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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.

Glycerius
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    I have only glanced at the paper but my immediate reaction is skepticism that forcing can be used this way. I'm reminded of Henry Pogorzelski's attempts to use methods from logic to prove Goldbach's Conjecture; this feels similar, although obviously Czelakowski is an accomplished logician so the claim seems to merit at least some credibility. They also claim in the paper to have proven the stronger claim that for any k, there are infinitely many primes p with p+2k prime. – JoshuaZ Oct 26 '22 at 17:16
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    The author published 4 papers titled with "Applications of Rasiowa–Sikorski Lemma in Arithmetic", one for the infinity of twin primes, one for the infinity of Mersenne primes, one for a generalization of the Dirichlet theorem (which I think is a step forward Schinzel's hypothesis H), the final one for the truth of Goldbach's conjecture. I would be surprised if all of the above famous problems are solved by the "forcing technique" at the same time. – Zerox Oct 26 '22 at 17:26
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    At a quick glance, I think either lemma 9.5 or 9.6 secretly assumes the twin prime conjecture. (At best, when unwound I think this might give a proof in a fairly strong theory that if for each $n$ the existence of $n$ twin primes is consistent with $\mathsf{PA}$, then so is the unboundedness of the set of twin primes - but that's trivial.) I imagine something similar is at play with the other claims @Zerox mentioned above. – Noah Schweber Oct 26 '22 at 17:30
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    Noah, can you explain your " but that's trivial" remark? I don't see this. After all, for each n it is consistent with PA+~Con(PA) that there are n successive levels of complexity ISigma_k that are all consistent, but the claim that this is true for unboundedly many n is refutable in the theory. – Joel David Hamkins Oct 26 '22 at 18:06
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    Sorry, I still don't get it. Suppose (working in ZFC, whatever) that there are nonstandard models of PA with n prime pairs, a separate model for each n. You seem to say it is trivial that there is a model of PA with unboundedly many. How? – Joel David Hamkins Oct 26 '22 at 18:21
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    Knowledgeable people are often, and understandably, eager to engage with these questions, but we literally have a close reason specifically devoted to closing "requests to check a work for correctness", which has historically been interpreted to include checking a preprint; so I think that this question is not appropriate for MO. – LSpice Oct 26 '22 at 18:29
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    @JoelDavidHamkins Whoops, I see the issue now. I retract the parenthetical, but I still suspect my guess at the mistake is correct. – Noah Schweber Oct 26 '22 at 18:38
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    @LSpice I'm not saying I disagree, but this isn't a preprint: it's been published in a to-my-understanding-solid journal. – Noah Schweber Oct 26 '22 at 18:39
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    @NoahSchweber, you are right; I'm not sure why I used the word "preprint", except possibly out of unthinking habit. The rest of my comment still stands, I think. – LSpice Oct 26 '22 at 19:18
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    Is this journal (Studia Logica) that serious? it appears that the author, J. Czelakowski, is in the editorial board. – YCor Oct 26 '22 at 21:08
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    @YCor At least, it once was a pretty decent journal. I think it might have fallen on hard times... – François G. Dorais Oct 26 '22 at 21:32
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    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 (2^2-42+5)((2+1)^20+8) prime). – Wojowu Oct 26 '22 at 22:06
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    @Wojowu, re, I think you meant to link "its value at $2$ is prime." – LSpice Oct 26 '22 at 22:15
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    @LSpice You're right, copy pasting is hard! – Wojowu Oct 26 '22 at 22:18
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    @Wojowu I have to confess curiosity as to how you came up with that example. – JoshuaZ Oct 26 '22 at 23:30
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    @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. – Wojowu Oct 26 '22 at 23:46
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    @Wojowu Something is missing from that olympiad statement because with $f(x) = (x-2)^2 + 1$ we find that $g(x) = f(x)(x + 1) = x^3 - 3x^2 + x + 5$. Is there an "there exists a $k$ such that" missing? – orlp Oct 27 '22 at 10:53
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    @orlp That's right, apologies. I took some grammatical corners to fit in the character limit, but it seems I lost some of the substance there too! – Wojowu Oct 27 '22 at 11:29
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    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". – Dima Pasechnik Oct 28 '22 at 09:24
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    @DimaPasechnik: I don't think there's any danger that at some point in the future a mathematician will come upon this paper and genuinely wonder whether the twin primes conjecture has been known all along. Further discussing it here just gives it unnecessary publicity. – Andy Putman Oct 28 '22 at 17:37
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    @AndyPutman - unnecessary publicity to what? To the apparent fact that the referees haven't done the job (or worse, that the paper was waved through)? IMHO it's very much necessary publicity... – Dima Pasechnik Oct 28 '22 at 20:42
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    @DimaPasechnik: Maybe I'm just more cynical than you, but plenty of wrong papers are published every year. This case probably involves corruption re the editorial board, but that is not exactly shocking either. – Andy Putman Oct 28 '22 at 21:02
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    @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). – Dima Pasechnik Oct 28 '22 at 22:19
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    @DimaPasechnik: Yes, but anyone who is even moderately knowledgable will know not to take this paper seriously. It fails all kinds of crackpot tests. I don't think it is a good use of MathOverflow to waste time on it. – Andy Putman Oct 28 '22 at 23:14
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    @AndyPutman - Bureaucrats are not knowledgeable, for them it might look legit. MathOverflow is hopefully influential enough to drive through a message that the text should be retracted. – Dima Pasechnik Oct 28 '22 at 23:35
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    @DimaPasechnik, I think, as I take @‍AndyPutman to be saying, the kind of people who would think to look to MO for an evaluation of the correctness of a paper—mathematicians, mostly—can figure out for themselves (and shouldn't be relying on MO to judge correctness anyway!), and the kind of bureaucrats who have no idea that a paper published in a journal on which its author serves as an editor aren't going to know to look on MO, either. – LSpice Oct 29 '22 at 00:45
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    I think at this point, with James Hanson's answer, the question has achieved all it can, and should be either closed, have his answer accepted, or both. – Will Sawin Oct 29 '22 at 01:31
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    @LSpice how many number theorists are able to evaluate details of this paper? Only these who took several courses in logic... – Dima Pasechnik Oct 29 '22 at 07:36
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    @Will I concur on the "both" option, but waiting on some side-channel input – David Roberts Oct 29 '22 at 10:19
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    I’m voting to close this question because it's a real edge case for MO and sets a difficult precedent about checking published dubious claims of big conjectures. In this instance, there was a good outcome: an error was found and the paper, in fact two papers, have been or will soon be retracted. Thank you everyone who spent time looking into this, but I think we can safely call the case closed, and don't need any more answers on this one. – David Roberts Oct 30 '22 at 12:58
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    @Dima I agree it was good to get some clarity and closure on this question, but now the public record will be corrected, and the papers are retracted, I don't think the question needs to remain open. If you have very strong views on this, meta is the place to discuss further. – David Roberts Oct 30 '22 at 13:00
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    @DavidRoberts - I think this was a legitimate question, just as questions on details in published texts are legitimate. – Dima Pasechnik Oct 31 '22 at 10:12
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    @Dima I know, it's a real edge case. If it were a weird niche journal that's obviously unsound, I'm sure this would be closed quickly and not re-opened. The question is where to draw the line. That the biggest theorem in number theory in decades was published in a (?mid-tier?) logic journal is already a flag, despite the journal being otherwise perfectly legitimate. Were the question "where's the inevitable mistake?" I'd be less bothered, even if that seems unfair. But we will see what the community thinks about the question now. – David Roberts Oct 31 '22 at 11:00

