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It seems like the article "The Twin Primes Conjecture is True in the Standard Model of Peano Arithmetic: Applications of Rasiowa–Sikorski Lemma in Arithmetic (I)" by Janusz Czelakowski published in Studia Logica yesterday, claims to have proven that the twin prime conjecture holds in the standard model of Peano arithmetic using the technique of forcing.

This seems like a very significant achievement (if the claim is not erroneous) but I am by no means an expert in logic or number theory, and therefore I'm not qualified enough to understand and evaluate the contents of this paper. So I would appreciate others' inputs on whether this claim has merit.

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    $\begingroup$ @LSpice I'm not saying I disagree, but this isn't a preprint: it's been published in a to-my-understanding-solid journal. $\endgroup$ Oct 26, 2022 at 18:39
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    $\begingroup$ I've looked at the author's other papers, and the main theorem in at least one of them is simply incorrect: in "An application of Rasiowa-Sikorski Lemma in arithmetic (III)" it is proven that if a polynomial has nonnegative integer coefficients, and takes at least one prime value at an integer $a\geq 2$, then it takes infinitely many prime values. But the reducible polynomial $((x-2)^2+1)((x+1)^{20}+8)$ has nonnegative coefficients and [its value at 2 is prime](is (2^2-4*2+5)*((2+1)^20+8) prime). $\endgroup$
    – Wojowu
    Oct 26, 2022 at 22:06
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    $\begingroup$ @JoshuaZ An old olympiad problem I've once read says that if a polynomial $f(x)$ is positive on $\mathbb R_{\geq 0}$, then $g(x)=f(x)(x+1)^k$ has nonnegative coefficients. Starting with $f(x)=(x-2)^2+1$, I found that value $k=20$ works. Of course the value of $g$ at $2$ is then $3^{20}$ which is definitely not prime, but I crossed my fingers and hoped that adding a small enough value to $(x+1)^k$ factor won't break nonnegativity. Turns out $3^{20}+8$ is prime, and it worked out. This should work in much greater generality, to generate arbitrarily many prime values for a reducible polynomial. $\endgroup$
    – Wojowu
    Oct 26, 2022 at 23:46
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    $\begingroup$ I think it should be reopened, and I voted so. It's extremely important for maths research that incorrect published claims are refuted, it has nothing to do with "checking correctness of preprints". $\endgroup$ Oct 28, 2022 at 9:24
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    $\begingroup$ @AndyPutman - I'm not trying to be judgemental here, my concern is merely ease of doing research. Not having to sift through erroneous papers is important. Erroneous papers many be damaging to whole areas (papers of D.Biss being a case of the latter). $\endgroup$ Oct 28, 2022 at 22:19

3 Answers 3

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The error in the paper is in the proof of Theorem 7.2. The proof of Theorem 7.2 is immediately suspicious because of how vague it is in places and because of how lofty the expository text before and after it is. In the proof, the author claims that because we can identify the set of variables $(v_i)_{i \in \mathbb{N}}$ with the natural numbers, the induction scheme

  • $(\beta)$ If $\mathbf{P} \Vdash \varphi(0)$ and for every variable $v_i$, $\mathbf{P} \Vdash \varphi(v_i)\Rightarrow \mathbf{P} \Vdash \varphi(S(v_i))$, then for every variable $v_i$, $\mathbf{P} \Vdash \varphi(v_i)$.

is just an instance of ordinary induction in $\text{ZFC}$, but this is ridiculous. $0$ and $S(v_i)$ are not variables; they're terms. Furthermore, even if the assumptions of Theorem 7.2 were enough to ensure that for every $i$, $\textbf{P}\Vdash S(v_i) \approx v_j$ for some $j$ (and they're not), that would in no way ensure that $\mathbf{P}\Vdash S(v_i) \approx v_{i+1}$.


That said, we need to step through a fair amount of the paper if we want to show conclusively that Theorem 7.2 is wrong. After all, maybe the assumptions of the theorem are inconsistent or otherwise overly strong and the error is really elsewhere in the paper. What makes this really tedious though is the forcing machinery, which only manages to make the proof more confusing and technical. It's pretty clear given the forcing posets being used (discrete posets and singleton posets) that the forcing can't really be doing anything that couldn't be described more simply in some other way.

