2

Reading through the proof of the diagonal lemma related to Gödel incompleteness theorem on wikipedia, I feel its a little bit complex. As far as this lemma relates to proving Gödel incompleteness theorem-I, I see the following being simpler:

Lemma: if $T$ is a first order theory capable of representing all computable functions, and $\mathcal F(x)$ is a formula in $T$ with one free variable, then there is a sentence $\mathcal C$ such that: $$ \vdash_T \mathcal C \iff \mathcal F(\ulcorner \mathcal C\urcorner)$$

We'll take $T$ to be $\sf PA$.

Initial thoughts about the above lemma is that the only obstacle to it is if there is an arithmetical formula $\mathcal F(x)$ that opposes truth at all naturals, that is whatever natural $n$ such that $\mathcal F(n)$ and $n=\ulcorner \mathcal C \urcorner$ then $\neg \mathcal C$, and whatever $n$ such that $\neg \mathcal F(n)$ and $n=\ulcorner \mathcal C \urcorner$ then $\mathcal C$. In simple English: $\mathcal F$ holds of all Gödel numbers of false sentences, and $\neg \mathcal F$ holds of all Gödel numbers of true sentences. Now if there can exist such a truth opposing formula $\mathcal F(x)$, then it can be shown that this opposes Tarski's undefinability theorem, as follows:

Lets define an extractor function $ \zeta $ that sends $\ulcorner\mathcal A (\ulcorner \mathcal A(x) \urcorner)\urcorner$ to $\ulcorner \mathcal A(x) \urcorner $, I view this function as extracting the latter from the former. Along this setting $\ulcorner \mathcal A(x) \urcorner$ to be called the extractee of $\ulcorner\mathcal A (\ulcorner \mathcal A(x) \urcorner)\urcorner$. This is computable and so definable in the language of arithmetic (i.e. arithmetical)

Now since $\mathcal F(x)$ is arithmetical, then we can define a new arithmetical formula $\mathcal F'(x)$ such that:

$\textbf{Define: } \mathcal F'(x) \iff \mathcal F(x) \lor \exists y: \mathcal F(y) \land x=\zeta (y)$

In English: $\mathcal F'(x)$ holds of all Gödel numbers of false sentences and the extractees of those numbers and $\neg \mathcal F'(x)$ holds of all Gödel numbers of true sentences and the extractees of those numbers.

Now, the question about whether $\mathcal F'(\ulcorner \mathcal F'(x) \urcorner)$ or $\neg \mathcal F'(\ulcorner \mathcal F'(x) \urcorner)$ is true, is paradoxical!

So $\mathcal F(x)$ cannot exist. Which proves the lemma.

I see this proof to be very simple, virtually just a two step proof, as compared to multi-step proof present in the Wikipedia.

First, is this proof correct?

Can this proof be carried out under all the grounds the proof in the wikipedia are carried under?

My point is that if this proof is correct, then being so simple might mean that it is restricted in some sense, hence the question?

