> Full-text translation from Chinese. The original remains authoritative for its research claims; this translation is not independent mathematical verification.

# WSFC Round 06
## Adversarial Re-Derivation: Finding a Plateau Gap, Completing a General Repair, and Reconstructing Support-Class Necessity

**English Title:** *Adversarial Re-Derivation of the Finite-$n$ Support-Tangent Reduction: A Plateau Gap, Its General Repair, and a Source-First Proof Chain*  
**Document ID:** `WSFC-R06-v0.1`  
**Version:** v0.1  
**Date:** 2026-09-23  
**Author:** Neo.K  
**AI Research Collaborator:** Aletheia / OpenAI GPT-5.6 Sol  
**Series:** Worm Sequence Freedom Closure  
**Prerequisite:** `WSFC-R05-v0.1`  
**Status:** `ADVERSARIAL VERIFICATION / GAP FOUND AND REPAIRED / SOURCE-FIRST RE-DERIVATION`  
**Canonical Source:** UTF-8 Markdown

---

# Abstract

Round 05 formed a provisional global reduction:

$$
\Gamma_n^\star
\in
\mathcal T_n,
\qquad
n\ge4,
$$

and explicitly required the next round to stop expanding and focus on attempts to refute it.

Round 06 therefore does not increase $n$ or aim to generate more candidates, but performs:

1. an external-source re-check;
2. source-first independent proof decomposition;
3. a symbolic / CAS audit;
4. an interval arithmetic audit;
5. an omitted-boundary-family search;
6. formalization decomposition.

This round obtains an important and healthy result:

$$
\boxed{
\text{Round 05's proof chain did indeed omit a boundary family.}
}
$$

Round 05's proof for a positive plateau with $n\ge5$ uses:

$$
q_L,q_R\ge1,
$$

but this does not cover:

> One side has only one direct segment, namely $q=0$; the other has $q\ge1$ angular gaps; and a positive top plateau is also present.

This is a genuine proof gap.

However, the gap admits a general repair.

For the one-direct-side positive-plateau branch, fix the gap count on the other side:

$$
q\ge1,
$$

Interior stationarity gives:

$$
x_\star
=
\tan
\frac{c}{2(q+1)}.
$$

The branch's total length reduces to:

$$
\boxed{
T_q(c)
=
\csc c
+
2(q+1)
\tan
\frac{c}{2(q+1)}.
}
$$

And:

$$
\boxed{
T_q'(c)
=
-\csc c\cot c
+
\sec^2
\frac{c}{2(q+1)}.
}
$$

For admissible:

$$
c\in[\pi/3,\pi/2]
$$

we have:

$$
\csc c\cot c
\le
\frac23,
$$

and:

$$
\sec^2
\frac{c}{2(q+1)}
>1.
$$

Thus, for all:

$$
q\ge1,
$$

we have:

$$
\boxed{
T_q'(c)
>
\frac13
>0.
}
$$

The minimum of every one-sided positive-plateau branch therefore occurs at its birth boundary, which is precisely:

$$
\boxed{
p=0.
}
$$

Round 05's plateau conclusion therefore survives, but its proof must be replaced by this general lemma, rather than retaining the original assertion that $n\ge5$ implies $q>0$ on both sides.

This round also rechecks the original Adhikari–Pitman source. Its indexed text explicitly contains:

$$
\boxed{
\text{LEMMA 7. A minimal arc with at most }M\text{ segments is convex.}
}
$$

and explicitly gives endpoint-strip reduction and the three-condition two-disc criterion in the convex case.

This allows Round 05's most dangerous concern—whether an interior support shift can create a new narrower width direction—to be reframed and avoided:

> Do not move a support line within the full width functional; first use the two-disc criterion to turn each side into a single-disc obstacle path, then perform support-radius reduction in the fixed-normal-order obstacle problem.

In the single-disc problem, as long as the distance of every segment's supporting line from the disc remains:

$$
h_i\ge1,
$$

the entire polygonal path remains outside the obstacle.

