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It shows that the sentences of the form $\forall x, \neg Dem(x,sub(n,y,n))$ are true but unprovable, where $y$ is the Godel number mapped to the symbol $Y$ in arithmetic language, and $n$ is the Godel number of the sentence $\forall x, \neg Dem(x, sub(Y,y,Y))$

Suppose for some $y$, the sentence $\forall x, \neg Dem(x,sub(n,y,n))$ has the Godel number 1000. Then following is a true but unprovable property of the number 1000:

"The number 1000 can never be at the end of a sequence of natural numbers $x_{k}, k=1,2,3,....n$, such that every $x_{k}$ is either a Godel number of an axiom or of a sentence logically deducible from the axioms."

Properties like this, however, are a very narrow class of the properties of natural numbers. Is this even an arithmetic property? I'm asking this because this property does not involve operations of arithmetic like addition, multiplication. Instead, it involves decoding the number to a language of symbols and using rules of logical deduction. Arithmetic properties are something like '$\forall x, 10x\neq 5, x\in Z$'

Even if this is accepted as an arithmetic property, it is a very narrow class of arithmetic properties. These properties are also dependent on the map that we use for the one-to-one correspondence between integers and symbols (there are multiple maps possible). Does the Incompleteness theorem say anything about non- self referential properties of numbers?

Ryder Rude
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    It proves that there are sentences in the language of arithmetic that are unprovable. Following G's discovery, more "mathematical" unprovable sentences have been found; see Paris-Harrington. – Mauro ALLEGRANZA Apr 03 '20 at 06:35
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    If you look at the sentence all you see is a lot of arithmetic. If you didnt know how it was devised, how would you be able to tell that it was "self-referential"? – bof Apr 03 '20 at 06:38
  • @MauroALLEGRANZA Do we have proof that the unprovable and non self-referential truths are also infinite in number? Godel shows that we can keep producing the self-referential truths infinitely (and add them as axioms), hence our theory will never be complete. But maybe Peano Arithmetic can be extended (by adding axioms) so that it's able to prove at least all of the non self referential truths (like the one you linked)? – Ryder Rude Apr 03 '20 at 07:03
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    How do you define "self-referential"? I suppose one could devise some bizarre coding under which Goldbach's conjecture or $2+2=4$ were "self'-referential" statements. Anyway, if there is one unprovable "non-self-referential" true sentence, call it $\sigma$, then wouldn't $\sigma$, $\sigma\land\sigma$, $\sigma\land\sigma\land\sigma$, . . . be an infinitude of such sentences? – bof Apr 03 '20 at 21:57
  • @bof Yes, but if we add $\sigma$ as an axiom, we've proven all of $\sigma$,$\sigma \cap \sigma$, $\sigma \cap \sigma \cap \sigma$, etc. I'm not sure how we could write $2+2=4$ in the form $\forall x, \neg Dem(x, sub(n,y,n))$ – Ryder Rude Apr 04 '20 at 01:56

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As others have said, the very distinction you're trying to draw is fuzzy to the point of meaninglessness, and there are very natural examples of undecidable (with respect to any of the usual theories) sentences. However, there's a very nice fact which hasn't been mentioned yet and I think puts the final nail in the coffin:

For every "reasonable" theory $T$, there is a Diophantine equation $\mathcal{E}_T$ such that $\mathcal{E}_T$ has no solutions but $T$ cannot prove that $\mathcal{E}_T$ has no solutions.

It's hard to get more concrete and arithmetic than Diophantine equations!

Here as usual "reasonable" means "computably axiomatizable, consistent, and extending $Q$" (although even that is overkill: we can replace "extending $Q$" with "interpreting $R$" or even weaker hypotheses - see here, Section $4$). The result above is a more-or-less immediate consequence of the proof of the MRDP theorem: basically, we can assign to $T$ a Diophantine equation $\mathcal{E}_T$ such that putative solutions to $\mathcal{E}_T$ correspond to putative contradictions in $T$. Of course, the $\mathcal{E}_T$s so produced are truly god-awful, but they are honest-to-goodness Diophantine equations

Noah Schweber
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The first incompleteness theorem states that certain theories, under certain conditions, are incomplete.

To prove that something is incomplete, you need to provide a witness of that. But of course, if the example of incompleteness is something like the Goldbach Conjecture, or the Riemann Hypothesis, what would be the example once you add those statements to your theory? The proof needs to be robust.

Well, Gödel's proof is generic, it uses the fact that theories with certain properties can encode certain statement, and this is how the witness is produced. By creating these "self-referential statements". The second incompleteness theorem is slightly deeper, and it talks about a theory not being able to verify its own consistency, which is arguably a more interesting example.

But if your goal is to prove incompleteness, you need to be able to do it in a generic, constructive way. And since we decide, socially, what are "meaningful statements", it's hard to do that with "meaningful statements".

Some people fall into the trap of thinking that incompleteness is only with regards to "meaningless self-referential statements and consistency statements", but if we take a gander at set theory, we see this is very much not true. And while it is admittedly harder to find independence in the natural numbers, it is still not impossible. (And arguably, consistency statements are meaningful mathematical statement.)