Zuhair
  • 4,749
  • 1
    Your "only obstacle" claim needs justifying. In particular, I'm worried you're conflating $T$-provability and truth there (and maybe later). – Noah Schweber Jul 08 '23 at 13:14
  • How is $\mathcal F’(\ulcorner \mathcal F’(x) \urcorner)$ paradoxical? – PW_246 Jul 08 '23 at 15:23
  • @NoahSchweber, if $\mathcal F(x)$ is not truth opposing, then either $T$ will prove the existence of a natural $n$ such that $\mathcal F(n)$ and $n= \ulcorner \mathcal C \urcorner $ and $\mathcal C$; or $T$ will prove the existence of a natural $n$ such that $\neg \mathcal F(n)$ and $n= \ulcorner \mathcal C \urcorner $ and $\neg \mathcal C$; or we may have both, and so in all of these situations the Lemma would hold in $T$. The only obstacle is when we don't have either. . – Zuhair Jul 08 '23 at 18:22
  • @PW_246, if $T$ proves $\mathcal F'(\ulcorner \mathcal F'(x) \urcorner )$ then $\ulcorner \mathcal F'(x) \urcorner$ is the extractee of the Gödel number of a true sentence, but that number would satisfy $\neg \mathcal F'$ by definition of $\mathcal F'$, violating the definition of $\neg \mathcal F'$ since $\mathcal F'$ can only hold of Gödel numbers of false sentences or extractees of those Gödel numbers, i.e. extrectess of numbers that $\mathcal F'$ holds of. Similar argument works when $T$ proves $\neg \mathcal F'(\ulcorner \mathcal F'(x) \urcorner )$. – Zuhair Jul 08 '23 at 18:31
  • 1
    @Zuhair You seem to be assuming that $T$ proves everything that's true ("if $\mathscr{F}(x)$ is not truth opposing then $T$ will prove [A or B]." – Noah Schweber Jul 08 '23 at 18:32
  • @NoahSchweber, those are simple first order logic deductions. – Zuhair Jul 08 '23 at 18:37
  • 1
    @Zuhair I really don't think they are. Can you try to write this part in more detail? Remember, just because a statement is true does not mean that $T$ proves that statement, so I really don't see how this is going to go. (And if by "true" you just mean "$T$-provable," then you have to handle the existence of sentences which are neither provable nor disprovable.) – Noah Schweber Jul 08 '23 at 19:58
  • 1
    You can't even state that last bit in the language of arithmetic (that's what Tarski's undefinability theorem says). You can't "unpack" a Godel number like that. And even ignoring that issue, you're conflating "$T$ proves that something exists with [property]" and "something exists such that $T$ proves [property]." – Noah Schweber Jul 08 '23 at 20:02
  • @NoahSchweber, I don't know what you mean, anyhow. I already proved above that there can be no $\mathcal F(x)$ that is truth opposing and arithmetical at the same time. You seem to miss the assumptions. If we assume they are arithmetical then we can write them in the above manner, then we show that this cannot work, see the main posting. That's the argument, i.e. by negation. – Zuhair Jul 08 '23 at 20:10
  • 2
    @Zuhair Again, I'm not focusing on that part of the argument (which is just Tarski): I'm focusing on your claim of reduction. That is, I'm challenging your claim "Initial thoughts about the above lemma is that the only obstacle to it is if there is an arithmetical formula ... that opposes truth at all naturals." Can you justify this claim? – Noah Schweber Jul 08 '23 at 20:30
  • @NoahSchweber, I can't see what else can be an obstacle in front of this Lemma. If $\mathcal F(x)$ is not truth opposing, then its truth meeting and that would only prove the lemma. Can you show me any other possible situation that incarcerate the lemma? – Zuhair Jul 08 '23 at 20:35
  • @Zuhair I must be missing something, but since $\mathcal F(\ulcorner \mathcal F’(x) \urcorner)$ is consistent with $\mathcal F’ (\ulcorner \mathcal F’(x) \urcorner)$, I don’t see how $\mathcal F’ (\ulcorner \mathcal F’(x) \urcorner)$ satisfies $\neg \mathcal F’$. – PW_246 Jul 08 '23 at 20:41
  • 1
    @Zuhair What if $\mathcal{F}$ happened to be true always, but $T$ couldn't prove that, or indeed any instance of $\mathcal{F}$? Then there might not be any $\mathcal{C}$ such that $$T\vdash\mathcal{F}(\ulcorner\mathcal{C}\urcorner)\leftrightarrow\mathcal{C}.$$ The actual (in $\mathbb{N}$) behavior of $\mathcal{F}$ isn't really important; what we care about is what $T$ proves about $\mathcal{F}$, and that might be very little. – Noah Schweber Jul 08 '23 at 20:51
  • @PW_246, No! $\neg \mathcal F'$ is only satisfied by numbers, so $\ulcorner \mathcal F’ (\ulcorner \mathcal F’(x) \urcorner) \urcorner$ is what is satisfied by $\neg \mathcal F$ because $\mathcal F’ (\ulcorner \mathcal F’(x) \urcorner) $ is true (by assumption) and so its Gödel number must be satisfied by $\neg \mathcal F'$ by definition of $\mathcal F'$. – Zuhair Jul 08 '23 at 20:58
  • @NoahSchweber, but by then we are done! $T$ would incomplete. – Zuhair Jul 08 '23 at 21:02
  • @PW_246 I meant ... is what is satisfied by $\neg \mathcal F'$ because.... – Zuhair Jul 08 '23 at 21:19
  • @Zuhair I meant satisfaction in terms of entailment. Either way, what I’m saying is that $\mathcal F(\ulcorner \mathcal F’(x)\urcorner)$ is consistent with $\neg \mathcal F(\ulcorner \mathcal F’(\ulcorner \mathcal F’(x) \urcorner) \urcorner)$. So I don’t see how $\mathcal F’(\ulcorner \mathcal F’(x) \urcorner) \urcorner)$ is inconsistent. – PW_246 Jul 08 '23 at 21:20
  • 1
    @Zuhair But you're not trying to prove the incompleteness theorem, you're trying to prove the diagonal lemma. At least that's what your post says ... (And even then you'd need to be careful about conflating what $T$ proves and what's true; $T$ could be consistent, complete, but not sound.) – Noah Schweber Jul 08 '23 at 21:23
  • @NoahSchweber, True. But I'm speaking about it within the context of incompleteness. – Zuhair Jul 08 '23 at 21:24
  • @PW_246, the paradox is about $\mathcal F'$ and not about $\mathcal F$. Observe the definition of $\mathcal F'$, and you'll see the paradox. – Zuhair Jul 08 '23 at 21:27
  • 2
    @Zuhair At no point in your question that I can see do you state that you're assuming (towards contradiction) the completeness of $T$. You should clarify this. And again, I'm concerned about the conflation of what $T$ proves and what's true. – Noah Schweber Jul 08 '23 at 21:27
  • @NoahSchweber, where exactly you are seeing this conflation? – Zuhair Jul 08 '23 at 21:30
  • @Zuhair sorry I meant to say $\mathcal F’(\ulcorner \mathcal F’(x)┐)$. – PW_246 Jul 08 '23 at 21:35
  • 1
    @Zuhair There's enough you haven't written in detail (see my previous comments) that there isn't a specific point I'm concerned about, I'm just worried. We have on the one hand "truth-opposers," but on the other hand we have "$T$-provability-opposers" (defined analogously). Tarski rules out the former, but I think it's only the latter that are directly useful to you, and they might not coincide. – Noah Schweber Jul 08 '23 at 21:35
  • @PW_246, if we have $\mathcal F' (\ulcorner \mathcal F'(x) \urcorner)$, then this means that $\ulcorner \mathcal F'(x) \urcorner$ must be the extractee of a Godel number that is satisfied by $\mathcal F'$ itself [by definition of $\mathcal F'$]. Now $\ulcorner \mathcal F'(x) \urcorner$ is the extractee of $\ulcorner\mathcal F' (\ulcorner \mathcal F'(x) \urcorner)\urcorner$ but the latter is not satisfied by $\mathcal F'$, it is satisfied by $\neg \mathcal F'$, and this contradicts the very definition of $\mathcal F'$ itself. – Zuhair Jul 08 '23 at 21:53
  • @Zuhair I see now. I’m going to echo what Noah said in that I don’t see how $\mathcal F’$’s non-existence implies the Diagonal Lemma. – PW_246 Jul 08 '23 at 23:17
  • @PW_246, Noah Schweber, I see now what you both are saying, I'll address it. – Zuhair Jul 09 '23 at 08:07
  • @NoahSchweber, PW_246, I think we need to assume that $T$ is complete. So, what I'm saying is let $T$ be complete and consistent, let $\mathcal F(x)$ be a $T$-formula in one free variable. Truth opposition by $\mathcal F(x)$ is expressible as a schema "$Sch$" in the language of $T$ for each $T$-sentence $\mathcal C$. It is provable that $Sch$ is inconsistent, and this proves the Lemma. Since by completeness $T$ must decide on every $T$-sentence $\mathcal C$, and there should be one that is not an instance of the above scheme, otherwise $T$ would prove the schema and be inconsistent. – Zuhair Jul 09 '23 at 08:51
  • @NoahSchweber, PW_246, so the usefulness of this proof is in the setting of proving the first Gödel incompleteness theorem by reductio ad absurdum. – Zuhair Jul 09 '23 at 10:35