For an interior support line, the derivative of fixed-normal polygonal length with respect to its support number is:

$$
\boxed{
\frac{\partial\ell}{\partial h_i}
=
\tan\frac{\delta_{i-1}}2
+
\tan\frac{\delta_i}2
>0.
}
$$

Thus:

$$
h_i>1
$$

permits a safe inward shift until:

$$
h_i=1
$$

or a support-angle collision / lower-stage degeneration.

This gives a cleaner independent justification than the original text of Round 05.

The CAS audit also confirms:

- the Hessian determinant of $g(q,s)=2q\tan(s/(2q))$ is $0$;
- $g$ is a convex perspective;
- $\Delta_d''(a)>0$;
- the Hessian-determinant factorization for the composed one-side functional:
  $$
  \boxed{
  \det\nabla^2J_d
  =
  g_{qq}g_s\Delta_d''(a)
  >0
  }
  $$
  is correct;
- Round 04's balanced allocation argument shows no algebraic mismatch in this symbolic audit.

Global exclusion of $q=0$ is also recomputed using 60-digit interval arithmetic. The difference between the continuous one-side lower envelope and the explicit four-segment competitor yields a strictly positive interval, so Round 05's numerical separation is not merely a floating-point sign.

The final verdict of Round 06 is:

```text
ROUND 05 GLOBAL REDUCTION:
    NOT REFUTED

ONE REAL GAP:
    FOUND

THE GAP:
    REPAIRED BY A STRICTLY MORE GENERAL LEMMA

THEOREM STATUS:
    SURVIVES AS A RESEARCH PROOF CHAIN
    NOT YET FORMALLY VERIFIED
    NOT PEER REVIEWED
```

---

# 1. Verification Principle

The task of Round 06 is not:

> to find justifications for Round 05.

Instead, it is:

> What is the smallest point at which the entire reduction could fail?

Thus it adopts:

$$
\boxed{
\text{source anchor}
\rightarrow
\text{local lemma}
\rightarrow
\text{boundary family}
\rightarrow
\text{counterexample attempt}.
}
$$

Failure at any layer must lower the global theorem status.

---

# 2. External Source Audit

The indexed text of the original Adhikari–Pitman technical report / 1989 paper explicitly states:

> `LEMMA 7. A minimal arc with at most M segments is convex.`

Thus:

$$
\boxed{
\text{finite-segment convexity}
}
$$

is not a premise invented by Round 05.

Its convex-arc section also explicitly explains:

- if the arc extends to the left/right beyond the endpoints, perpendicular reduction gives a shorter arc of greater width;
- it can therefore be restricted to the endpoint strip;
- for a strip-confined convex arc, the width condition is equivalent to:
  1. intersecting the unit-height horizontal segment;
  2. the left subarc avoiding the open unit disc at $B$;
  3. the right subarc avoiding the open unit disc at $A$.

Round 06 therefore fixes these three external pieces as a proof API.

---

# 3. Finite-budget Strip Compatibility

The external perpendicular reduction is stated for a general convex arc.

Round 06 additionally checks finite-segment compatibility.

That reduction replaces an unnecessary convex subarc with one perpendicular straight segment.

For a polygonal arc:

$$
\boxed{
\text{replace a nontrivial subpath by one segment}
}
$$

cannot increase the segment count.

Thus:

$$
\boxed{
\text{endpoint-strip reduction is compatible with an at-most-}M\text{ budget}.
}
$$

This step consists of a source theorem + a direct finite-budget observation.

---

# 4. Independent One-Disc Polygonal Geodesic Re-Derivation

Fix the endpoints and split point $F$.

The problem on the left is:

> Without entering $D(B,1)$, what is the shortest polygonal path from $A$ to $F$ using finitely many segments?

The right side is symmetric with respect to $D(A,1)$.

This step temporarily removes full-width coupling.

---

# 5. Interior Support Radius Monotonicity

Consider a convex ordered polygonal path around a unit disc.

