Independent derivations for the consecutive-gap conjunction

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Mathematical examination and independent derivations: Plasma AI. Examination date: 2026-10-15 UTC.

Read depth: the complete selected L165 statement, conventions and proof; the complete supplied L163 and L164 statement/convention interfaces; their selected standing metadata; and the supplied E1221 formulation text. The premise proofs and original publications are outside this examination. The report states the dependency reading and verdict. The exact claim, proof, premises, and catalog formulation are retained for reconstruction.

This is analytic evidence. None of the examples below is used to infer a universal asymptotic claim from finitely many cases. The universal inputs are precisely the two premise interfaces; the deductions and attacks below are independent checks of their composition. No executable mathematical check is needed to reproduce these finite identities and arguments.

1. Object convention and its extreme cases

Fix positive integers $r\le n$. List the first $n$ labeled points in cyclic order, using arrival order to break a position tie. Choose their ordered real lifts within one turn and extend by $y_{i+n}=y_i+1$. Put

$$g_i=y_{i+1}-y_i,\qquad s_i=\sum_{j=0}^{r-1}g_{i+j}=y_{i+r}-y_i.$$

The $g_i$ are nonnegative, periodic with period $n$, and sum to $1$ over one period. This includes the closing gap: replacing all lifts by their residues before subtraction would lose it. For a fixed gap index $k$ and each $j\in\{0,\ldots,r-1\}$ there is exactly one start $i$ modulo $n$ for which $i+j=k$. Hence, counting multiplicities,

$$\sum_{i=1}^n s_i =\sum_{j=0}^{r-1}\sum_{i=1}^n g_{i+j} =r.$$

Consequently the minimum and maximum of these $n$ spans satisfy

$$0\le m_n^{(r)}\le\frac rn\le M_n^{(r)}\le1.$$

In particular $M_n^{(r)}>0$, so the specified infinite ratio at zero minimum never involves $0/0$. The last bound follows because at most one complete period of nonnegative gaps is summed. Nothing here removes a zero gap or identifies two starting labels.

Changing the circular cut cyclically permutes the spans. Changing the names of labels at the same position does not change their position list; the prescribed tie rule also fixes their order. Thus the extrema are well-defined under the stated convention.

At $n=r$, every span is exactly $1$. Both scaled deviations are zero and the ratio is $1$. This would refute a positive bound at every $n\ge r$, but the claim is about limits as $n$ tends to infinity for fixed $r$. The single prefix $n=r$ has no bearing on those limits. It also exposes an invalid convention change: the shortest circle distance between the identical endpoint residues would be zero, whereas the lifted span is a complete turn.

For a sequence whose every term is the same point, and $n>r$, the gap list has one gap $1$ and $n-1$ gaps $0$. Exactly $r$ of the $n$ spans equal $1$; the others equal $0$. Thus

$$M_n^{(r)}=1,\quad m_n^{(r)}=0,\quad nM_n^{(r)}-r=n-r,\quad r-nm_n^{(r)}=r, \quad M_n^{(r)}/m_n^{(r)}=+\infty.$$

This admissible repeated-point sequence does not invalidate the conjunction. For every fixed positive $c$, $c\sqrt{\log r}\le r$ for all sufficiently large $r$, so it does not contradict the lower-deviation assertion either. The check does not claim that every repeated-point sequence behaves this way.

2. A nested uniform-grid attack and finite extrema

Define a single infinite sequence of distinct points as follows. Start with $0$. Once the points $j/2^k$, $0\le j<2^k$, have been inserted, append in increasing order all their gap midpoints

$$\frac{2j+1}{2^{k+1}},\qquad 0\le j<2^k.$$

Every stage is finite; the next stage completes the next finer grid. Newly inserted odd dyadic fractions have not occurred at an earlier stage, so this defines an element of the distinct-point family without any unproved existence claim.

