Auditing an AI-Generated Mathematical Proof: A Correction to a Greedy Conditioning Lemma in Quantum Parallel Repetition

# A Correction to a Greedy Conditioning Lemma in Quantum Parallel Repetition
Source: [https://arxiv.org/html/2608.14673](https://arxiv.org/html/2608.14673)
## Auditing an AI\-Generated Mathematical Proof: A Correction to a Greedy Conditioning Lemma in Quantum Parallel Repetition

Krzysztof SienickiChair of Theoretical Physics of Naturally Intelligent Systems \(NIS\), Podkowa Leśna, Poland, European Union

\(3 August 2026\)

###### Abstract

Chapter 6 of OpenAI’s*Ten Advances in Mathematics and Theoretical Computer Science*claims an exponential parallel\-repetition theorem for all finite two\-player, one\-round entangled games\. Early in the proof, the chapter uses a quantitative greedy conditioning lemma\. The lemma is meant to select a small set of coordinatesDDsuch that, after conditioning on winning every coordinate inDD, a randomly chosen remaining coordinate is won with average probability at least1−δ1\-\\delta\. The statement is correct, but the proof as printed contains a polarity error\. Its continuation test is written in terms of average success, while the next step requires a coordinate with large conditional failure probability\. That implication is false, and even simple examples can leave the printed procedure without a valid next move\.

This note gives an explicit counterexample, identifies the intended continuation condition, and supplies a complete corrected proof\. The repair is local: it leaves the statement of the lemma and the parameters used later in the chapter unchanged\. It should not, however, be read as an independent verification of the main parallel\-repetition theorem\. More broadly, the example shows how a mathematically plausible AI\-generated argument can hide a small but decisive reversal between complementary events\.

Keywords:AI\-generated mathematics; proof auditing; parallel repetition; entangled games; conditioning; mathematical error correction\.

## 1Introduction

OpenAI’s*Ten Advances in Mathematics and Theoretical Computer Science*presents ten substantial results in mathematics and theoretical computer science, reported as having been produced by an internal OpenAI model\[[1](https://arxiv.org/html/2608.14673#bib.bib1)\]\. One of these results is an exponential parallel\-repetition theorem for arbitrary finite two\-player entangled games\.

OpenAI has also described a separate initiative involving model\-generated proof attempts in the article[“Our First Proof submissions”](https://openai.com/index/first-proof-submissions/)\[[2](https://arxiv.org/html/2608.14673#bib.bib2)\]\. That article is useful background, but it concerns a different initiative and is not the publication page for the work examined here\.

Chapter 6, “Exponential Parallel Repetition for All Two\-Player Entangled Games,” studies a finite two\-player, one\-round gameGG\. A referee sends questions to two noncommunicating players, who may share an entangled state, and then decides whether their answers are accepted\. The entangled value of the game is denoted byω∗​\(G\)\\omega^\{\*\}\(G\)\. InG⊗nG^\{\\otimes n\}, the referee runsnnindependent copies of the game and accepts only when the players win every copy\. The chapter claims exponential decay wheneverω∗​\(G\)=1−ε<1\\omega^\{\*\}\(G\)=1\-\\varepsilon<1\.

The proof follows a conditioning\-and\-rounding strategy related to earlier work\[[4](https://arxiv.org/html/2608.14673#bib.bib4),[5](https://arxiv.org/html/2608.14673#bib.bib5),[3](https://arxiv.org/html/2608.14673#bib.bib3)\], together with a postselection\-stable quantum sampleability argument\. One of its preliminary ingredients is Lemma 3\.1, called the “quantitative greedy conditioning” lemma\. Starting from a repeated\-game strategy that wins allnncoordinates with probabilityϑ\>0\\vartheta\>0, the lemma seeks a small setDDsuch that, conditional on winning the coordinates inDD, the average success probability on a remaining coordinate is close to one\.