3 Answers3

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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 $j$.

James Hanson
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    Wow! Kudos for thoroughness. I was dreading doing this myself. – Noah Schweber Oct 29 '22 at 00:11
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    @NoahSchweber Please don't encourage my poor life choices. /s – James Hanson Oct 29 '22 at 00:20
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    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? – Alec Rhea Oct 29 '22 at 02:57
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    @AlecRhea: For this paper, I rather doubt that it was refereed in a serious way. – Andy Putman Oct 29 '22 at 03:34
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    @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. – David Roberts Oct 29 '22 at 05:18
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    Most likely there was no referee. As I mentioned in a comment, the author is an editor in the journal. So, referees, if any, couldn't be anonymous. The most plausible explanation is that the author accepted his own article (or that other editors accepted it under pressure of their colleague). – YCor Oct 29 '22 at 07:01
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    @YCor The paper says “presented by Jacek Malinowski” (who is the Editor in Chief of the journal). I take it to mean that the editor handling the paper was Malinowski, and being the big boss, I have a hard time believing he was under pressure. Though it might well have been a favour out of friendship or something like that. – Emil Jeřábek Oct 29 '22 at 09:15
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    @DavidRoberts Perhaps I am naïve, but I would have hoped that in case of something as high profile as a claimed proof of the twin prime conjecture, a serious referee would ask the editor for more time rather than accept the paper after a light reading. – Emil Jeřábek Oct 29 '22 at 09:19
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    @Emil oh, I agree! – David Roberts Oct 29 '22 at 09:40
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    @AlecRhea ''MathOverflow is a more reliable peer-review system than established journals?'' I guess this depends on the referee, how carefully it is refereed etc. The idea that something must be correct because it was formally published in a good journal is a misconception that people typically get out of once they leave undergraduate level. – Hollis Williams Oct 29 '22 at 11:23
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    @HollisWilliams What does an assumption of correctness for published work have to do with anything? – Alec Rhea Oct 29 '22 at 11:26
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    I actually think that "MathOverflow" might be more reliable than an established journal. In the traditional peer-reviewed publication process an editor chooses referees who then agree to look at the paper or not. Possibly only one referee looks at the paper halfheartedly. With results that generate a lot of interest it might be better to have several people in a restricted forum like this one look at the results. – Stefan Geschke Oct 29 '22 at 19:15
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    What I don't understand is why someone would submit a result like this to Studia Logica and not the Annals of Math or something similar. This fact already raises some doubts in the result. – Stefan Geschke Oct 29 '22 at 19:16
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    @StefanGeschke, in what sense is MathOverflow a restricted forum? – LSpice Oct 29 '22 at 22:16
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    @JamesHanson "We do not do this thing because it is permitted. We do it because we have to. We do it because we are compelled." - W. J. Kovacs. (Also, bravo) – Yemon Choi Oct 30 '22 at 12:54
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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