Write $N=2^k$. At every prefix $N\le n<2N$ all gaps have length either $1/N$ or $1/(2N)$. For any $r\le n$ it follows directly that

$$\frac{r}{2N}\le m_n^{(r)}\le M_n^{(r)}\le\frac rN, \qquad nm_n^{(r)}\ge\frac r2, \qquad nM_n^{(r)}<2r, \qquad \frac{M_n^{(r)}}{m_n^{(r)}}\le2.$$

These inequalities hold at every prefix, with its appropriate $N$. Therefore both sequence families are nonempty and, for each positive integer $r$,

$$r\le\Lambda_r^{\mathcal F}\le2r,\qquad \frac r2\le\lambda_r^{\mathcal F}\le r,\qquad 1\le\mu_r^{\mathcal F}\le2, \qquad \mathcal F\in\{\mathcal A,\mathcal D\}.$$

The witness establishes finiteness if the outer extrema are understood as real numbers. Individual sequence limsups may still be infinite, as in Section 1; the proof's limit manipulations are valid for those too. These auxiliary bounds are a well-definedness check, not an optimality claim and not a replacement for either native premise.

At the complete-grid prefixes $n=N\ge r$, every $r$-span is $r/N$. Thus both deviations vanish and the ratio is $1$ along infinitely many prefixes of this one infinite sequence. This is a concrete attempted counterexample to an incorrectly strengthened all-prefix or liminf claim.

To check whether it attacks the actual limsups, inspect $n=3N/2$, after the first $N/2$ midpoints have been inserted, taking $N\ge2r$. The first half of the circle contains a run of $N$ gaps of length $1/(2N)$; the second half contains a run of $N/2$ gaps of length $1/N$. Each run contains $r$ consecutive gaps. Therefore

$$M_{3N/2}^{(r)}=\frac rN,\qquad m_{3N/2}^{(r)}=\frac{r}{2N},$$

and at these infinitely many prefixes

$$nM_n^{(r)}-r=\frac r2,\qquad r-nm_n^{(r)}=\frac r4,\qquad M_n^{(r)}/m_n^{(r)}=2.$$

For this example the liminf of each deviation is zero and the liminf ratio is one; the respective limsups are at least $r/2$, at least $r/4$, and exactly two. No monotonicity in $n$ can be assumed. The attempt fails against L165 because its limit directions are precisely the ones that retain the intermediate-prefix excursions. No inference about other sequences follows from this example.

3. Exact premise composition

Write L163's supplied universal interface as

$$(R):\quad \exists r_R\ \forall r\ge r_R\ \forall X\in\mathcal A, \quad\limsup_n\frac{M_n^{(r)}}{m_n^{(r)}} \ge1+\frac{\log r}{200r}.$$

L164 supplies one $c$ and one threshold for both inequalities:

$$(O):\quad \exists c>0\ \exists r_O\ \forall r\ge r_O\ \forall X\in\mathcal A,\quad \begin{cases} \limsup_n(nM_n^{(r)}-r)\ge c\sqrt{\log r},\\ \limsup_n(r-nm_n^{(r)})\ge c\sqrt{\log r}. \end{cases}$$

All displayed $r$ and $n$ are integers in the positive range, and the limits use $n\ge r$. Here $(R)$ is L163 and $(O)$ is L164; their proofs are not claimed as part of this derivation. Their supplied conventions agree on every object used here, including repetitions, tie order, lifted forward spans and natural logarithms. The counting-arc convention in those interfaces is not an additional counting lemma consumed by this composition.

Increase the supplied thresholds to at least $2$ if necessary, which only restricts the eventual range, and choose $r_0=\max(r_R,r_O)$. The constant $c$ is the one in $(O)$. This gives one pair of absolute constants valid for both families and all their members. An onset time for approaching a limsup may depend on $X$ and $r$; no common onset time is asserted.

