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Subject and independence

The reviewer is an independent reader working in a fresh context from the review assignment alone. The reviewer took no part in writing the page, the source card, or any neighboring reconstruction, and the page's author is identified here only by role. The charge is refutation.

Subject: path wiki/research/erdos_354/yu_chen_lemma_2_1_reconstruction.md as it stood at 2026-09-28T05:03:27Z, read whole as of that time.

Artifact: the seventeen-page PDF retained under the library card Yu and Chen (2026), the manuscript dated 13 September 2026 on its title page (the file's own metadata records a later build, and the card's provenance paragraph pins the bytes). Physical pp. 2--4 were read in full from the text layer with layout preserved. Physical p. 3, which carries the definition of $h$, Lemma 2.1 and identity (2.1), was also read from a page image rendered at 150 dots per inch, and physical p. 1 (title, date, main theorem) from an image at 110 dots per inch. The page count of 17 was confirmed from the file's metadata.

Allowed material actually read: the page; the card's provenance paragraph; the guidance sections "Whole-claim report" and "Audit checklist" (the shared list and the repository list) of the verification guide, "Source fidelity" of the evidence guide, and the mathematics-authoring guide. The page cites no other reconstruction as an input, so none was read. The problem page E0354 has no "Statement" heading, and the region before its assessment section mixes in standing text, so it was not read; the lemma is self-contained and needs nothing from it.

Exposures: two, disclosed here. First, the card's _index.md was printed whole while its provenance paragraph was being read, so its read-status, standing, bears-on and results text passed before the reviewer; that text concerns the manuscript's main theorem and its acceptance evidence, not Lemma 2.1, and played no part in this review. Second, a heading-only scan of E0354 returned, besides its headings, two body fragments that begin with a hash character (a line naming a pull-request number and a line listing issue numbers); they were not read further and played no part.

Independent computation: the reviewer wrote a short enumeration, not retained. It listed every nonempty subset of $\mathbb Z/d\mathbb Z$ for $1\le d\le12$ (8178 subsets) and tested identity (2.1) together with the page's three intermediate claims: the complement formula, the decomposition of the union's complement into the shortened sets, and the identification of the union's maximal runs with the nonempty shortened sets. No failure. This is base-case coverage, not a proof; the verdict rests on the derivations below.

Restatement

Fix an integer $d\ge1$ and a nonempty subset $X$ of the cyclic group $\mathbb Z/d\mathbb Z$. Call an arc $\{s,s+1,\ldots,s+r-1\}$ of $r\ge1$ cyclically consecutive residues a missing run of $X$ when none of its residues lies in $X$. Since $X$ is nonempty, a missing run has at most $d-1$ residues. Put $h(X)=0$ when $X$ is the whole group, and otherwise let $h(X)$ be the largest number of residues in a missing run of $X$. With $X+1=\{x+1:x\in X\}$, the claim is that for every such $d$ and $X$,

$$h\bigl(X\cup(X+1)\bigr)=\max\bigl(0,\,h(X)-1\bigr).$$

Conventions: runs are measured by their number of residues; arcs wrap around the circle; the modulus $d$ is a positive integer, which the source leaves implicit ("around the circle"). No constant, limit or uniformity enters; the identity is exact for each $d$ and each $X$.

Checklist

• Quantifiers and scope: pass. The page's "for every nonempty $X$" matches the source's "for a nonempty $X$"; the modulus is arbitrary in both. The two boundary cases, $X$ full (both sides $0$) and $h(X)=1$ (the union becomes full), are argued separately on the page.
• Circularity: pass. Nothing equivalent to the identity is assumed; the proof computes the union's complement residue by residue.
• Model and convention changes: pass. The page's definitions of missing run, maximal missing run and $h$ reproduce the source's "number of residues in its longest consecutive missing run around the circle" and its "$h(X)=0$ when $X$ is full". "Let $d\ge1$" is a reading of the source's implicit modulus, not a change (F4).
• Finite and statistical overreach: inapplicable. The page uses no finite cases or averages as proof. The reviewer's enumeration is recorded above as base-case coverage only.
• Uniformity: inapplicable. There are no constants, error terms, limits or infinite families; the identity is exact.
• Extremal conclusions: pass. $h$ is a maximum over the finitely many maximal missing runs, a nonempty family when $X$ is not full, computed in the lemma's own unit, the residue count; the page shows the maximum for the union is attained by a shortened set of length $h(X)-1$ when $h(X)\ge2$.
• Consequences and composition: pass. Each "hence", "therefore" and "so" was checked separately; see Weakest steps. No external clause is consumed.
• Computation: inapplicable. The page contains no computation.
• Reproduction: inapplicable. The page states no rerun command and no coverage claim.
• Source and verdict fidelity: pass. The page quotes nothing; its characterization "the source gives the lemma three sentences" matches the three-sentence proof on physical p. 3; the locators (Lemma 2.1, identity (2.1), physical p. 3, which is also the printed page 3, seventeen pages, manuscript dated 13 September 2026) are correct; the standing sentence claims only an author-recorded reconstruction.

