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The incompleteness theorem says that certain theories+deduction system contain at least one sentence (the Gödel sentence "$G$"), which can't be proven (in the system in which it holds).

(i) Is this theorem (incompleteness theorem) a statement formulated within the system of which the statement is about or is the theorem formulated in a meta language?

(ii) As soon as the theorem is established, is there readily the implication that "$\neg G$" is also not provable? And again, is this then a statement of the meta language?

(iii) In case that a mega language is crucial, what are the minimal requirements for it's logic?

(The thread here, "Is it always possible to decide if either a statement or its negation is provable in a given axiomatic system?" is related.)

For me this question is kind of a follow up to Aftermath of the incompletness theorem proof. I don't understand the notion of "A sentence $p$ is true in $\mathbb{N}$" if $p$ is neither an axiom nor provable by a deduction system. My ansatz was to establish "$G$ not provable","$\neg G$ not provable" while $G\lor\neg G$ is true, which would directly imply that one of them ($G$ or $\neg G$) is proven to be true and not provable in the investigated system. If the proposed conditions are always satisfied for the Gödel sentence and the meta language, then I could comprehend the formulation "it's true but unprovable", because "true" wouldn't come from outside.

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Nikolaj-K
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    (i) The unprovable statement in is within the system, the unprovability is within the meta-system. (ii) Yes, that's the idea. Unprovable statements whose negation is provable are trivial to obtain "$\exists x: x\neq x$." – Michael Greinecker Aug 15 '12 at 19:04
  • Isn't the point of the Gödel sentence that it can't be assigned a truth value? Assigning a truth value to $\neg G$ would also assign one to $G$. – axblount Aug 15 '12 at 19:05
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    In fact, any statement of the pure form '$X$ is not provable' can only be proven by a metasystem, because a statement's unprovability implies the consistency of the original system (an inconsistent system will gladly prove anything), and so by Godel's second cannot be shown within the system. Instead, the canonical way of internally formulating unprovability is '$\mathrm{Con}(S)\implies \neg B_S(G)$', where $B_S$ is the provability predicate for the systrem $S$. – Steven Stadnicki Aug 15 '12 at 19:13
  • @MichaelGreinecker: Okay, that should answer (i). I don't get why you bring the example for (ii), does the existence of such provable statements help me say anything about $\neg G$? – Nikolaj-K Aug 15 '12 at 19:24
  • @axblount: "Assigning a truth value to ¬G would also assign one to G." Well, yeah if I argue in the meta language and in it I can talk about the statement $G$ and additionally $G\lor\neg G$ holds there, then you're right. – Nikolaj-K Aug 15 '12 at 19:26
  • @axblount: Given an $L$-structure $M$, every sentence of $L$ is unambiguously assigned a ruth value. Of course, we often do not know what that truth value is. – André Nicolas Aug 15 '12 at 19:38
  • @AndréNicolas: If we do not know it and it's one of the sentences for which we can't compute it, what does "unambigously" mean? Guess: Are there statement which say "All the consistent meta-languages which are strong enough to provide a computable truth value, will provide the same truth value"? – Nikolaj-K Aug 15 '12 at 19:44
  • @NickKidman: Unambiguously means that the definition of the meaning of truth in $M$ is by a simple induction on the length of the sentence, and the uniqueness is an immediate consequence of the definition. Consequence in what system? A very weak system. But I have never been interested in meta. Any conventional mathematical tool is OK for analyzing the relationship between formal theories and their models. – André Nicolas Aug 15 '12 at 20:00
  • @AndréNicolas: Okay, do you have a short reference, if only Stanford encylopedia or Wikipedia? – Nikolaj-K Aug 15 '12 at 20:17
  • @NickKidman: Last time I taught the course, I used Mathematical Logic by Joseph Shoenfield . A bit old-fashioned, but very careful. I am sure many books will do. If I have time, will try to locate a respectable-looking set of notes on the Internet. But I am not good at looking, I think it would make a good question on MSE. Since you are seriously interested in the subject, you need a structured development, not subjective "informal interpretations." – André Nicolas Aug 15 '12 at 20:48
  • The big breakthrough with $G$ is that you can express $G$ in $T$. You don't then try to arrive at $G$ by following axioms and rules of inference. Its very existence is sufficient. Suppose you can prove $G$. Well, $G$ says "you can't prove $G$", so that's a problem. Now suppose you can prove $¬G$. $¬G$ says "you can prove $G$", so that's also a problem. Either way, you've got a problem. Now, if you can, by some other method, prove that your theory $T$ is consistent, then you rule out proving either $G$ or $¬G$. So the consistency of $T$ implies that both $G$ and $¬G$ are unreachable... – crf Aug 15 '12 at 21:23
  • But if $G$ is unprovable then "G is unprovable" is true. So you've proven that there exists a statement which is true and unprovable. – crf Aug 15 '12 at 21:24
  • @crf: Thanks, though regarding what you point out in the last sentence, I'm aware of how to conclude "$G$ is true" within the meta language. This is basically my rant at the beginning of the previous thread. Problem is that by this reasoning the phase "the Gödel sentence is true" solely relies on the truth understanding of the meta-logic dealing with the theory. However, there seems to be another truth understanding around which makes people say "$p$ is true in $\mathbb{N}$, even if not provable". – Nikolaj-K Aug 15 '12 at 21:54
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    On one level, it's Platonism – some people believe there is one "true" model of arithmetic, called $\mathbb{N}$. On another level, we're just thinking about provability in a bigger system. As I said before, don't read too much into the word "true". – Zhen Lin Aug 16 '12 at 01:10

