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First the statement as I understand it:

Given some consistent logic (axioms + rules of inference) which can encode enough arithmetic to do addition and multiplication, we can always find propositions for which neither the proposition itself nor its negation is provable.

But if the logic has two non-isomorphic models, then this is rather trivial, or? For then just choose a proposition which is true in one model, but false in the other...

Or does Gödel means by a true proposition a proposition that is true in every model of the logic?

StefanH
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    If a statement is true in EVERY model, it is provable. – Peter Aug 03 '17 at 20:04
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    In Gödel's completeness theorem (his dissertation), he proves that a statement true in every model is a theorem (provable statement) of the predicate calculus. Note that two non-isomorphic models can be elementarily equivalent, that is satisfying the same statements of first-order logic. – hardmath Aug 03 '17 at 20:08
  • Yes, thanks that is completeness, and also the converse holds, so provability and logical validity are equivalent. But then I cannot make much sense of Gödels Incompleteness-Theorem, for then simply the fact that we have two different models gives the result, and if we do not have two different models it cannot hold either... – StefanH Aug 03 '17 at 20:08
  • It is "intuitively true." Basically, we encode the notion of proof in number theory. If the statement is false, the internal model of proofs is incorrect in some way - it could allow some odd proofs. – Thomas Andrews Aug 03 '17 at 20:08
  • @ThomasAndrews I do not understand? – StefanH Aug 03 '17 at 20:10
  • The result is stronger than simply "incompleteness" of the Peano axioms or some close variation. Gödel proves that this arithmetic system is hereditarily incomplete, meaning that no formal extension of it is complete. – hardmath Aug 03 '17 at 20:10
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    https://www.youtube.com/watch?v=O4ndIDcDSGc –  Aug 03 '17 at 20:12
  • @hardmath Your reference to elementarily equivalent might make sense, but then I wonder why Gödel do not refer to such propositions, saying "then there exists a proposition of first order logic such that it, or its negation is not provable." – StefanH Aug 03 '17 at 20:13
  • Wait! Gödel is about Peano arithmetic, and this is not just first order, so the completeness result does not apply here... – StefanH Aug 03 '17 at 20:13
  • "... unless arithmetic is inconsistent." Yes, Gödel constructs (modulo an improvement by JB Rosser) just such a proposition. – hardmath Aug 03 '17 at 20:14
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    "But if the logic has two non-isomorphic models, then this is rather trivial, or?" Well yes (modulo pedantry about elementary equivalence), but then you're assuming the conclusion. It is not trivial to determine whether a particular theory has non-equivalent models or not; the incompleteness theorem gives a (surprising) sufficient condition for this to happen. – hmakholm left over Monica Aug 03 '17 at 20:15
  • I'm not sure what you mean by "not just first order". Gödel is working with a formal first-order theory (of arithmetic) essentially the same as Peano's axioms. – hardmath Aug 03 '17 at 20:16
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    The incompleteness theorems apply in a more general context than just for Peano arithmetic, but this may add a bit to the confusion: In the context of arithmetic, to be "true in a model" is not the same as being "true", which in turn is not the same as being "true in every model". A sentence that is true in every model (of the appropriate theory) is provable, this is Gödel's completeness theorem. We say that a sentence is true to mean that it is "true in the standard model", that is, $\mathbb N$ with the usual interpretation of $+,\times$, etc. – Andrés E. Caicedo Aug 03 '17 at 20:16
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    Naturally, if a sentence is true in every model, it is true. (Implicitly, we are assuming soundness, that is, that $\mathbb N$ is indeed a model of the Peano axioms.) If we work with a true theory (not necessarily just Peano arithmetic), meaning one for which $\mathbb N$ is a model, the same designations apply. If we work with something else (such as Peano arithmetic $\mathsf{PA}$ plus the statement that $\mathsf{PA}$ itself is inconsistent), then of course it is only relevant to discuss truth-in-a-model rather than truth. – Andrés E. Caicedo Aug 03 '17 at 20:17
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    @hardmath: Actually, in Gödel's original article, he was working with a higher-order theory system from Principia Mathematica. Later on it has become more common to present the argument as applied to first-order theories. – hmakholm left over Monica Aug 03 '17 at 20:19
  • This is close to a duplicate of the recent https://math.stackexchange.com/questions/2380885/understanding-the-definition-of-completeness-of-formal-theorysand-godels-famous, but not quite enough for me to apply the dupehammer alone. – hmakholm left over Monica Aug 03 '17 at 20:23
  • @HenningMakholm I looked it up, and it looks like he states the Peano Axioms, hence is logic just has a single model (as PA is categorical) which gives that we cannot have two models, so this conclusion could not be drawn... – StefanH Aug 03 '17 at 20:24
  • @StefanH: He states a second order variant of Peano's axioms at the bottom of page 177, but I can't find anywhere in the paper where he explicitly considers what today we call first-order Peano Arithmetic. – hmakholm left over Monica Aug 03 '17 at 20:33
  • @HenningMakholm Me too, so as for second order PA we just have the single model $\mathbb N$ and hence truth in the standard model and truth in every model means the same. – StefanH Aug 03 '17 at 20:37
  • @StefanH: In the 1931 article, however, Gödel is not concerned with "truth in every model", but just with provability in his system P. (He doesn't even bother to define what a model of P or an extended theory would mean). – hmakholm left over Monica Aug 03 '17 at 20:39
  • @HenningMakholm How can he talk about "truth" without reference to (at least) some model???? Isn't that a semantic notion... – StefanH Aug 03 '17 at 20:56
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    Note that the OP's own statement above of Gödel's Incompleteness Thm. makes no mention of "truth" or "true propositions". – hardmath Aug 03 '17 at 20:59
  • @StefanH: He seems to take it for granted that an arithmetical proposition is inherently either true or false (speaking about "inhaltlich richtig"). But this does not, as far as the 1931 article goes, depend on a notion of the "actual integers" being an interpretation among other possible interpretations. – hmakholm left over Monica Aug 03 '17 at 21:24
  • Guess I got it now, thanks to all! Maybe some of you might consider writing an answer so I can mark this question as answered... – StefanH Aug 03 '17 at 21:32