Weakest steps

W1. The complement of $X$ splits into pairwise disjoint maximal runs (page: "every missing run is contained in a unique maximal one"). The page asserts this without argument and later relies on it, both for "the disjoint union of the shortened sets" and for taking $h(X)$ over maximal runs alone. Re-derivation: let $\{s,\ldots,s+r-1\}$ be a missing run. Walking backward from $s-1$, the first residue in $X$ is met within $d-1$ steps because $X$ is nonempty; call the residue after it $s_0$. Walking forward from $s+r$, the first residue in $X$ is met likewise; call it $e$. Every residue from $s_0$ to $e-1$ lies outside $X$, so $\{s_0,\ldots,e-1\}$ is a missing run containing the given one, with $s_0-1\in X$ and $e\in X$: it is maximal. If two maximal runs share a residue $y$, each contains no element of $X$ and has its predecessor in $X$, so each starts at the residue after the first element of $X$ met walking backward from $y$; likewise each ends at the residue before the first element of $X$ met walking forward from $y$. So they coincide, and distinct maximal runs are disjoint. Composition: this is what makes the union's complement a disjoint union of shortened sets and makes $h(X)$ the largest length of a maximal run.

W2. No two shortened sets merge, and each is a maximal run of the union (page: "Two shortened sets never join into a longer run ... So the maximal missing runs of the union are exactly the shortened sets"). Re-derivation: for a maximal run $R=\{s,\ldots,s+r-1\}$ with $r\ge2$, the shortened set $S_R=\{s+1,\ldots,s+r-1\}$ has predecessor $s$ and successor $s+r$. The successor lies in $X$ by maximality of $R$, and the predecessor lies in $X+1$ because $s-1\in X$; both lie in the union. So $S_R$ is a missing run of the union whose two outside neighbors are in the union, which is a maximal missing run of the union in the page's sense. Since the union's complement is exactly the disjoint union of the nonempty $S_R$ (W1 with the page's residue-by-residue criterion $y\notin X$, $y-1\notin X$), and each $S_R$ is already bounded on both sides by residues of the union, the maximal missing runs of the union are precisely the nonempty $S_R$. The page's sentence proves the forward bound $s+r\in X$ and relies on its earlier line "fails for $y=s$" for the backward bound; its phrase "of the next" reads correctly when "the next" means the next nonempty shortened set in cyclic order (F2). The degenerate case of a single maximal run, $X$ a singleton and $r=d-1$, gives $s+r=s-1$, the one element of $X$; the shortened set then has $d-2$ residues bounded by $s-1$ and $s$, which is consistent with the identity.

W3. Reading off $h$ of the union (page: "their largest length is $h(X)-1$ when $h(X)\ge2$; when $h(X)=1$ every shortened set is empty"). Re-derivation: $h(X)$ equals the largest $r$ over maximal runs, because every missing run sits inside a maximal one of at least its length (W1). If $h(X)\ge2$, some $S_R$ is nonempty, the union is not full, and by W2 $h$ of the union is the largest $r-1$ over maximal runs with $r\ge2$, which is $h(X)-1$. If $h(X)=1$, every maximal run has one residue, every $S_R$ is empty, the union's complement is empty, and $h$ of the union is $0=\max(0,1-1)$ by the full-circle convention. With the page's opening case, $X$ full and the union full, the three cases exhaust the possibilities.

Strongest attack

The attack aimed at the wrap-around structure of the circle: make two shortened sets fuse across a boundary, or make a shortened set fail to be maximal in the union, so that $h$ of the union exceeds $h(X)-1$; or make the union's complement contain a residue outside every shortened set, so that the decomposition fails. Candidates tried by hand: $X$ a singleton (one run of $d-1$ residues whose two boundary neighbors coincide); a run of length one wedged between a single element of $X$ and a longer run, so that "the next" shortened set is empty; the alternating set in even modulus, where every run has length one; and $d=1$, $d=2$. In every case the residue $s+r\in X$ immediately after a shortened set and the residue $s\in X+1$ immediately before it prevent fusion and force maximality, and the criterion $y\notin X$, $y-1\notin X$ places every missing residue of the union inside a shortened set. The exhaustive enumeration for $d\le12$ found no counterexample to the identity or to any intermediate claim. The attack failed because the page's two boundary facts, $s-1\in X$ and $s+r\in X$, are exactly the definition of a maximal run, and the page uses both.