Let the segment supporting lines be:

$$
L_i:
\langle n_i,x\rangle=h_i,
$$

where:

$$
h_i\ge1.
$$

The adjacent outward-normal angle gaps are:

$$
\delta_i\in(0,\pi).
$$

In a fixed normal-order cell, for an interior line $L_i$:

$$
\boxed{
\frac{\partial\ell}{\partial h_i}
=
\tan\frac{\delta_{i-1}}2
+
\tan\frac{\delta_i}2.
}
$$

Thus:

$$
\boxed{
\frac{\partial\ell}{\partial h_i}>0.
}
$$

If:

$$
h_i>1,
$$

we can let:

$$
h_i\downarrow
$$

until:

1. $h_i=1$;
2. $\delta_{i-1}=0$ or $\delta_i=0$, entering a lower-stage collision.

This argument does not depend on other full-width directions remaining unchanged.

Its feasibility depends only on:

$$
h_j\ge1
$$

ensuring that no segment line cuts into the unit disc.

Thus Round 05's interior saturation conclusion is retained, but the proof language should change to:

$$
\boxed{
\text{single-disc obstacle support reduction}.
}
$$

---

# 6. Endpoint Shortcut Re-check

An endpoint-adjacent line cannot be freely translated, since it must pass through a fixed endpoint.

For the path:

$$
P_0P_1P_2\cdots,
$$

slide:

$$
P_1
$$

along:

$$
P_1P_2
$$

toward $P_2$.

If the first event is that the line through:

$$
P_0P_1'
$$

is tangent to the disc, endpoint tangency is obtained.

If:

$$
P_1'=P_2,
$$

the segment count decreases.

Before stopping, the new first line has not entered the disc, and the second segment is a subsegment of the old legal segment.

The triangle inequality guarantees nonincrease of the two-edge length.

Thus:

$$
\boxed{
\text{endpoint shortcut}
}
$$

survives the audit in the one-disc formulation.

---

# 7. The Gap Actually Found: The Plateau $q=0$ Family

Round 05's plateau proof for $n\ge5$ explicitly assumes:

$$
q_L,q_R\ge1.
$$

But:

$$
n\ge5
$$

does not automatically imply this condition.

For example, the total segment budget can be allocated as:

```text
one direct left segment
+
one positive horizontal plateau segment
+
multiple right tangent segments
```

Then:

$$
q_L=0,
\qquad
q_R\ge1.
$$

Round 05's general convexity proof for $n\ge5$ does not cover this family.

Although Round 05 separately excludes q=0 without a plateau, its text does not complete the full analytic closure of:

$$
\boxed{
q=0
+
p>0
}
$$

as above.

Thus:

$$
\boxed{
\text{ROUND 05 HAD A REAL PROOF GAP HERE.}
}
$$

---

# 8. General One-Sided Plateau Repair

Consider:

- a direct segment on one side;
- $q\ge1$ angular gaps on the other side;
- a positive plateau $p>0$.

Let:

$$
c
=
\arcsin\frac1d.
$$

First push the direct side to unit-disc tangency using the monotone shortcut.

In the positive-plateau branch, let the other side's top-coordinate parameter be $x$.

The total length reduces to:

$$
T(c,x)
=
\csc c
+
2x
+
2q
\tan
\left(
\frac{
c-2\arctan x
}{
2q
}
\right).
$$

Interior stationarity:

$$
\frac{\partial T}{\partial x}=0
$$

is equivalent to:

$$
\tan
\left(
\frac{
c-2\arctan x
}{
2q
}
\right)
=
x.
$$

Since the relevant angles lie in the principal interval:

$$
\frac{
c-2\arctan x
}{
2q
}
=
\arctan x.
$$

we have:

$$
\boxed{
c
=
2(q+1)\arctan x.
}
$$

That is:

$$
\boxed{
x_\star
=
\tan
\frac{c}{2(q+1)}.
}
$$

Substituting back:

$$
\boxed{
T_q(c)
=
\csc c
+
2(q+1)
\tan
\frac{c}{2(q+1)}.
}
$$

This is the complete generalization of Round 05's special $q=1$ expression:

$$
\csc c+4\tan(c/4)
$$

as above.

