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Can you find a division by zero error in the following short proof of Gödel's incompleteness theorem?

First a little background. $\text{G($a$)}$ returns the Gödel number of the formula $a$. As in Gödel's proof the primitive recursive funtion $\text{Z}(a)$ returns the Gödel number of the number $a$ of the formal system. Example: $\text{Z}(2) =\text{G}(\underline{2})=\text{G}(S(S(0)))$.

The primitive recursive function $\text{Sb($a$, $v_1$|$\text{Z}(y)$)}$ corresponds to the substitution of the formal system's number $y$ in the place of free variable $v_1$ in the formula that corresponds to number $a$. Example: $\text{Sb($\text{G}(v_1=0)$, $v_1$|$\text{Z}(2)$)}=\text{G}(v_1=0[v_1|\underline{2}])=\text{G}(\underline{2}=0)$.

The primitive recursive relation $x\text{B}y$ says that number $x$ corresponds to a sequence of formulas that proves the formula corresponding to number $y$. $\text{Provable}$ is then equivalent to $\exists x: x\text{B}y$.

And now to the actual proof. We assume that in the system $P$ primitive recursive relations are representable.

One of the following relations hold for any number $x$ corresponding to a formula with free variable $v_1$, either the formula corresponding to number $x$ with its free variable substituted by $y$ is provable or not:

$\text{Provable}(\text{Sb}(x,v_1\space|\text{Z}(y)))$ or $\neg \text{Provable}(\text{Sb}(x,v_1\space|\text{Z}(y)))$

Let formal system formula $r=\text{$\neg$Provable}(\text{Sb}(v_1,v_1\space|\text{Z}(v_1))$, where $v_1$ is the free variable. The formula is definable in $P$ since it's a primitive recursive relation with an added quantifier. Choose $x=G(r)$, $y=G(r)$. Now consider the first case:

$\text{Provable}(\text{Sb}(x,v_1\space|\text{Z}(x)))\Rightarrow P\vdash r[v_1|\underline{\text{G}(r)}] \Rightarrow P \vdash \neg \text{Provable}(\text{Sb}(\underline{x},v_1\space|\text{Z}(\underline{x})))\Rightarrow \neg \text{Provable}(\text{Sb}(x,v_1\space|\text{Z}(x)))$

The last line comes from assuming that $P$ is consistent and doesn't prove falsities. And as such, a contradiction! Therefore the second case $\neg \text{Provable}(\text{Sb}(x,v_1\space|\text{Z}(x)))$ must hold, and $r[v_1|\underline{\text{G}(r)}]$ is true but unprovable!

Dole
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  • Not sure about thew technical details, but isn't this just the 'last paragraph' of any proof of Godel's First Incompleteness Theorem? That is, after all the technical set-up, don't all proofs conclude with something like: "G says that G cannot be proven. So, if G can be proven, then (since system is sound), G is true, and hence G cannot be proven. Contradiction! So, G cannot be proven .. and G is therefore true." – Bram28 Aug 05 '18 at 19:05
  • @Bram28 No, this version of Gödel's theorem would require Rosser's improvement or some other additional concepts and is a much longer proof. Please don't down-vote or give a reason for doing so and I will make the appropriate changes. – Dole Aug 05 '18 at 19:10
  • I wasn't the downvote ... like I said, the technical details are a little beyond me, so I would certainly not feel confident to do that .. in fact, do you think I should remove my earlier comment as it it really missing the point by missing the technical details? – Bram28 Aug 05 '18 at 19:21
  • @Bram28 I don't mind the comment at all, it was a fair observation. Sorry, I was talking to whoever downvoted... – Dole Aug 05 '18 at 19:22
  • After a quick read, I don't see what the improvement is. You assume that $P$ is sound ("assuming that P ... doesn't prove falsities") which is in fact a stronger assumption than what Godel used (and which Rosser removed), so Rosser's improvement isn't relevant since you're not proving the improved version. Meanwhile, your assumption that $P$ respresents p.r. functions is subsuming a good deal of technical work. Can you clarify in what way this is simpler (or indeed different from, once all technical details are filled in) the usual proof? – Noah Schweber Aug 05 '18 at 19:29
  • @NoahSchweber In the proof consistency is assumed, not $\omega$-consistency as in Gödel's. I am not claiming the proof is an improvement (over Rosser's), just a neat (fake?) proof. – Dole Aug 05 '18 at 19:36
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    No, you've assumed much more than mere consistency: "$P$ doesn't prove falsities" is in fact stronger than $\omega$-consistency, and much stronger than consistency. Indeed, one of the consequences of GIT is that PA + "PA is inconsistent" is consistent even though it proves clearly false statements (assuming PA is consistent of course). – Noah Schweber Aug 05 '18 at 19:37
  • "Let formal system formula r=¬Provable(Sb(v1,v1 |Z(v1)), where v1 is the free variable." What does it mean ? The syntax of $Sb(x,v_1 |Z(y)))$ is subst in the formula $x$ with free var $v_1$ the numeral (a term) $Z(y)$. In $r=¬Provable(Sb(v_1,v_1 |Z(v_1))$, we have to subst the numeral for $v_1$ in place of var $v_1$ into formula $v_1$. But $v_1$ alone is not a formula. – Mauro ALLEGRANZA Aug 05 '18 at 19:39
  • (Also, I think language like "mind-blowing" and "amazingly short" is rather off-putting.) – Noah Schweber Aug 05 '18 at 19:46
  • @MauroALLEGRANZA Those are actually functions on numbers and the argument is all numbers. That's why G(formula) is substituted for the number to represent the formula. The $v_1$ in the middle is not an argument at all, just makes it easier to visualize I think. – Dole Aug 05 '18 at 19:57
  • Assuming that the operation of Subst makes sense in that case, the subst of $Z(v_1)$ in place of $v_1$ into $v_1$ will result in $Z(v_1)$. Thus the formula will be $r=\lnot Provable(Z(v_1))$ and we can say taht $r$ is true, because we cannot "prove" a number (in the formal theory, the provable formulas a not terms). – Mauro ALLEGRANZA Aug 05 '18 at 20:07
  • See this post regarding the way the $\text {Sb}$ function works. – Mauro ALLEGRANZA Aug 05 '18 at 20:09
  • @MauroALLEGRANZA Interesting observation, I would say that $v_1$ is not a number so you can't evaluate it in that way. And that in the proof the expression has the free variable substituted by number of the formal system. – Dole Aug 05 '18 at 21:03
  • $\text{Sb}$ is a numerical function that "maps" the operations on linguistic entity. Thus, the output value of $\text{Sb}$ is a number : the number of the formula resulting from starting with formula numbered $x$ with free var $v_1$ and replacing the free var with the numeral (a term of the language) $Z(y)$. – Mauro ALLEGRANZA Aug 08 '18 at 09:06
  • @MauroALLEGRANZA Yes, I do not mean that in the definition of r literally that function is used. Rather a representation of that function in the formal system. The $v_1$ is the free variable of the expression. – Dole Aug 08 '18 at 21:55