The statement of the lemma is valid\. The proof printed in Appendix A\.2, however, reverses success and failure in the condition that determines whether the greedy procedure should continue\. The purpose of this note is narrowly defined: to isolate that inference, show why it fails, state the correct condition, and prove the lemma in full\. We also indicate what the repair does and does not establish for the later argument\. No claim is made here that the rest of Chapter 6 has been independently verified\.

The version examined is the official OpenAI PDF dated 1 August 2026, available at[https://cdn\.openai\.com/pdf/ten\-proofs\-oai\.pdf](https://cdn.openai.com/pdf/ten-proofs-oai.pdf), with SHA\-256 checksum64b900d5fae6fe22f2ae1b8e3b712d20055194a6c81cf343a2455e5898ac7dd6\. The corresponding publication page is[https://openai\.com/index/ten\-advances\-in\-mathematics/](https://openai.com/index/ten-advances-in-mathematics/)\. The passage discussed below appears in Chapter 6, Appendix A\.2, on printed page 175 \(PDF page 177\)\. Throughout the note,log\\logdenotes the natural logarithm\.

## 2The conditioning problem

LetW1,…,WnW\_\{1\},\\ldots,W\_\{n\}be events in a probability space, withWiW\_\{i\}representing the event that coordinateiiis won\. ForD⊆\[n\]D\\subseteq\[n\], define

WD:=⋂j∈DWj,W\_\{D\}:=\\bigcap\_\{j\\in D\}W\_\{j\},withW∅=ΩW\_\{\\varnothing\}=\\Omega, and set

ϑ:=ℙ​\(W\[n\]\)\.\\vartheta:=\\mathbb\{P\}\(W\_\{\[n\]\}\)\.
SinceW\[n\]⊆WDW\_\{\[n\]\}\\subseteq W\_\{D\}, we always haveℙ​\(WD\)≥ϑ\\mathbb\{P\}\(W\_\{D\}\)\\geq\\vartheta\. Consequently, wheneverϑ\>0\\vartheta\>0, every conditional probability used below is well defined\.

The statement used in Chapter 6 may be written as follows\.

###### Lemma 1\(Quantitative greedy conditioning\)\.

Supposeϑ\>0\\vartheta\>0,0<δ<10<\\delta<1, and

log⁡\(1/ϑ\)δ<n\.\\frac\{\\log\(1/\\vartheta\)\}\{\\delta\}<n\.Then there existsD⊊\[n\]D\\subsetneq\[n\]such that

\|D\|≤log⁡\(1/ϑ\)δ,ℙ​\(WD\)≥ϑ,1n−\|D\|​∑i∉Dℙ​\(Wi∣WD\)≥1−δ\.\|D\|\\leq\\frac\{\\log\(1/\\vartheta\)\}\{\\delta\},\\qquad\\mathbb\{P\}\(W\_\{D\}\)\\geq\\vartheta,\\qquad\\frac\{1\}\{n\-\|D\|\}\\sum\_\{i\\notin D\}\\mathbb\{P\}\(W\_\{i\}\\mid W\_\{D\}\)\\geq 1\-\\delta\.\(1\)

In words, the lemma says that one can condition on winning only a limited number of coordinates and still arrange that a uniformly chosen unconditioned coordinate is won with probability at least1−δ1\-\\delta, on average\.

## 3The defective continuation criterion

The printed proof begins withD=∅D=\\varnothingand instructs the procedure to continue whenever

1n−\|D\|​∑i∉Dℙ​\(Wi∣WD\)\>δ\.\\frac\{1\}\{n\-\|D\|\}\\sum\_\{i\\notin D\}\\mathbb\{P\}\(W\_\{i\}\\mid W\_\{D\}\)\>\\delta\.\(2\)
It then calls for an indexi∉Di\\notin Dwhose conditional failure probability is greater thanδ\\delta:

ℙ​\(Wic∣WD\)\>δ⟺ℙ​\(Wi∣WD\)<1−δ\.\\mathbb\{P\}\(W\_\{i\}^\{c\}\\mid W\_\{D\}\)\>\\delta\\quad\\Longleftrightarrow\\quad\\mathbb\{P\}\(W\_\{i\}\\mid W\_\{D\}\)<1\-\\delta\.\(3\)
Such a coordinate would give

ℙ​\(WD∪\{i\}\)=ℙ​\(WD\)​ℙ​\(Wi∣WD\)<\(1−δ\)​ℙ​\(WD\)\.\\mathbb\{P\}\(W\_\{D\\cup\\\{i\\\}\}\)=\\mathbb\{P\}\(W\_\{D\}\)\\mathbb\{P\}\(W\_\{i\}\\mid W\_\{D\}\)<\(1\-\\delta\)\\mathbb\{P\}\(W\_\{D\}\)\.\(4\)
The difficulty is immediate:[Equation˜2](https://arxiv.org/html/2608.14673#S3.E2)does not imply[Equation˜3](https://arxiv.org/html/2608.14673#S3.E3)\. Knowing that average success exceedsδ\\deltasays nothing about whether any coordinate has failure probability greater thanδ\\delta\. The printed proof therefore relies on the invalid step

average success\>δ⟹some failure\>δ\.\\text\{average success\}\>\\delta\\quad\\Longrightarrow\\quad\\text\{some failure\}\>\\delta\.
The continuation test should instead be

1n−\|D\|​∑i∉Dℙ​\(Wic∣WD\)\>δ,equivalently1n−\|D\|​∑i∉Dℙ​\(Wi∣WD\)<1−δ\.\\frac\{1\}\{n\-\|D\|\}\\sum\_\{i\\notin D\}\\mathbb\{P\}\(W\_\{i\}^\{c\}\\mid W\_\{D\}\)\>\\delta,\\quad\\text\{equivalently\}\\quad\\frac\{1\}\{n\-\|D\|\}\\sum\_\{i\\notin D\}\\mathbb\{P\}\(W\_\{i\}\\mid W\_\{D\}\)<1\-\\delta\.\(5\)
Now the averaging argument works: if the average conditional failure probability is greater thanδ\\delta, then at least one remaining coordinate has conditional failure probability greater thanδ\\delta, exactly as required by[Equation˜3](https://arxiv.org/html/2608.14673#S3.E3)\. The most natural reading is that a complement sign was dropped in the printed condition:WiW\_\{i\}should have beenWicW\_\{i\}^\{c\}\.

## 4A counterexample to the printed procedure

###### Example 1\.

Letn=2n=2,δ=0\.1\\delta=0\.1, and letEEbe an event withℙ​\(E\)=0\.95\\mathbb\{P\}\(E\)=0\.95\. Set

Then

ϑ=ℙ​\(W1∩W2\)=0\.95,\\vartheta=\\mathbb\{P\}\(W\_\{1\}\\cap W\_\{2\}\)=0\.95,and

log⁡\(1/ϑ\)δ=log⁡\(1/0\.95\)0\.1≈0\.513<2=n\.\\frac\{\\log\(1/\\vartheta\)\}\{\\delta\}=\\frac\{\\log\(1/0\.95\)\}\{0\.1\}\\approx 0\.513<2=n\.\(6\)
AtD=∅D=\\varnothing, the average success probability is0\.950\.95, so the printed continuation condition is satisfied\. But

ℙ​\(Wic\)=0\.05<0\.1\\mathbb\{P\}\(W\_\{i\}^\{c\}\)=0\.05<0\.1for both coordinates\. There is therefore no coordinate that satisfies the next instruction\.

In fact, the procedure should already have stopped, because

12​∑i=12ℙ​\(Wi\)=0\.95≥0\.9=1−δ\.\\frac\{1\}\{2\}\\sum\_\{i=1\}^\{2\}\\mathbb\{P\}\(W\_\{i\}\)=0\.95\\geq 0\.9=1\-\\delta\.Thus the printed test forces the algorithm to continue at exactly the point where its desired conclusion has already been reached\.