We have the Lindenbaum-Tarski algebra of $\text{PA}$, written $\mathbf{B}_{\text{PA}}(L)$, which is the Boolean algebra of formulas in some fixed countable collection of variables modulo logical equivalence over $\text{PA}$. We write $[\varphi]_{\text{PA}}$ for the set of formulas that are logically equivalent to $\varphi$ over $\text{PA}$. The forcing posets $\mathbf{P}=(P,\subseteq)$ considered are certain families $P$ of non-empty subsets of $\mathbf{B}_{\text{PA}}(L)$ (page 14). But the poset considered in Theorem 7.2 is a singleton, which means that all of the forcing machinery isn't really doing anything in Theorem 7.2. Nevertheless, let's go through some of the paper and track what the assumption that $P = \{p\}$ means (where $p$ is some non-empty subset of $\mathbf{B}_{\text{PA}}(L)$).

On page 15 we get to the definition of a condition $p$ forcing an atomic formula $\sigma$. Again, since $P = \{p\}$, what this definition collapses to is just $p \Vdash \sigma$ if and only if $[\sigma]_{\text{PA}} \in p$. We then extend this to arbitrary formulas in the standard way, but again everything collapses:

  • $p \Vdash \neg \varphi$ if and only if $p \Vdash \varphi$ fails.
  • $p \Vdash \varphi \wedge \psi$ if and only if $p\Vdash \varphi$ and $p\Vdash \psi$.
  • $p \Vdash \exists x \varphi$ if and only if there is a variable $y$ such that $p \Vdash \varphi(x//y)$ (where $\varphi(x//y)$ is $\varphi$ with instances of $x$ substituted by $y$ and existing instances of $y$ in $\varphi$ changed to some fresh variable to avoid binding).
  • $p \Vdash \varphi \vee \psi$ if and only if $p\Vdash \varphi$ or $p \Vdash \psi$.
  • $p \Vdash \varphi \to \psi$ if and only if $p\Vdash\neg \varphi$ or $p\Vdash \psi$.
  • $p\Vdash (\forall x)\varphi$ if and only if $p\Vdash\varphi(x//y)$ for all variables $y$.

(The author mentions this simplification at the end of page 15 and the beginning of page 16.) Finally, we write $\mathbf{P}\Vdash \varphi$ to mean that $p \Vdash \varphi$ for all $p \in P$, which in our case is just equivalent to $p \Vdash \varphi$.

We say that $\mathbf{P}$ is compatible with equality axioms if

  • $[x \approx x]_{\text{PA}} \in p$ for some variable $x$,
  • whenever $[x \approx y]_{\text{PA}}$ and $[R(...,x,...)]_{\text{PA}}$ are in $p$, then $[R(...,y,...)]_{\text{PA}}$ is in $p$ for any relation symbol $R$, and
  • if $[x \approx y]_{\text{PA}} \in p$, then $[F(...,x,...)\approx F(...,y,...)]_{\text{PA}} \in p$ for any function symbol $F$.

This is essentially just what you need to ensure that $p$ forces the standard axioms of equality. (Transitivity and symmetry follow from special cases of the second bullet point.)

We say that $\mathbf{P}$ is standard if $\mathbf{P}$ is compatible with equality axioms and has that for any atomic formula $\sigma$, if $\text{PA}\vdash \neg \sigma$, then $[\sigma]_{\text{PA}} \notin p$. (Remember, we're assuming $P = \{p\}$.)