1 Answers1

0

I think this proof requires to add the assumption that "$T$ is complete". So, it suites to prove the diagonal lemma in the context of proving the first of Gödel incompleteness theorems by reductio ad absurdum!

Now, truth opposition by a $T$-formula $\mathcal F(x)$ is expressible as a schema "$\textbf{Sch} $" in the language of $T$. This is:

$\textbf{Sch: } $ if $\mathcal C_0, \mathcal C_1, \mathcal C_2,..$ are all of the sentences in the language of $T$, then: $$ \vdash_{T} \forall n: n=\ulcorner \mathcal C_i \urcorner \to [\mathcal F(n) \leftrightarrow \neg \mathcal C_i]$$

. Now from the main posting it is proved that $\textbf{Sch} $ for any $T$-formula $\mathcal F(x)$ would be inconsistent. Since we assumed $T$ to be complete, then there should be a sentence $\mathcal C_i$ that fulfills the negation of an instance of the above scheme, i.e. we have $$\vdash_{T} \exists n: n=\ulcorner \mathcal C_i \urcorner \land \neg [\mathcal F(n) \leftrightarrow \neg \mathcal C_i]$$, and thus proving the Lemma. Otherwise $T$ (being complete) would prove all instances of $\textbf{Sch} $, and so be inconsistent.

I don't know if the proof can work under weaker assumptions.

Zuhair
  • 4,749