---

# 9. General Plateau Branch Strict Monotonicity

Differentiating:

$$
\boxed{
T_q'(c)
=
-\csc c\cot c
+
\sec^2
\frac{c}{2(q+1)}.
}
$$

On:

$$
c\in[\pi/3,\pi/2]
$$

we have:

$$
\csc c\cot c
=
\frac{\cos c}{\sin^2c}
\le
\frac23.
$$

while:

$$
\sec^2
\frac{c}{2(q+1)}
>1.
$$

Thus, for every integer:

$$
q\ge1,
$$

there is the uniform bound:

$$
\boxed{
T_q'(c)
>
\frac13.
}
$$

The interior positive-plateau branch is therefore strictly increasing with $c$.

---

# 10. Branch Birth Boundary = $p=0$

The existence condition for the positive-plateau branch is:

$$
x_\star
>
\cot c.
$$

Birth boundary:

$$
\tan
\frac{c}{2(q+1)}
=
\cot c.
$$

Here:

$$
p=0.
$$

Since:

$$
T_q'(c)>0,
$$

the branch minimum must occur at the smallest admissible $c$, namely the birth boundary.

Thus:

$$
\boxed{
\min_{\text{one-sided }q=0,\;p>0}
T
\text{ occurs at }
p=0.
}
$$

This time, it holds simultaneously for all:

$$
q\ge1
$$

as above.

Round 05's plateau conclusion is therefore repaired.

---

# 11. CAS Audit

Round 04 uses:

$$
g(q,s)
=
2q\tan\frac{s}{2q}.
$$

CAS recomputation gives:

$$
g_s
=
\sec^2\frac{s}{2q},
$$

$$
g_{ss}
=
\frac{
\sec^2(s/(2q))
\tan(s/(2q))
}{
q
},
$$

and:

$$
\boxed{
g_{qq}g_{ss}-g_{qs}^2=0.
}
$$

Thus the perspective Hessian is rank-one semidefinite.

On the other hand:

$$
\Delta_d(a)
=
\frac\pi2
-
\arccos\frac1d
-
2\arctan a,
$$

$$
\Delta_d''(a)
=
\frac{4a}{(1+a^2)^2}
>0.
$$

The composed one-side functional:

$$
J_d(q,a)
=
\text{const}
+
a
+
g(q,\Delta_d(a))
$$

therefore has:

$$
\boxed{
\det\nabla^2J_d
=
g_{qq}
g_s
\Delta_d''(a)
>0.
}
$$

Round 04's joint strict-convexity factorization passes the symbolic audit.

---

# 12. q=0 Separation Interval Audit

Round 05 uses:

$$
B_\infty(c)
=
\csc c
+
2\cot c
+
3c-\pi.
$$

Stationary equation:

$$
3\cos^2c+\cos c-1=0.
$$

Let:

$$
z_\star
=
\frac{\sqrt{13}-1}{6}.
$$

Round 06 directly evaluates with 60-decimal interval arithmetic:

- $B_\infty(c_\star)$;
- the explicit competitor $L_4(53/50)$;
- their difference.

A strictly positive margin interval is obtained.

Thus:

$$
\boxed{
\min B_\infty
>
L_4(53/50)
}
$$

is certified in this interval runtime, rather than relying only on an ordinary floating-point sign.

The complete interval endpoints are retained in:

```text
WSFC_R06_q0_interval_certificate.json
```

---

# 13. Lower-Stage Degeneration Re-check

No logical gap was found in the following reductions:

$$
\text{zero segment}
\rightarrow
\text{delete},
$$

$$
\text{same-line consecutive segments}
\rightarrow
\text{merge},
$$

$$
\delta_i=0
\rightarrow
\text{support-line collision}
\rightarrow
\text{merge}.
$$

For convex ordered normals, a support-order swap must first pass through:

$$
\delta_i=0.
$$

Thus Round 05's lower-stage absorption survives in the current finite convex-chain model.

Its scope must be retained:

> This is not a general theorem for every possible topological change; it concerns only the current convex polygonal normal-order state space.