The lemma is stated for win events generated by an arbitrary repeated\-game strategy, so the events need not be independent\. Nothing in the example turns on independence, however\. Its only role is to disprove the purely probabilistic implication used in the printed greedy step\.

## 5Corrected statement and proof

###### Lemma 2\(Corrected quantitative greedy conditioning\)\.

LetW1,…,WnW\_\{1\},\\ldots,W\_\{n\}be events and let

ϑ=ℙ​\(W\[n\]\)\.\\vartheta=\\mathbb\{P\}\(W\_\{\[n\]\}\)\.Supposeϑ\>0\\vartheta\>0,0<δ<10<\\delta<1, and

log⁡\(1/ϑ\)δ<n\.\\frac\{\\log\(1/\\vartheta\)\}\{\\delta\}<n\.Then there existsD⊊\[n\]D\\subsetneq\[n\]satisfying the three conclusions in[Equation˜1](https://arxiv.org/html/2608.14673#S2.E1)\.

###### Proof\.

Begin withD=∅D=\\varnothing, and write

qD:=1n−\|D\|​∑i∉Dℙ​\(Wi∣WD\)\.q\_\{D\}:=\\frac\{1\}\{n\-\|D\|\}\\sum\_\{i\\notin D\}\\mathbb\{P\}\(W\_\{i\}\\mid W\_\{D\}\)\.
BecauseW\[n\]⊆WDW\_\{\[n\]\}\\subseteq W\_\{D\}, we have

ℙ​\(WD\)≥ϑ\>0,\\mathbb\{P\}\(W\_\{D\}\)\\geq\\vartheta\>0,soqDq\_\{D\}is well defined at every stage of the construction\.

If

qD≥1−δ,q\_\{D\}\\geq 1\-\\delta,stop\. If not, then

Hence at least onei∉Di\\notin Dsatisfies

ℙ​\(Wi∣WD\)<1−δ\.\\mathbb\{P\}\(W\_\{i\}\\mid W\_\{D\}\)<1\-\\delta\.
Add this coordinate toDD\. By[Equation˜4](https://arxiv.org/html/2608.14673#S3.E4), the probability of the conditioning event then falls by a factor strictly smaller than1−δ1\-\\delta\.

Afterk≥1k\\geq 1additions, iterating this estimate gives

ϑ≤ℙ​\(WD\)<\(1−δ\)k\.\\vartheta\\leq\\mathbb\{P\}\(W\_\{D\}\)<\(1\-\\delta\)^\{k\}\.\(7\)
Taking logarithms and using

−log⁡\(1−δ\)≥δ,\-\\log\(1\-\\delta\)\\geq\\delta,we obtain

k<log⁡\(1/ϑ\)−log⁡\(1−δ\)≤log⁡\(1/ϑ\)δ\.k<\\frac\{\\log\(1/\\vartheta\)\}\{\-\\log\(1\-\\delta\)\}\\leq\\frac\{\\log\(1/\\vartheta\)\}\{\\delta\}\.\(8\)
Since

log⁡\(1/ϑ\)δ<n,\\frac\{\\log\(1/\\vartheta\)\}\{\\delta\}<n,the procedure cannot add allnncoordinates\. ThusD⊊\[n\]D\\subsetneq\[n\]\.