The conjunction is a conjunction of three statements about limsups. It is not a limsup of a conjunction, minimum, or vector assertion. For example, the scalar sequences $a\mathbf1_{n\text{ even}}$ and $a\mathbf1_{n\text{ odd}}$, with $a>0$, both have limsup $a$ although their pointwise minimum is always zero. These are a diagnostic for a false common-subsequence inference, not proposed circle-span sequences. L165 makes no such inference.

The ratio bound is an essential independent input. From a deviation of size $a=c\sqrt{\log r}$ and the mean inequality alone, one would obtain bounds such as $1+a/r$ or $r/(r-a)$, when the latter is defined. These are only of order $1+\sqrt{\log r}/r$, not the claimed logarithmic improvement. The proof actually invokes $(R)$ for that improvement; it does not derive it by combining unrelated extrema or different times from $(O)$.

4. Tail identities, outer extrema, and adversarial optimization

Fix $r\ge r_0$, and abbreviate

$$a=c\sqrt{\log r},\quad b=\frac{\log r}{200},\quad u_n=nM_n^{(r)},\quad v_n=nm_n^{(r)},\quad q_n=\frac{M_n^{(r)}}{m_n^{(r)}}.$$

Then $u_n\ge r$, $0\le v_n\le r$, and $1\le q_n\le+\infty$. For every sufficiently large integer $K$, the exact tail identities are

$$\sup_{n\ge K}(u_n-r)=\sup_{n\ge K}u_n-r, \qquad \sup_{n\ge K}(r-v_n)=r-\inf_{n\ge K}v_n.$$

Taking limits of tails gives

$$\limsup_n(u_n-r)=\limsup_n u_n-r, \qquad \limsup_n(r-v_n)=r-\liminf_n v_n.$$

The second identity is between finite bounded quantities. The first remains valid when the limsup is positive infinity because the subtracted number $r$ is finite; there is no subtraction of two infinite quantities.

For each $X\in\mathcal A$ define its three limiting values

$$U_X=\limsup_n u_n,\qquad V_X=\liminf_n v_n,\qquad Q_X=\limsup_n q_n.$$

The premise interfaces and the tail identities imply, pointwise in $X$,

$$U_X\ge r+a,\qquad V_X\le r-a,\qquad Q_X\ge1+b/r.$$

For any nonempty family $\mathcal F\subseteq\mathcal A$, every member of the first set of values is at least $r+a$, every member of the second is at most $r-a$, and every member of the third is at least $1+b/r$. By the defining order properties of infimum and supremum, respectively,

$$\inf_{X\in\mathcal F}U_X\ge r+a,\qquad \sup_{X\in\mathcal F}V_X\le r-a,\qquad \inf_{X\in\mathcal F}Q_X\ge1+b/r.$$

No member attaining an extremum is selected. In particular, taking approximate optimizers separately for every $r$ cannot evade a bound that already holds for every sequence at that fixed $r$. An extremum strictly below its asserted lower bound would require a member below that bound; a supremum strictly above its asserted upper bound would require a member above that bound.

The most threatening attempted invalidation was to let the chosen sequence have arbitrarily late excursions, thereby making every finite observation look better. The following exact scalar diagnostic shows what this could break. For positive integers $k$ set

$$f_k(n)=\begin{cases}0&n<k,\\a&n\ge k.\end{cases}$$

Then $\inf_k\limsup_n f_k(n)=a$, but $\limsup_n\inf_k f_k(n)=0$. Likewise, if $0<a\le r$ and $h_k(n)=r-f_k(n)$, then $\sup_k\liminf_n h_k(n)=r-a$, whereas $\liminf_n\sup_k h_k(n)=r$. These numerical families are not asserted to be realizable by circle sequences. They isolate exactly the invalid exchange that would make the attack work. L165 retains the outer extrema after the limits, so it uses the first expression in each comparison. The direct order argument above rules out this failure even for realizable delayed families.