1 Answers1

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Once all technical details are filled in,$^1$ this seems to be just the usual argument.

In particular, from your comment "in the proof consistency is assumed, not $\omega$-consistency as in Gödel's," I think what you're trying to do is optimize the hypotheses on $P$ without invoking Rosser's trick, but you haven't done so. The dangerous phrase is

The last line comes from assuming that $P$ is consistent and doesn't prove falsities.

You're not just assuming that $P$ is consistent, but also that it is sound. This is a very strong assumption, in fact strictly stronger than Godel's assumption of $\omega$-consistency.$^2$

Just in case, let me summarize the standard argument. As far as I can tell yours is ultimately the same, but let me know if I'm missing something:

  • We start by proving that every p.r. function is representable in our theory and that the relevant operations/relations are p.r. We have some flexibility here. In particular, you'll often see instead an argument that proves that every computable function is representable and then invokes Church's thesis; however, that last bit isn't truly rigorous. Personally I think that the representability of all computable functions is fundamental and worth proving even on its own.

  • We then prove the diagonal lemma. You've essentially done what Godel did here and simply proved the specific case we need - the general diagonal lemma wasn't proved until Bernays, a few years later - but in my opinion there's no reason to not prove it; it's important, it's short, and we're in the neighborhood.

  • We can now write a sentence asserting its own $P$-unprovability, which because of the soundness of $P$ must be $P$-unprovable and hence true.

$^1$Namely, the first bulletpoint in my summary.

$^2$Incidentally both of these are stronger than what you actually need to run the argument with literally no change, namely $\Sigma_1$-soundness. This is indeed strictly stronger than consistency, since (by GIT) $PA$ + "$PA$ is inconsistent" is consistent (unless PA is inconsistent) but not $\Sigma_1$-sound. And of course any amount of soundness implies consistency, since $0=1$ is false.