When the procedure stops,

qD≥1−δ,q\_\{D\}\\geq 1\-\\delta,whileW\[n\]⊆WDW\_\{\[n\]\}\\subseteq W\_\{D\}still guarantees

ℙ​\(WD\)≥ϑ\.\\mathbb\{P\}\(W\_\{D\}\)\\geq\\vartheta\.All three conclusions in[Equation˜1](https://arxiv.org/html/2608.14673#S2.E1)therefore hold\. ∎

## 6Comparison with the printed proof

Most of Appendix A\.2 survives unchanged\. In particular, the proof correctly uses

W\[n\]⊆WD,W\_\{\[n\]\}\\subseteq W\_\{D\},the bound

ℙ​\(WD\)≥ϑ,\\mathbb\{P\}\(W\_\{D\}\)\\geq\\vartheta,the identity

ℙ​\(WD∪\{i\}\)=ℙ​\(WD\)​ℙ​\(Wi∣WD\),\\mathbb\{P\}\(W\_\{D\\cup\\\{i\\\}\}\)=\\mathbb\{P\}\(W\_\{D\}\)\\mathbb\{P\}\(W\_\{i\}\\mid W\_\{D\}\),and the multiplicative decrease in[Equation˜4](https://arxiv.org/html/2608.14673#S3.E4)\. The logarithmic estimate in[Equation˜8](https://arxiv.org/html/2608.14673#S5.E8)is also correct\.

The only defective part is the initial continuation test\. Replacing[Equation˜2](https://arxiv.org/html/2608.14673#S3.E2)by[Equation˜5](https://arxiv.org/html/2608.14673#S3.E5)repairs the proof; no change to the statement of the lemma is needed\.

## 7Effect on the parallel\-repetition argument

Later in the chapter, the authors set

m=n−\|D\|,p=ℙ​\(WD\),m=n\-\|D\|,\\qquad p=\\mathbb\{P\}\(W\_\{D\}\),and

q=1m​∑i∉Dℙ​\(Wi∣WD\),q=\\frac\{1\}\{m\}\\sum\_\{i\\notin D\}\\mathbb\{P\}\(W\_\{i\}\\mid W\_\{D\}\),and introduce a further parameterη\\etameasuring the information cost per remaining coordinate\.

The conditioning lemma is used to ensure

control\|D\|\|D\|, and retain the lower bound

The subsequent rounding argument aims for an inequality of the form

ω∗​\(G\)≥q−Bqs​η1/12\.\\omega^\{\*\}\(G\)\\geq q\-B\_\{\\mathrm\{qs\}\}\\eta^\{1/12\}\.\(9\)
The corrected proof delivers exactly the conclusions stated in the lemma\. It preserves the bound on\|D\|\|D\|and therefore leaves the later definitions and estimates involvingpp,qq, andη\\etaintact\. It does not imply that the corrected procedure selects the same setDDas the printed procedure; in some cases, as the counterexample shows, the printed procedure does not even specify a valid next step\.

This local error therefore does not, by itself, refute the main theorem\. It shows only that the conditioning argument must be repaired before the rest of the proof can be assessed\. The later sampleability, correlated\-sampling, state\-alignment, and rounding arguments remain separate questions requiring specialist verification\.

## 8Relevance to the auditing of AI\-generated mathematics

The mistake is elementary, but that is precisely why it is instructive\. It is not a difficult operator inequality or a subtle point of quantum mechanics\. It is a reversal between the complementary eventsWiW\_\{i\}andWicW\_\{i\}^\{c\}\.

The surrounding proof still sounds convincing: choose a coordinate with substantial failure probability, condition on winning it, reduce the mass of the conditioning event by a factor below1−δ1\-\\delta, and stop when the remaining coordinates have sufficiently high average success\. The flaw lies in a single displayed condition that does not match this logic\.

This is a useful reminder that fluent mathematical prose is not the same as a valid deduction\. A careful audit must ask, at every step, whether the stated hypothesis really guarantees the object selected or the inequality used next\. Here the mismatch becomes clear as soon as one places average success, average failure, and individual failure side by side\.

The two\-coordinate example also illustrates the value of testing an elementary probabilistic lemma outside the advanced theory in which it appears\. At the same time, one counterexample cannot support broad empirical claims about AI\-generated proofs in general\. It establishes something more modest: a plausible proof template can conceal a local reversal of complementary predicates, and that reversal can invalidate the procedure as written\.