A second attack targeted the locators and the source's own three sentences, looking for a proof step on the page that the source does not support or a strengthened conclusion. The page proves exactly (2.1) under exactly the source's hypothesis, with the source's implicit modulus made explicit as $d\ge1$; the physical page, the lemma label, the equation number, the page count and the manuscript date all match the held PDF.

Premises

• Yu and Chen (2026), Lemma 2.1 and identity (2.1), physical p. 3, with the definition of $h$ at the top of the same page: the subject itself, not an import. Source held; read at full depth on pp. 2--4 in the text layer and on p. 3 from the page image. Interface: for a nonempty $X\subseteq\mathbb Z/d\mathbb Z$, $h(X\cup(X+1))=\max(0,h(X)-1)$.
• No imported theorem. The page names none and uses none; the proof is elementary set manipulation on the cyclic group.
• Explicit assumptions: $X$ nonempty (the source's hypothesis, retained); $d\ge1$ (the page's reading of the source's implicit modulus, F4). No local claim is consumed, and there is no batch acceptance order.

Findings

F1. Severity: suggested. Location: "every missing run is contained in a unique maximal one". Defect: the fact is asserted without argument, although the proof later relies on it for the disjointness of the shortened sets and for taking $h(X)$ over maximal runs; the page's global label "the proof below writes them out" covers the expansion of the source's three sentences, and this fact is the page's own supplied premise for that expansion. Witness: the source (physical p. 3) says only "Each missing run of length $r$ is shortened to $\max(0,r-1)$" and never that runs lie in unique maximal ones. Proposed replacement: "every missing run is contained in a unique maximal one: extend it in each direction to the first residue of $X$, which exists because $X$ is nonempty, and two maximal runs sharing a residue both start after the same element of $X$ and end before the same one".

F2. Severity: suggested. Location: "the first residue $s'+1$ of the next". Defect: when the next maximal run in cyclic order has length one, its shortened set is empty and $s'+1$ lies in $X$, so the sentence names a residue outside every shortened set; the conclusion holds because the separating residue $s+r\in X$ is the immediate successor of $s+r-1$ and because the predecessor $s\in X+1$ of each shortened set was shown two paragraphs earlier. Witness: $d=6$, $X=\{0,2,5\}$ has the maximal runs $\{1\}$ and $\{3,4\}$; the shortened set of $\{1\}$ is empty and $s'+1=2$ lies in $X$. Also $d=5$, $X=\{0,4\}$, where the only run is $\{1,2,3\}$ and "the next" shortened set is the same set. Proposed replacement: "Each nonempty shortened set $\{s+1,\ldots,s+r-1\}$ is preceded by $s\in X+1$ and followed by $s+r\in X$, both in the union; so no two shortened sets are adjacent, and each is a maximal missing run of the union."

F3. Severity: note. Location: desc, "shortens its longest missing run by exactly one". Defect: the sentence presumes a missing run exists; for the full circle both sides of (2.1) are $0$ and nothing shortens. Witness: the $\max(0,\cdot)$ in (2.1), physical p. 3. Proposed replacement: "shortens its longest missing run by exactly one and leaves a full circle full".

F4. Severity: note. Location: "Let $d\ge1$". Defect: the source never bounds the modulus; positivity is implied by "around the circle" (physical p. 3) and by the manuscript's use of $d=D_n$, a greatest common divisor of positive integers (physical p. 2). The page states the bound as setup rather than as its own reading. Proposed replacement: "Let $d\ge1$ (the source leaves the modulus implicit; a circle needs a positive one)".

Verdict

Source fidelity: faithful. The statement, its hypothesis, its quantifiers, the definition of $h$ with the full-circle convention, and every locator match the held PDF at physical p. 3.

The argument as reconstructed: sound. Every deduction follows from what precedes it; the two suggested findings concern an unproved elementary fact (F1) and the wording of one sentence (F2), neither of which breaks a step.

Limitations: the review covers Lemma 2.1 alone and not the manuscript's later uses of $h$ or of the lemma; the reviewer's enumeration for $d\le12$ is base-case coverage; the two exposures disclosed above concern the manuscript's main theorem and its standing, not this lemma.

This focused review assigns no tier and changes no status.