Noah Schweber
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  • Just to be clear, this is not my proof ;). Also a quick question. By soundness you mean: $P\vdash \phi \Rightarrow P\models \phi$. This is valid for first order logic, so then so is the proof? Also what do you mean by $\sum_1$-sound or "any amount" of soundness? – Dole Aug 05 '18 at 21:19
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    @Dole There are two different notions of soundness. A logic is sound if provability implies entailment (what you've written). However, there's also the separate notion of soundness of a theory. When we have a "standard model" in mind (like $\mathbb{N}$) a theory is sound (with respect to that model) if it only proves sentences true in that model. A theory is $\Sigma_1$-sound if all the $\Sigma_1$ sentences it proves are true in the intended model. – Noah Schweber Aug 05 '18 at 21:25
  • Scweber Aaaah! This is so confusing. Aren't you now assuming that the meta system is sound then, since it's assumed to tell the truth? And if it uses exact same or equivalent axioms to reason about the numbers as $P$, then should not $P$ also be assumed to be sound? What if the meta system itself is not considered sound and uses the same axioms for numbers as $P$... but then maybe it just proved a falsity? Maybe the meta-system inconsistent and proved a falsity anyway... Sorry if this seems like a totally stupid observation, but it is confusing! – Dole Aug 05 '18 at 22:24
  • @Dole Let's call our meta-system $M$. Then the usual proof of GIT shows that $M$ proves "If $P$ is consistent and sound, then $P$ is incomplete." $M$ doesn't need to reason about its own soundness at all: $M$ says, "If $P$ is consistent, then if $P$ is sound $P$ can't prove that $P$ is inconsistent." Taking $P=M$ wouldn't be a problem. There's an important point here: when we prove a statement in a theory, we're just doing that - we're not proving the statement + "the theory is sound" in the theory. (cont'd) – Noah Schweber Aug 05 '18 at 22:34
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    The important point is that a theory can prove things - even about itself - without being able to prove itself sound. As a good example, take the theory PA + "PA is inconsistent." Obviously this theory doesn't think itself sound, since it doesn't even think itself consistent. But it can still prove things about itself - e.g. its own inconsistency! – Noah Schweber Aug 05 '18 at 22:37
  • Initially when you generously provided these comments I was not sure what to think and decided to give it a rest. Logicians could work on their language. "PA thinks itself consistent" or "proves things about itself", more accurately means the meta-language makes the observation. We must indeed assume BOTH $P$ and $M$ sound, or that they prove same things regarding PRRs. This seems to lead to a contradiction when $P=M$, as the unprovable formula is provable in $M$, but it's the same system, so likely there is an error in the proof. – Dole Dec 23 '18 at 15:05
  • @Dole I think you've still misunderstood me: it doesn't lead to a contradiction when $P=M$, since $M$ might not prove the $P$-unprovable sentence. $M$ can only prove "If $P$ is consistent and sound, then $P$ is incomplete." At no point do we need to assume that $M$ is sound in order to prove, in $M$, things about $P$. ("The important point is that a theory can prove things - even about itself - without being able to prove itself sound.") – Noah Schweber Dec 23 '18 at 16:32
  • Thanks for the comment. If $P$ is consistent and sound, then $M$ is consistent and sound, as they are the same system. Therefore outside of $M$ we can make the conclusion that it lies or is inconsistent. – Dole Dec 23 '18 at 18:06
  • @Dole What? What statement does $M$ make that is incorrect? All $M$ says is that if $P$ is consistent and sound, then $P$ is incomplete. – Noah Schweber Dec 23 '18 at 18:20
  • $M$ makes the statement that there is a sentence in $M$ that is unprovable if $M$ is sound and consistent, but is "true" (or in other words, provable in $M$). – Dole Dec 23 '18 at 18:34
  • @Dole You're conflating "true" with "provable in $M$." $M$ - even if sound and consistent (indeed, especially then) - need not prove all true sentences. – Noah Schweber Dec 23 '18 at 18:35
  • Scweber In this case by "true" I literally meant provable in $M$, as the above proof shows the sentence to be provable in $M$ by contradiction. Further, the mere possibility of $M$ proving the sentence seems enough for the purpose. – Dole Dec 23 '18 at 18:47
  • @Dole "the above proof shows the sentence to be provable in M by contradiction." Um, what? Again, all $M$ can prove is that if $P$ is sound and consistent then it's incomplete. $M$ may not know that $P$ is sound and consistent, and conflating "true" with "provable (in $M$)" is exactly the source of your confusion. I don't want to discuss this in the comments any further; I think you should ask a separate question about this if you're still confused. – Noah Schweber Dec 23 '18 at 18:49
  • I am aware of that. If you apply the argument to the system $M$ itself as you brought up, we can conclude that $M$ is stating that an unprovable statement in $M$ (given the consistency etc.) is provable in $M$ which means $M$ lies or is inconsistent. The argument no longer takes place in $M$ but in a meta system to $M$ of course. Apologies for the extended comments, I find this topic quite interesting. – Dole Dec 23 '18 at 18:59
  • @Dole " If you apply the argument to the system M itself as you brought up, we can conclude that M is stating that an unprovable statement in M (given the consistency etc.) is provable in M which means M lies or is inconsistent." No, you can't. Try to write it down in detail - where do you get provability in $M$ from? At no point in the above have we concluded that any statement, other than "IF $M$ is consistent/sound, then $M$ is incomplete" is provable in $M$. The hypothesis - which $M$ doesn't know - is not removable. – Noah Schweber Dec 23 '18 at 19:02
  • Try to write out a proof in $M$ - and if you think you get one, ask about it here. (And I do mean write out - don't just say "applying the same logic" or similar, but write down the actual steps $M$ needs to go through and keep track of what assumptions on $M$ we need in order to guarantee that $M$ can go through those steps.) – Noah Schweber Dec 23 '18 at 19:09
  • Thank you for the chat, perhaps I might ask something after thinking about it more. Clarification to readers: the proof does not take place in $M$, but in a meta system to the meta-system, $M_2$. – Dole Dec 23 '18 at 19:26