## 9Conclusion

The quantitative greedy conditioning lemma in Chapter 6 of*Ten Advances in Mathematics and Theoretical Computer Science*is stated correctly but proved incorrectly in the published text\. Condition[Equation˜2](https://arxiv.org/html/2608.14673#S3.E2)does not justify choosing a coordinate whose conditional failure probability exceedsδ\\delta\. The appropriate continuation criterion is[Equation˜5](https://arxiv.org/html/2608.14673#S3.E5)\.

Once this condition is corrected, the greedy proof goes through\. Each added coordinate reduces the probability of the conditioning event by a factor strictly below1−δ1\-\\delta, while that probability remains bounded below byϑ\\vartheta\. The number of additions is consequently bounded by[Equation˜8](https://arxiv.org/html/2608.14673#S5.E8), and the procedure stops with the required average conditional success probability\.

The repair is local\. It preserves the statement of the lemma and the quantitative parameters used later in the chapter\. It removes this particular obstruction from the Chapter 6 argument, but it should not be mistaken for an independent verification of the deeper parallel\-repetition theorem\.

> The broader lesson is that the gap between advanced AI\-generated mathematics and ordinary human mathematical practice is becoming harder to locate\. The successful parts can be highly sophisticated, while the mistakes may be strikingly familiar: small, local, and entirely human in character\.

## Acknowledgment

We thank Félix de la Poterie\-Sienicki \(McGill University, Montréal, Canada\) for drawing our attention to the relevance of the OpenAI article to our research\.

This note was prepared as an independent audit of the argument in Chapter 6 of\[[1](https://arxiv.org/html/2608.14673#bib.bib1)\]\. The authors were not involved in producing the source publication\.

## References

- \[1\]OpenAI,*Ten Advances in Mathematics and Theoretical Computer Science*,OpenAI publication, 1 August 2026\.Chapter 6, “Exponential Parallel Repetition for All Two\-Player Entangled Games,” Appendix A\.2, printed p\. 175 \(PDF p\. 177\)\.Official publication page:[https://openai\.com/index/ten\-advances\-in\-mathematics/](https://openai.com/index/ten-advances-in-mathematics/)\.Official PDF:[https://cdn\.openai\.com/pdf/ten\-proofs\-oai\.pdf](https://cdn.openai.com/pdf/ten-proofs-oai.pdf)\.SHA\-256:64b900d5fae6fe22f2ae1b8e3b712d20055194a6c81cf343a2455e5898ac7dd6\.
- \[2\]OpenAI,*Our First Proof submissions*,20 February 2026\.Separate OpenAI initiative; not the publication page for\[[1](https://arxiv.org/html/2608.14673#bib.bib1)\]\.[https://openai\.com/index/first\-proof\-submissions/](https://openai.com/index/first-proof-submissions/)\.
- \[3\]I\. Dinur, D\. Steurer, and T\. Vidick,A parallel repetition theorem for entangled projection games,*Computational Complexity*24\(2015\), 201–254\.DOI:[10\.1007/s00037\-015\-0098\-3](https://doi.org/10.1007/s00037-015-0098-3)\.
- \[4\]T\. Holenstein,Parallel repetition: simplification and the no\-signaling case,*Theory of Computing*5\(2009\), 141–172\.DOI:[10\.4086/toc\.2009\.v005a008](https://doi.org/10.4086/toc.2009.v005a008)\.
- \[5\]H\. Yuen,A parallel repetition theorem for all entangled games,in*43rd International Colloquium on Automata, Languages, and Programming \(ICALP 2016\)*, LIPIcs55, Article 77, 2016\.DOI:[10\.4230/LIPIcs\.ICALP\.2016\.77](https://doi.org/10.4230/LIPIcs.ICALP.2016.77)\.