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Source. Y. Yu and K. Chen, Erdős Problem 354(i): Strong Completeness of Two Dyadic Floor Sequences, manuscript of 13 September 2026: the unnumbered Theorem under "Theorem and scope", physical p. 1; Section 6 "Event spacing by well-founded descent", p. 7; Section 7 "Reduction to the finite-event contradiction", p. 7; and the closing paragraph of Section 11, p. 14. In the seventeen-page PDF held by its library source card, Yu and Chen (2026); the statement is filed on the card's Theorem page. The remaining sections are reconstructed on the linked pages of this folder.

Standing. This is an author-recorded reconstruction of an unrefereed manuscript. It is not an independent review, changes no status and assigns no tier. The problem page's recorded answer to the first question rests on a different, site-accepted proof; this reconstruction adds no acceptance evidence for either.

Definitions

$A_{\alpha,\beta}$, complete and strongly complete sets, the normalized pair, its weights, conversions, events, $K_n$, $P_n$ and $h_n$ are as on the normalization page; $C_M=16(M+1)^2$ is the constant of (5.2) on the Theorem 5.1 page; bounded event spacing is as on the bounded-spacing page.

Statement

Theorem (p. 1). Let $\alpha,\beta>0$ with $\alpha/\beta$ irrational. For every finite $F\subseteq\mathbb Z$ there is an integer $H$ such that every integer $m\ge H$ is a sum of distinct elements of $A_{\alpha,\beta}\setminus F$. In particular $A_{\alpha,\beta}$ is strongly complete, and every sufficiently large integer is $\sum_{s\in S}\lfloor2^s\alpha\rfloor+\sum_{t\in T}\lfloor2^t\beta\rfloor$ for some finite $S,T\subset\mathbb N$.

Proof

Step 1: normalization

Fix $F$. The normalization on the normalization page (items 1--3) gives $u,v\ge0$ such that the pair $2^u\alpha$, $2^v\beta$ is normalized ($N<M<2N$, $N\ge2$), has irrational ratio in $(1,2)$, and has all its values above $\max(F\cup\{0\})$. By its reduction, it suffices to show that every sufficiently large integer lies in $\bigcup_nP_n$ for this normalized pair. Suppose, for a contradiction, that the normalized sequence is incomplete. By item 5 its event set is infinite.

Step 2: incompleteness forces bounded event spacing (Section 6)

Call two consecutive events $n<m$ (both in the event set, none between) a qualifying pair if $m-n\ge2n+C_M$. For such a pair the conversions at indices $n,\ldots,m-2$ are zero and the conversion at index $m-1$ is nonzero, so the hypotheses of Section 3 on the Theorem 5.1 page hold at layer $n$ with $\ell=m-n$, and (5.2) gives $K=2^\ell\ge K_*$. Theorem 5.1 then yields

$$h_t\le\max(0,h_n-1)\qquad(t\ge m+3),$$

and completeness if $h_n\le1$.

Claim: only finitely many pairs qualify. Otherwise let $(n_1,m_1)$ qualify. Since the sequence is incomplete, $h_{n_1}\ge2$, and $h_t\le h_{n_1}-1$ for all $t\ge m_1+3$. Choose a qualifying pair $(n_2,m_2)$ with $n_2\ge m_1+3$; then $h_{n_2}\le h_{n_1}-1$, and again $h_{n_2}\ge2$ by incompleteness. Repeating produces a strictly decreasing sequence $h_{n_1}>h_{n_2}>\cdots$ of integers that are all at least $2$, which is impossible. (The argument uses only the permanent bound after each qualifying pair, not monotonicity of $h_t$ from layer to layer.)

Hence there is $n_1$ such that every pair of consecutive events $n<m$ with $n\ge n_1$ satisfies $m-n<2n+C_M$, that is, $m<3n+C_M$. Let $t_1$ be an event with $t_1\ge n_1$ and put $n_0=\max(t_1,C_M)$. For every integer $n\ge n_0$, let $n'$ be the largest event $\le n$ (it exists and $n'\ge t_1\ge n_1$) and $m$ its successor event (it exists because the event set is infinite). Then $m>n$ and $m<3n'+C_M\le3n+n=4n$. So every $n\ge n_0$ has an event in $(n,4n]$: the sequence has bounded event spacing with $R=4$.

Step 3: the contradiction

The normalized pair has irrational ratio and is assumed incomplete, so (BG) on the bounded-spacing page applies and says that it does not have bounded event spacing. This contradicts Step 2. Hence the normalized sequence is complete: there is $H$ such that every integer $m\ge H$ lies in some $P_n$.

Step 4: back to the original set (Section 7)

By the reduction on the normalization page, every $m\ge H$ is a sum of distinct elements of $A_{\alpha,\beta}\setminus F$, because the retained tails consist of pairwise distinct elements of $A_{\alpha,\beta}$ above $\max(F\cup\{0\})$. As $F$ was arbitrary, $A_{\alpha,\beta}$ is strongly complete. With $F=\emptyset$, reading each represented value with its index gives the indexed statement: a sum over distinct indices of the two floor sequences, which is the "That is" clause of Problem 354.

Dependency map

• Normalization, interlacing, infinite events, prefix bounds, reduction: Sections 1 and 7.
• Finite lemmas: 2.1, 2.2, 2.3.
• Exact block, certificate, mesh connection, permanent descent and the length constant: Theorem 5.1.
• Finite-event decay (FE) and (FE-R): Section 8.
• Window lemma and digit budget (DB): Section 9.
• Good rationals and sparse windows (10.2): Section 10.
• Bounded-spacing contradiction (BG): Section 11.

External inputs. Dirichlet's approximation theorem (windows page, imported). Finite data. The mask certificate of Appendix A, rechecked by the folder's evidence.

Scope. Base exactly $2$ and the irrational-ratio hypothesis. The rational-ratio cases of Hegyvári's conjecture and the variable-base second question of the problem are not addressed by this argument; the source says the same (pp. 1--2). The source's Lean formalization and its axiom audit are the source's own account and were not built or read here.