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In the question as asked, your mistake is the following:

if the logic has two non-isomorphic models, then this is rather trivial, or? For then just choose a proposition which is true in one model, but false in the other...

Just because $\mathcal{A}\not\cong\mathcal{B}$ does not mean that there is a sentence true in one and false in the other! There are non-isomorphic structures which are elementarily equivalent. One easy way to see this is just by a cardinality argument: in a given language $L$, there are only $2^{\vert L\vert+{\aleph_0}}$-many sets of sentences in the language $L$ but a proper class of structures, so we have to be able to find non-isomorphic structures with the same theory. More satisfyingly, the compactness theorem (or rather, its corollary the upwards Lowenheim-Skolem theorem) tells us that for any structure $\mathcal{A}$ there is a nonisomorphic, elementarily equivalent structure $\mathcal{B}$. And in particular, if $T$ is a complete theory with an infinite model, then $T$ has lots of nonisomorphic but elementarily equivalent models (since any two models of $T$ are elementarily equivalent since $T$ is complete). (Exercise: "has non-elementarily-equivalent models" is exactly the same as "is incomplete.")

So Godel's incompleteness theorem says far more than just that there are non-isomorphic models of any "reasonable" theory of arithmetic: it says that no reasonable theory of arithmetic even pins down the true natural numbers up to elementary equivalence!

It's worth considering, at this point, some examples of complete theories.

  • The theory of dense linear orders without endpoints is computably (indeed, finitely) axiomatizable and is complete; this is a standard exercise in most first-semester logic classes.

  • A more interesting example is the following. If we consider the structure $(\mathbb{N}; +)$, it turns out that Godel fails: the theory Presburger arithmetic is complete (it axiomatizes $Th(\mathbb{N}; +)$). It's also decidable, since Presburger arithmetic is computably axiomatized.

  • Similarly, it turns out that the theory of real closed fields is complete (and computably axiomatized); this is due to Tarski.

You also asked about truth (although truth doesn't even show up in your statement of Godel's theorem). The issue here - as in your other question - is that you are conflating two notions of truth. The truthy version of Godel's incompleteness theorem says

For any "reasonable" theory $T$, there is a true sentence $\varphi$ which is not provable in $T$.

"True" here means "true in the structure $(\mathbb{N}; +, \times)$;" this has nothing to do with truth in all models of $T$. A better statement of the "truthy" incompleteness theory is:

For any "reasonable" theory $T$, there is a sentence $\varphi$ such that $\varphi$ is not provable in $T$ but $(\mathbb{N}; +,\times)\models\varphi$.

This makes it very clear what's going on, and why Godel's completeness theorem is irrelevant. Again, it boils down to the fact that any reasonable theory of arithmetic $T$ will have lots of non-elementarily-equivalent models; the sentence $\varphi$ above is not true in all of them, hence (by Completeness) not provable in $T$, but is true in the standard model.

(Keep in mind that "reasonable" here means, among other things, "computably axiomatizable"; of course the set of all sentences true in $(\mathbb{N}; +,\times)$ is a complete theory, but Incompleteness doesn't apply to it since it's not computably axiomatizable.)

Noah Schweber
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  • I realised for the first time that there are two notions of complete around, one meaning that logical validity and provability are the same, the other that for every statement either it or its negation is provable (quite a strong assertion I find), and Gödels incompleteness refers to the last, and Gödels completeness to the first (quite confusing that two notions of completeness are associated with two famous theorems of one person). – StefanH Aug 04 '17 at 13:25
  • @StefanH See also Tarski's definition of truth and Tarski's theorem on the undefinability of truth! – Alex Kruckman Aug 04 '17 at 13:43
  • Ok, I will check that out. But by the way, how you know there are $2^{|L| + \aleph_0}$ sentences and what does $2^{|L| + \aleph_0}$ mean? – StefanH Aug 04 '17 at 15:18
  • @StefanH Whoops, that was a bad typo - there are $2^{\vert L\vert+\aleph_0}$-many sets of sentences (that is, theories) in the language $L$. This notation is cardinal exponentiation - if you're not familiar with cardinals (e.g. what "$\aleph_0$" means), start there and you'll quickly figure things out. The point is that there are only set-many sets of sentences, but class-many structures; so there will be lots of instances of structures which are non-isomorphic but elementarily equivalent. (In fact, we'll be able to find a proper class of elementarily equivalent non-isomorphic structures!) – Noah Schweber Aug 04 '17 at 17:19