This is all we need to understand the statement of Theorem 7.2, which claims that if $\mathbf{P} = (\{p\},\subseteq)$ is standard, then $\mathbf{P}\Vdash \mathrm{Ind}(x;\varphi)$ for every formula $\varphi$ (where $\mathrm{Ind}(x;\varphi)$ is $\forall \bar{z}[\varphi(0,\bar{z}) \wedge \forall x(\varphi(x,\bar{z})\to\varphi(S(x),\bar{z})) \to \forall x\varphi(x,\bar{z})]$, which is induction for the formula $\varphi(x,\bar{z})$). There is a typo in the statement of Theorem 7.2, but it's clear from the proof that the statement is meant to be $\mathbf{P} \Vdash \mathrm{Ind}(x;\varphi)$, not $\mathbf{P} \Vdash \mathrm{Ind}(x;\sigma)$.

The argument (suppressing the other free variables) proceed by showing that $\mathbf{P} \Vdash \mathrm{Ind}(x;\varphi)$ if and only if the following holds:

  • $(\beta)$ If $\mathbf{P} \Vdash \varphi(0)$ and for every variable $y$, $\mathbf{P} \Vdash \varphi(y)\Rightarrow \mathbf{P} \Vdash \varphi(S(y))$, then for every variable $y$, $\mathbf{P} \Vdash \varphi(y)$.

As discussed above, $(\beta)$ does not work.

Let's see a concrete example of Theorem 7.2 failing. Fix an enumeration $(v_i)_{i \in \mathbb{N}}$ of our variable symbols. From now on we'll write $\varphi(y)$ for $\varphi(x//y)$, where $x$ is established by context to be the relevant free variable of $\varphi$. Let $p$ be $$\{[\sigma]_{\text{PA}}: \text{PA} \cup \{v_0 \approx 0\}\vdash \sigma,~\sigma~\text{atomic}\}.$$ It is easy to check that $\mathbf{P} = (\{p\},\subseteq)$ is standard. Consider the formula $$\varphi(v_1) = \exists v_2(v_2 + v_2 \approx v_1 \vee S(v_2 + v_2) \approx v_1),$$ i.e., "$v_1$ is either even or odd."

First, let's see that $\mathbf{P} \Vdash \varphi(0)$ (i.e., $p\Vdash \varphi(0)$). We have that $[v_0+v_0\approx 0]_{\text{PA}} \in p$, so $p \Vdash v_0+v_0 \approx 0$. Therefore $p \Vdash v_0+v_0\approx 0 \vee S(v_0+v_0) \approx 0$ and $p \Vdash \exists v_2( v_2+v_2\approx 0 \vee S(v_2+v_2) \approx 0)$.

Now fix a variable $v_i$. There are two cases. Either $i = 0$ or $i \neq 0$.

If $i = 0$, then we have that $[S(v_0+v_0) \approx S(v_0)]_{\text{PA}} \in p$, so $p \Vdash S(v_0+v_0)\approx S(v_0)$ and $p \Vdash v_0 + v_0 \approx S(v_0) \vee S(v_0+v_0) \approx S(v_0)$. Therefore $p \Vdash \exists v_2(v_2+v_2 \approx S(v_0) \vee S(v_2+v_2) \approx S(v_0)$, i.e., $p \Vdash \varphi(S(v_0))$.

If $i \neq 0$, then I claim that $p \not \Vdash \varphi(v_i)$ (i.e., $p \not \Vdash \exists v_2(v_2 + v_2 \approx v_i \vee S(v_2+v_2) \approx v_i)$). Fix a variable $v_j$. Since $i \neq 0$, we have that $v_j+v_j \not \approx v_i\wedge S(v_j+v_j)\not\approx v_i$ is consistent with $\text{PA} \cup \{v_0 \approx 0\}$ (even if $j = 0$), so $[v_j+v_j \approx v_i]_{\text{PA}} \notin p$ and $[S(v_j+v_j)\approx v_i]_{\text{PA}} \notin p$. Since we can do this for any $j$, we have that $p \not \Vdash \varphi(v_i)$.

So in any case, we have that if $p\Vdash \varphi(v_i)$, then $p\Vdash \varphi(S(v_i))$, but as we just established, $p \not \Vdash \varphi(v_1)$, contradicting Theorem 7.2. Incidentally, this also shows that the assumptions of Theorem 7.2 are not enough to ensure that for every $i$, $\mathbf{P} \Vdash S(v_i)\approx v_j$ for some $j$.