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There are two things to be aware of here.

The first one is that the Godel sentence is independent of the theory. It's specifically constructed so that neither G nor not G is a theorem.

The second one is that if one considers a particular model of the theory, then either G is true under that interpretation, or not G is true.

The "unprovable but true" version of Godel's incompleteness theorem only makes sense when the speaker is implicitly referring to a particular model -- usually the particular set of natural numbers that is implicitly part of the theory of formal logic.

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    $\neg G$ won't be a theorem of the relevant theory $T$ so long as $T$ is $\omega$ consistent. But if $T$ is $\omega$ inconsistent we can have $T \vdash \neg G$. – Peter Smith Aug 16 '12 at 07:40
  • @PeterSmith: No we can't. Gödel needed an assumtion of $\omega$-consistency to show that neither $G$ nor $\neg G$ is provable, but Rosser later gave a stronger construction that needs only consistency. – hmakholm left over Monica Aug 16 '12 at 12:57
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    @HenningMakholm Sorry, I'm afraid you are doubly wrong. (i) it is a standard textbook result that $T = PA + \neg\mathit{Con(PA)}$ proves $\neg G_T$. See §25.6 of my book for the proof. – Peter Smith Aug 16 '12 at 17:42
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    @HenningMakholm (ii) Sure, $T$ won't prove the negation of its own Rosser-sentence. But that's irrelevant, as $T$'s Gödel sentence and its Rosser sentence are not equivalent in $T$. – Peter Smith Aug 16 '12 at 17:45
  • @PeterSmith: I'm assuming that by $G$ we mean the particular unprovable sentence that's produced by the proof of the strongest available form of the incompleteness theorem (i.e. Rosser's one). If you think that letter should be reserved for one produced by Gödel's original construction, then naturally our conclusions about it won't agree. – hmakholm left over Monica Aug 16 '12 at 19:05
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    @HenningMakholm The original poster said "(the Gödel sentence "G")", and that is what I certainly assumed was under discussion. After all, when people talk of "the Gödel sentence" they surely don't usually mean "the Rosser sentence"! – Peter Smith Aug 16 '12 at 19:18
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    @PeterSmith: My impression is that "Gödel sentence" is usually used as an imprecise term for "some sentence that is systematically constructed to be independent of a particular axiomatic theory, following techniques similar to those introduced by Gödel". After all, each author seems to have his own particular way to fill in the details of the construction (understandably enough, because they want it to match the particular formalization of FOL they work with). – hmakholm left over Monica Aug 16 '12 at 19:24
  • So if you start out with an $\omega$-inconsistent theory and follow Gödel's original steps (which only work under the assumption of $\omega$-consistency), the result wouldn't even be a "Gödel sentence" in the fuzzy sense I'm used to. – hmakholm left over Monica Aug 16 '12 at 19:27
  • @Henning: That's certainly how I meant the term (and had previously seen it used). I had not known people ever used the phrase "Godel sentence (of T)" in a situation where the sentence is actually decidable by T, as Peter suggests! –  Aug 16 '12 at 19:31
  • @HenningMakholm I agree talk of Gödel sentences can be used quite widely for sentences produced by variations of Gödel's construction, or more generally for fixed points of $\neg\mathsf{Prov}$. But these are still to be contrasted with Rosser sentences which have interestingly different properties. So I think your over-wide usage is probably out of step with the usual practice. – Peter Smith Aug 16 '12 at 19:34
  • @Hurkyl Use "canonical Gödel sentence for $T$" to mean the result of applying Gödel's 1931 construction (or close variants) to $T$ to get a sentence $G_T$. Then canonical Gödel sentences (though they will always code up 'I am not provable', be fixed points for $\neg\mathsf{Prov}_T$ etc.) may not be Gödel sentences in the Henning/Hurkyl sense. Agreed. That gets us all singing from the same hymn-sheet! But I suspect most would say that Gödel sentences (unlike Rosser seentences) can get decided in $\omega$-inconistent theories, but are Gödel sentences nonetheless! – Peter Smith Aug 16 '12 at 19:52
  • I am also more familiar with the usage in which arithmetics with inconsistent proof predicates will prove $(\forall x)\operatorname{Pvbl}(x)$ (that's the definition of an inconsistent proof predicate) and thus will prove the negations of their own Goedel sentences, as Goedel sentences are constructed as a sort of fixed point for $\lnot \operatorname{Pvbl}$. – Carl Mummert Aug 20 '12 at 15:00
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To take your three questions in turn:

(i) The incompleteness theorem for theory $T$ is a theorem about what can't be proved in $T$. If $T$ is e.g. a pure theory of arithmetic, its language is about numbers, not about $T$-proofs. The theorem won't even be stateable in $T$'s language: rather is stated in e.g. mathematical English.

Of course, by the trick of Gödel coding we can produce in $T$ a sentence which codes up the claim that if $T$ is consistent then a Gödel sentence for $T$ is unprovable, to get $\mathsf{Con} \to \neg\mathsf{Prov(\overline{\ulcorner{G}_{\mathit{T}}\urcorner})}$. And that sentence will itself be provable in $T$ on very modest assumptions. So we might say that $T$ can itself prove the incompleteness theorem for $T$: but really that is rather careless talk. To repeat, if $T$ is a theory of arithmetic whose interpreted language is about numbers, then its theorems are strictly speaking about numbers and numerical properties, not about proofs.

(ii) No. There are consistent theories $T$ with a Gödel sentence $\mathsf{G}_T$ such that $T$ proves $\neg\mathsf{G}_T$. For example take $T$ to be the consistent but $\omega$-inconsistent theory you get by adding to PA the negation of a standard Gödel sentence for PA. (The unprovability of the negation of the Gödel sentence does indeed require we are dealing with an $\omega$-consistent theory.)

(iii) The proof of Gödel's theorem doesn't require excluded middle: it goes through intuitionistically. (It only involves the intuitionistically acceptable version of reductio.)