Asaf Karagila
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  • "but if we take a gander at set theory, we see this is very much not true". How can we say statments about Sets are true or false independent of our theory? We don't have a univeral model for sets. For example, people argue both in favor and against of continuum hypothesis. – Ryder Rude Apr 03 '20 at 10:22
  • Do we have proof that the unprovable and non self-referential truths are also infinite in number? Godel shows that we can keep producing the self-referential truths infinitely (and add them as axioms), hence our theory will never be complete. But maybe Peano Arithmetic can be extended (by adding axioms) so that it's able to prove at least all of the non self referential truths. – Ryder Rude Apr 03 '20 at 10:22
  • Ryder, the "universal model" you speak of for arithmetic depends on your set theoretic universe. But yes, this is why we normally avoid talking about "true and false" in the case of set theory. But the question "Is ZFC consistent?" is really a question about the natural numbers, for example. – Asaf Karagila Apr 03 '20 at 10:41
  • Ryder, do you consider a consistency statement to be "self-referential"? What is a non-referential statement? – Asaf Karagila Apr 03 '20 at 11:02
  • No, I accept that the consistency statement is a very powerful result. Can we say that there might exist a mathematical system which can prove all truths except for the self-referential truths, that the first theorem talks about, and its own consistency? – Ryder Rude Apr 03 '20 at 11:10
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    The issue here is not with the lack of such theory, but with its lack of usefulness. Taking "all true statements" is such theory, but it's not useful, since we can't recognise the statements which are true via an algorithm. But you seem to try to say "let's sweep this incompleteness nonsense under the carpet", which I have seen a few professional mathematicians trying to do (you're not in bad company, per se). But this is just missing the point, and failing to understand that (1) we haven't even agreed on what is "interesting and non-referential", and [...] – Asaf Karagila Apr 03 '20 at 11:22
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    [...] (2) we don't even know which of the statements we would like to think of as "interesting and non-referential" are even true, so how can we know which axioms are needed to prove them? And then you get research programmes like Harvey Friedman's work, where he looks for statements which are natural, but are equivalent to consistency statements, and are certainly not provable (not even if you assume ZFC). What do you do then? If we can't agree on a base-theory, how do we agree on "a theory that fixes incompleteness"? – Asaf Karagila Apr 03 '20 at 11:24
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Properties like this, however, are a very narrow class of the properties of natural numbers. Is this even an arithmetic property?

Yes, this is achieved via Gödel's encoding.

Does the Incompleteness theorem say anything about non- self referential properties of numbers?

What would be your definition of "self referential properties of numbers"? The most striking example of an unprovable sentence is given by Gödel's second theorem : let $F$ be your system, the sentence $Cons(F)$ which states that $F$ is consistent, eg $Cons(F) := \neg Dem("0 = 1")$, is not provable by $F$.

Olivier Roche
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  • Can the $Dem(x,z)$ predicate be computed by only performing arithmetic operations (addition, subtraction, multiplication, division) on $x$ and $z$?. Also, $Cons(F)$ is a property of the theory $F$, not of our model. The question is about the First Incompleteness theorem. – Ryder Rude Apr 03 '20 at 06:56
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  • Yes, that is the point of the whole story. 2. So, you are asking for a proof of the first incompleteness theorem that doesn't use a "self-referential property"? .
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  • Oh, and 3. please ask only one question at a time :) – Olivier Roche Apr 03 '20 at 07:10
  • I think the computation of $Dem(x,z)$ involves decoding $x$ and $z$ into symbols, and seeing if the steps in the proof $x$ logically follow to the sentence $z$. Is this not true? – Ryder Rude Apr 03 '20 at 07:12
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    @RyderRude If reading the wikipedia article linked in my answer isn't enough for you, I warmly recommand this book, which covers your (pertinent) questions. – Olivier Roche Apr 03 '20 at 07:16
  • Thanks I'll read it. But still, even if $Dem$ is purely arithmetical, the First Incompleteness theorem only talks about the unprovability of self-referential sentences – Ryder Rude Apr 03 '20 at 07:30
  • Well, any sentence $f$ is logically equivalent to a self referential sentence anyway : consider the sentence $\phi$ stating $f \wedge \big(Bew(\phi) \vee \neg Bew(\phi)\big)$. – Olivier Roche Apr 03 '20 at 07:46
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    @RyderRude I don't think the distinction between "self-referential" and "non-self-referential" statements is clear-cut, or even meaningful. Try to devise a rigorous but non-trivial definition of "non-self-referential sentence". – Z. A. K. Apr 03 '20 at 07:50
  • @Z.A.K. For now, let's just name $\forall x, \neg Dem(x,sub(n,y,n))$ as self referential sentences. First Incompleteness theorem proves the unprovability of these sentences. – Ryder Rude Apr 03 '20 at 08:10
  • @RyderRude I edited my question as to show you that $Cons(F)$ is a sentence that is not self-referential. – Olivier Roche Apr 03 '20 at 08:19
  • @OlivierRoche I don't know about $Bew$. Where can I read up on that? – Ryder Rude Apr 03 '20 at 08:21
  • @RyderRude Oops $Bew$ is the german equivalent of $Dem$, let me fix that... – Olivier Roche Apr 03 '20 at 08:30
  • Godel's encoding is very questionable, because it makes variables have the same type as constants, since numbers are constants. How can a number vary? If the number 17 is a variable, even a propositional variable, how can 17 be possibly true, or 17 be possibly false? If you wrote a Godel encoding for a formula like CpCqp, to evaluate all possibilities of that formula, we would have to let 17 equal truth at one spot, and then 17 at another spot. But, this isn't logical to do since every number is a constant. Thus, the encoding is not valid and WAS NEVER valid to begin with. – Doug Spoonwood Apr 03 '20 at 17:44
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    @RyderRude Doug Spoonwood's comment above is completely incorrect, FYI - based on previous conversation I'm not going to engage on it, but I think it's important to say this explicitly given that you're studying this material. – Noah Schweber Apr 03 '20 at 18:11