The same reasoning handles nonattainment: values can approach a bound strictly from one side without crossing it, but their infimum or supremum cannot cross it either. There is no ratio of separately optimized numerator and denominator anywhere in the inference; $Q_X$ is formed from the actual paired spans of one sequence before its infimum is taken.

5. Subclass restriction and divergence

For the nonempty subclass $\mathcal D\subseteq\mathcal A$, restriction can only increase an infimum and decrease a supremum. Applied separately to $U_X,V_X,Q_X$, this yields exactly

$$\Lambda_r^{\mathcal D}\ge\Lambda_r^{\mathcal A},\qquad \lambda_r^{\mathcal D}\le\lambda_r^{\mathcal A},\qquad \mu_r^{\mathcal D}\ge\mu_r^{\mathcal A}.$$

The second sign reverses once more upon subtracting from $r$, so all three claimed deviations are bounded below for the subclass as well. A reversed infimum/supremum sign would be fatal; none occurs. This argument establishes inequalities only and does not establish equality of the two families' extrema.

All quantities here use a fixed $r$ until the final step. For any real target $T>0$, all integers $r$ larger than $r_0$ and sufficiently larger than both $\exp((T/c)^2)$ and $\exp(200T)$ satisfy $a>T$ and $b>T$. The three extremal deviations therefore exceed $T$ for every such $r$ in either family. This proves divergence to positive infinity through the integers. No monotonicity of the extrema in $r$, no interchange of limits in $r$ and $n$, and no fixed optimizing sequence across different $r$ are needed.

6. Transfer to exactly the supplied formulation

In the supplied E1221 modified Statement together with its Notes, the circle has circumference $1$, the first $n$ terms are cyclically ordered, and the observable is a forward span of $r$ consecutive gaps. The Notes identify the displayed maximum and minimum as the largest and smallest such spans. This specifies extrema over the $n$ starting labels, despite the displayed $\max_r$ and $\min_r$ subscripts. The phrase concerning consecutive points is resolved by the supplied $r$-gap convention, not by changing it to an $(r-1)$-gap or shortest-distance convention.

The outer family is all infinite sequences, with no distinctness restriction. Repeated points remain labeled, giving zero gaps when appropriate; the same cyclic sum is retained. Therefore, under the supplied conventions, each of its three limiting functionals and each outer extremum agrees with the $\mathcal A$ version above. The requested expressions are

$$\Lambda_r-r,\qquad r-\lambda_r,\qquad r(\mu_r-1),$$

exactly those bounded in Sections 4 and 5. In mean-normalized notation the first two are $r(\Lambda_r/r-1)$ and $r(1-\lambda_r/r)$. They are not the printed $r(\Lambda_r-1)$ and $r(1-\lambda_r)$.

This comparison needs no theorem or claim about the cited publications. It establishes correspondence with the supplied modified mathematical question only. It does not establish that this is the wording, intended normalization, scope, or historical interpretation of an original publication. The historical inequalities, quotations, bibliographic claims and source-credit assertions within the catalog snapshot are not premises of this deduction and were not independently certified.

7. Reconstruction and limits

To reproduce this analytic examination from an ordinary clone containing the filed package, read the four linked mathematical snapshots and the selected dependency records identified by the report. Recompute the cyclic double sum in Section 1 and the two types of dyadic prefixes in Section 2. Check every quantifier in the two interfaces of Section 3, derive the two exact tail identities of Section 4, and then apply the defining order properties of infimum and supremum. Finally compare the catalog's modified Statement and Notes term by term with Section 6. The separate report gives the complete checklist and current-at-examination premise standing.

The dyadic sequence, constant repeated sequence, and scalar diagnostic families are specified in full in this file; no private data or numerical output is needed. The scalar diagnostics disprove particular invalid inference rules, not L165. The derivation establishes the conjunction from the two native interfaces, without re-proving them, improving their constants, establishing optimality or attainment, equating sequence-family extrema, or performing any kernel or historical-source fidelity audit.