Peter Smith
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  • In your (ii), you add $\neg G_T$ to be part of the peano axioms. Doesn't that imply that the sentence $G_T$ is false and so it can't be a Gödel sentence, as that sentence is supposed to be "true, while unprovable"? And yeah, meanwhile I'm less sure if the things I asked in (iii) have a significant impact. – Nikolaj-K Aug 16 '12 at 07:45
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    In (ii), the $G_T$ you start out with would not be a Gödel sentence for the extended theory; in other words $G_{T+\neg G_T}$ is different from $G_T$, and would still be independent of $T+\neg G_T$. So neither $T$ nor $T+\neg G_T$ has the property you claim. – hmakholm left over Monica Aug 16 '12 at 12:54
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    @HenningMakholm No, $PA + \neg G_{PA} \vdash \neg G_{PA + \neg G_{PA}}$. That's consistent with Gödel, because $PA + \neg G_{PA}$ is not $\omega$-consistent. As I say, a standard textbook result. – Peter Smith Aug 16 '12 at 17:49
  • @NickKidman No, adding $\neg G_{PA}$ as an axiom doesn't make it true! The resulting theory $PA + \neg G_{PA}$ is indeed unsound, which is why we can derive falsehoods in it (like the negation of its own canonical Gödel sentence!). – Peter Smith Aug 16 '12 at 17:53
  • @PeterSmith: By "unsound" and "falsehoods", do you mean merely that it doesn't have $\mathbb N$ as a model? – hmakholm left over Monica Aug 16 '12 at 19:00
  • @HenningMakholm I'm not sure what "merely" means here! I take it that our theories $PA$ and $PA + \neg G$ are written in a formalized language with a built-in standard interpretation, where "$0$" refers to zero, "+" means addition, and so on. Thus construed, some of the theorems of $PA + \neg G$ are plain false. – Peter Smith Aug 16 '12 at 19:26
  • @PeterSmith: That's more of a philosophical discussion, I think. I'm Platonist enough to think that "being true in $\mathbb N$" means something, but formalist enough to think that the ideas of formal language or formal theory do not inherently include that a language or theory come with a built-in intended interpretation. And certainly if one appends something like $\neg G$ to PA, then I won't consider it implied that the resulting theory is to be interpreted in $\mathbb N$. – hmakholm left over Monica Aug 16 '12 at 19:33
  • @Peter: Everything I've learned about formal logic has carefully made the distinction between syntax and semantics. It is very strange to me to see someone assert in this context that a formal theory has a "built-in interpretation"! –  Aug 16 '12 at 19:35
  • @Hurkyl English has syntax and semantics, yes? Polish has syntax and semantics, yes? $L_A$, the language of first-order arithmetic, has syntax and semantics, yes? Formalizing a language doesn't empty it of meaning! Of course, for various technical purposes, we can be interested in keeping syntax fixed and spinning the semantics: but that doesn't mean that e.g. $L_A$ doesn't come with a built-in semantics. Or that's what my local friendly logicians think ... :-) – Peter Smith Aug 16 '12 at 20:00
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    @Peter: I emphatically disagree with the assertion $L_A$ that has semantics: it is an inherently a purely syntactic object! Formalizing a language is the thing we do when we specifically want to manipulate a language at a purely syntacic level! (I disagree that pure syntax is "devoid of meaning" but that's a side topic.) Of course, our disagreement only emphasizes why it's important to separate syntax and semantics, even in English. While we are both making statements according to the same syntax, we are clearly applying very different semantics! –  Aug 16 '12 at 20:56
  • Regarding your third point "(iii) The proof of Gödel's theorem doesn't require excluded middle: it goes through intuitionistically.", I read this section of the Wikipedia article on prenex normalform and it says "Gödel's proof of his completeness theorem for first-order logic presupposes that all formulae have been recast in prenex normal form.". So does this mean the proof is very different intuitionistically? I guess this stuff matters w.r.t. the Gödel number coding. – Nikolaj-K Jun 09 '13 at 17:23
  • @NickKidman You are confusing Gödel's incompleteness theorem and Gödel's completeness theorem which, despite the confusing similarity of their labels, are about quite different things. – Peter Smith Jun 09 '13 at 18:44
  • @PeterSmith: Okay, so the incompleteness theorem doesn't need bringing it to normal form? What I really don't get is this: In the incompleteness theorem, the proof scheme of a logic capcable of expressing PA gets encoded by function from (Gödel)Number to 0 or 1, or vice versa, right? But can't there be a totally alternative way of proving theorems within that logic (which happens to know PA), which doesn't relate to the numbers at all? – Nikolaj-K Jun 10 '13 at 08:13
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    @NickKidman I suspect you need to read a textbook or two that gives a good presentation of Gödel's incompleteness theorems. There are plenty out there. For some recommendations see http://www.logicmatters.net/students/tyl/ – Peter Smith Jun 10 '13 at 08:34