The official Springer pages for the articles (paper I, paper II) both have the official retraction statement (I presume since November 2022, when they were retracted) as follows:

The Editor-in-Chief has retracted this article because, after publication, it became apparent that there is an error in the proof of Theorem 7.2. The author agrees with this retraction.

The online version of this article contains the full text of the retracted article as Supplementary Information.

David Roberts
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    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. – YCor Oct 30 '22 at 13:10
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    @YCor no, it was in an email forwarded to me, that came direct from the EiC. – David Roberts Oct 30 '22 at 13:25
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    "it was in an email forwarded to me" - but in this case, can this be called "public announcement"? – Alex M. Oct 30 '22 at 16:58
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    @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. – David Roberts Oct 30 '22 at 21:35
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    @DavidRoberts So… MathOverflow is also now a de-facto venue for official journals to publicly announce their errors (that we arguably caught)? For the record I have reffed math papers before, I’m familiar with how the current system “works”. – Alec Rhea Oct 31 '22 at 01:13
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    @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. – David Roberts Oct 31 '22 at 05:21
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    @DavidRoberts I was being slightly tongue-in-cheek -- more seriously, I think this sequence of events is interesting because MathOverflow seems to be taking on a role as the final stage of review. – Alec Rhea Oct 31 '22 at 06:26
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    There were more I could have linked sans the comment size restriction; I find these developments amusing in light of the mathematicians on here who insist that MathOverflow will always just be a one-off Q&A site, not destined to take on a larger role in the mathematical universe. Peer review is one of the main purposes of journals -- MathOverflow naturally supplanting them as the apex of the review process (and arguably also providing a better forum for dissemination of new mathematics, their other main task, for free) indicates to me that interesting things are coming down the pipes. – Alec Rhea Oct 31 '22 at 06:30
  • @PeterTaylor I only found https://link.springer.com/article/10.1007/s11225-022-10017-2 and https://link.springer.com/article/10.1007/s11225-022-10015-4 . Where are the other two? – Emil Jeřábek Oct 31 '22 at 10:56
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    There are two more preprints on his ResearchGate profile, but that’s not something for Studia Logica to retract. – Emil Jeřábek Oct 31 '22 at 11:00
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    @EmilJeřábek, the second comment on the question says that the author had four papers published in the series. Maybe the other two are in the pipeline and about to get rejected or withdrawn. – Peter Taylor Oct 31 '22 at 11:36
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    The same announcement now appears at the bottom of the studialogica.org home page. – FredH Nov 02 '22 at 00:18
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    @FredH thanks for checking! – David Roberts Nov 02 '22 at 03:33
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    @FredH unfortunately, nobody ever visits this website. On the contrary, the Springer website contains no mention that the paper is flawed. – Daniil Kozhemiachenko Nov 04 '22 at 21:57
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    @DaniilKozhemiachenko the notice is now posted at https://www.springer.com/journal/11225 also, though the papers do not seem to have been formally retracted yet by Springer – David Roberts Nov 08 '22 at 15:31
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    @DavidRoberts most importantly, the page with the paper itself does not contain any disclaimers. – Daniil Kozhemiachenko Nov 12 '22 at 11:41
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    Both papers are now clearly marked as retracted on their respective pages: https://doi.org/10.1007/s11225-022-10017-2, https://doi.org/10.1007/s11225-022-10015-4. – Emil Jeřábek Jan 23 '23 at 10:55
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    However, the author, J. Czelakowski, still appears now as an associate editor to this journal (link to editorial board at Springer site). – YCor Mar 25 '23 at 11:53
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    @YCor Noted, but that's also a separate issue to what was raised in the original question, and one for the editorial board to sort out. – David Roberts Mar 25 '23 at 21:15
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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.

Will Sawin
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