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    $\begingroup$ Wow! Kudos for thoroughness. I was dreading doing this myself. $\endgroup$ Oct 29, 2022 at 0:11
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    $\begingroup$ @NoahSchweber Please don't encourage my poor life choices. /s $\endgroup$ Oct 29, 2022 at 0:20
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    $\begingroup$ So… MathOverflow is a more reliable peer-review system than established journals? And a way more efficient one (2 day turnaround time instead of weeks/months/years)? Am I missing something here? $\endgroup$
    – Alec Rhea
    Oct 29, 2022 at 2:57
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    $\begingroup$ @AlecRhea: For this paper, I rather doubt that it was refereed in a serious way. $\endgroup$ Oct 29, 2022 at 3:34
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    $\begingroup$ @Alec the paper went from submission to acceptance in about 2 months, according to the journal abstract page. Often referees (including me) just put a paper on the "todo" pile, and only look at it when they get some time and inclination. Perhaps it took two months to open the pdf, look at it for an hour, and then say 'accept'. The incentive to get this question here out of the way is high, especially now an erroneous paper is published in a serious journal. The incentive to the referee (such as it was) would have been different. $\endgroup$
    – David Roberts
    Oct 29, 2022 at 5:18
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The Editor-in-Chief has released a statement on behalf of the journal retracting the papers, as follows.

Public announcement

Recently two articles on the applications of Rasiowa-Sikorski Lemma to arithmetic were published online in Studia Logica without proper examination and beyond reasonable standards of scholarly rigor. As it turned out, they contained an irreparable mistake and, consequently, have been retracted from the journal’s website. The papers will not appear in print.

I want to thank all our readers who alerted us to this unfortunate incident. I feel responsible for the reputation damage caused by these publications, and I want to offer my sincere apologies to the scientific community and the author.

Studia Logica editors have examined the journal's review procedure to ensure that a similar situation will not happen again.

Jacek Malinowski
Studia Logica
Editor-in-Chief

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    $\begingroup$ Is there any public place where this message can be found? The Springer link to the paper, at this time (2022, Oct 30, 13:10 GMT), has no indication of any retraction. $\endgroup$
    – YCor
    Oct 30, 2022 at 13:10
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    $\begingroup$ "it was in an email forwarded to me" - but in this case, can this be called "public announcement"? $\endgroup$
    – Alex M.
    Oct 30, 2022 at 16:58
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    $\begingroup$ @AlexM. The text is exactly as I was given, and it was explicitly requested from the EiC to spread it as widely as possible. Given that it was the weekend, I can imagine there might be a little lag time on getting the almighty Springer to change something on their website. I don't know what's going on with SL's own website. This is a public venue, and it's an announcement, not sure what the problem is. $\endgroup$
    – David Roberts
    Oct 30, 2022 at 21:35
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    $\begingroup$ @AlecRhea No, I just thought to share the fact that now the journal is agreeing with the consensus here, that the paper is obviously incorrect, and that it will not be physically published. No doubt this will be posted in official places soon, I just had a scoop, that's all. It's not the only place this has landed online. I'm fairly certain it's not just due to MO that the paper get retracted—the people emailing the editors and complaining about the loss of journal reputation would have helped a lot. $\endgroup$
    – David Roberts
    Oct 31, 2022 at 5:21
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    $\begingroup$ The same announcement now appears at the bottom of the studialogica.org home page. $\endgroup$
    – FredH
    Nov 2, 2022 at 0:18
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From my reading, the only facts about the concept of twin primes used in the argument are that there exist pairs of numbers $n$ and $n+2$ (in the discussion after equation (8.4)) and there exist primes (in the discussion after equation (8.5)).

In other words, the argument would work equally well to prove there are infinitely many pairs of primes separated by a gap of size 1, or infinitely many numbers that are both even and odd (since both even and odd numbers exist).

I conjecture that somehow a hypothesis of the form "every finite subset of these conditions may be satisfied" has been transposed with a hypothesis of the form "every one of these conditions may be satisfied" but I am not familiar enough with the logical techniques used to pinpoint where.

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