proof - here the standardness is important, because otherwise the "proof" might have nonstandard length and hence not be a real proof - and we then have a proof that $PA$ implies $1 = 0$. Since $PA$ proves $1 \neq 0$, this would make $PA$ inconsistent.

Reese Johnston
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  • Thank you for your answering! – maimai Jun 02 '17 at 10:09
  • But I can't understand enough. I want to derive [Con(PA) is true in the standard model], so in order to use proof by contradiction, I have to assume [Con(PA) is not true in the standard model] but not [Con(PA) is not true in a meta-theory] precisely? But in your answer, it seems that the latter is assumed... – maimai Jun 02 '17 at 10:36
  • @maimai If you don't assume that $Con(PA)$ holds in the metatheory, then there is no "standard model", because $PA$ is inconsistent and therefore has no models at all. My argument works like this: we suppose for contradiction that $Con(PA)$ fails in the standard model. We then deduce that $PA$ is actually inconsistent (so $Con(PA)$ fails in the metatheory), which we consider to be a contradiction because we're operating under the assumption that $PA$ is consistent (in the metatheory). – Reese Johnston Jun 02 '17 at 16:41
  • I'm so sorry about very basic misunderstanding! I've understand it at last. About it, please see comments of my question. So now, I can understand "Provided PA is consistent, Con(PA) is evaluated true at the standard model" and your proof by contradiction. – maimai Jun 03 '17 at 14:14
  • But, different problems occur. I can't understand your statements [the standardness is important, because otherwise the "proof" might have nonstandard length and hence not be a real proof ]. – maimai Jun 03 '17 at 14:14
  • As long as I understand, any given proof,it would be coded by a natural number n, but the natural number n code "the proof itself" but not "the length of the proof". I'm sorry about bothering you still. – maimai Jun 03 '17 at 14:14
  • @maimai You're correct that $n$ codes the proof rather than the length, but the coding techniques we usually assume use a fixed library of (standard) numbers to represent the symbols. Since nonstandard numbers are too big to be constructed in a standard number of steps from standard pieces, any nonstandard number that codes a proof codes a proof that has nonstandard length (either because it includes a nonstandard number of steps or because one of the sentences it uses has a nonstandard number of symbols). – Reese Johnston Jun 03 '17 at 14:41
  • First, in general,it is possible that a proposition P is true in a model A, but not in other model B. So, it is possible that Con(PA) is true in the standard model,but not in a nonstandard model. But,this argument isn't interesting and doesn't require your explanation. – maimai Jun 04 '17 at 09:34
  • Indeed, your explanation contains more connotation. and this just make your explanation interesting, intuitive and attractive. Probably, it can expressed as follows. (In the following, I assumed a nonstandard model is fixed.) – maimai Jun 04 '17 at 09:35
  • Con(PA) is originally defined to have the property

    (1) Con(PA) is true in the standard model iff PA is consistent(in usual sense).

    But, consequently Con(PA) doesn't only have the property (1), but also have the property

    (2) Con(PA) is true in the non-standard model iff PA is nonstandard-consistent (i.e. standard natural numbers replaced nonstandard in usual sense ).

     – maimai Jun 04 '17 at 09:35
  • Is this thought right? Did you mean such thing? – maimai Jun 04 '17 at 09:36
  • @maimai Not really. $Con(PA)$ isn't defined to have anything to do with standardness or nonstandardness - the statement "$Con(PA)$" literally is "there does not exist a number $n$ which codes a proof of $0 = 1$ from $PA$". Importantly, we can't say anything about standardness inside of a model - there is no way to formally express the notion of "standard" in the language of $PA$. Also, at no point in my answer did I move into a nonstandard model, so no, I haven't said anything about (2) (though it's true). My answer was proving that (1) holds, which is what you asked in your question. – Reese Johnston Jun 04 '17 at 20:33
  • But, Con(PA) is just "a string of symbols". Although Con(PA) means nothing itself, we can say "Con(PA)" literally is "there does not exist a number n which codes a proof of 0=1 from PA" because we refer to some model. – maimai Jun 05 '17 at 04:49
  • Of course, there is no way to formally express the notion of "standard" in the language of PA, but PA is object-theory but not meta-theory here. So, can we refer to "standard"? Otherwise, how does Con(PA) -which is just a "string of symbols"- code "PA is actually consistent"? – maimai Jun 05 '17 at 04:50
  • @maimai $Con(PA)$ is a string of symbols in the metatheory, not in $PA$. In order to interpret it in $PA$ and ask whether it is true in a model, we must first replace it with its definition in $PA$, which is "there does not exist a number $n$ which codes a proof of $0=1$ from $PA$". Importantly, this is a sentence in the language of $PA$ - that's why it makes sense to ask whether it's provable from $PA$. $Con(PA)$ most certainly does not code "$PA$ is actually consistent", which is a meta-theoretical statement and cannot be expressed in the language of $PA$. – Reese Johnston Jun 05 '17 at 04:57

  • Con(PA) most certainly does not code "PA is consistent" Probably not, rather Con(PA) certainly does code "PA is consistent" Of course, that is a meta-theoretical statement, but, that can be expressed in the language of PA. That's just the meaning of "coding"!

     – maimai Jun 05 '17 at 12:15
  • Otherwise, ___Con(PA)___ is true in the standard model doesn't follow from " ___PA___ is consistent". that's why your answer is right, I think. – maimai Jun 05 '17 at 12:18
  • @maimai $Con(PA)$ codes "$PA$ is consistent", not "$PA$ is 'actually' consistent". $Con(PA)$ is interpreted within whichever model we're in at the moment, not meta-theoretically. It expresses the idea that a number with a certain property does not exist; if that number did exist, and happened to be standard, then it would represent a "real" proof of a contradiction. – Reese Johnston Jun 05 '17 at 15:32
  • Hmm, I'm not convinced. But, I'm sure that your answer to my question is right. Moreover, your explanation is interesting. So, we'd better stop this too long argument. – maimai Jun 06 '17 at 20:49
  • I'm sorry about bothering you and, I appreciate your kindness very much! – maimai Jun 06 '17 at 20:49