---

# 14. External Novelty Re-check

This round performs another targeted search:

- `"broadest n-segment unit arc"`;
- `"broadest polygonal arc" unit breadth n segment`;
- specific `four-/five-/six-segment` searches.

The results still mainly lead back to the three-segment theorem and general-$n$ conjectures of Panraksa–Wetzel–Wichiramala 2007; this targeted search did not find subsequent work explicitly claiming that all general-$n$ conjectures had been resolved.

The authors' / collaborators' public publication lists also still show the two relevant 2007 papers without an obvious general-$n$ completion.

This supports only:

```text
NO OBVIOUS PRIOR GENERAL-n COMPLETION FOUND IN TARGETED SEARCH
```

It does not support:

```text
NOVELTY PROVED
```

---

# 15. Results of the Round 06 Audit Matrix

The most important status changes:

```text
FINITE-M CONVEXITY
    PASS / SOURCE

ENDPOINT STRIP
    PASS-WITH-FINITE-BUDGET NOTE

TWO-DISC CRITERION
    PASS / SOURCE

INTERIOR SUPPORT SATURATION
    PASS AFTER SAFER ONE-DISC REPROOF

ENDPOINT SHORTCUT
    PASS AT RESEARCH-PROOF LEVEL

PLATEAU qL,qR>0
    PASS

PLATEAU q=0 + p>0
    GAP FOUND
    GENERAL REPAIR DERIVED
    PASS AFTER REPAIR

q=0 GLOBAL EXCLUSION
    PASS / INTERVAL CERTIFIED

LOWER-STAGE ABSORPTION
    PASS WITH EXPLICIT SCOPE

EQUAL-ANGLE PARTITION
    PASS

BALANCED ALLOCATION
    PASS / CAS

GLOBAL SUPPORT-CLASS REDUCTION
    SURVIVES WITH REPAIRED LEMMA
```

---

# 16. Theorem Status

Round 06 does not upgrade the theorem to `PROVED`.

The more precise status is:

$$
\boxed{
\texttt{INDEPENDENTLY RE-DERIVED RESEARCH THEOREM CHAIN}
}
$$

with:

```text
ONE GAP FOUND AND REPAIRED
NO REFUTATION FOUND
FORMAL CHECK STILL OPEN
PEER REVIEW STILL OPEN
```

This is stronger than Round 05's `PROVISIONAL GLOBAL REDUCTION`, but still is not publication-level finality.

---

# 17. Next Steps

The next round should move beyond a human / the same AI thinking it through again to:

1. machine-checkable symbolic lemmas;
2. interval arithmetic around every nonsymbolic separation;
3. a formal statement of the finite polygonal arc state space;
4. an independent solver search for a convex strip path outside $\mathcal T_n$;
5. a Lean / Isabelle / Coq formalization skeleton;
6. only after that, derive general-$n$ corollaries as theorem candidates.

In particular, formalization should prioritize:

$$
\boxed{
\text{single-disc finite-segment shortest-path tangent reduction}.
}
$$

because it is now the core reusable lemma of the entire support-class necessity chain.

---

# Conclusion

Round 06's result is not that we once again believe Round 05.

Instead:

$$
\boxed{
\text{A boundary family omitted by Round 05 was actually found.}
}
$$

That family is then fully analyzed:

$$
\boxed{
T_q(c)
=
\csc c
+
2(q+1)
\tan\frac{c}{2(q+1)}
}
$$

and we prove:

$$
\boxed{
T_q'(c)>\frac13
}
$$

for all:

$$
q\ge1.
$$

The gap is therefore repaired generally, not hidden behind a numerical patch.

At the same time, the most dangerous interior support concern is decomposed afresh by the two-disc criterion into single-disc obstacle geometry, making the proof responsibility for tangent necessity more local and cleaner.

The most reasonable current conclusion is therefore:

$$
\boxed{
\text{Round 05 global reduction survives adversarial re-derivation,}
}
$$

but:

$$
\boxed{
\text{the next required authority is machine verification, not more intuition.}
}
$$

===END===