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

# WSFC Round 26
## Elasticity Kernel Closure: Computer-Assisted $K>0$ and Strictness in the 2007 Conjecture (6)

**English Title:** *Elasticity Kernel Closure: Computer-Assisted Positivity and Strict Endpoint-Distance Monotonicity*  
**Document ID:** `WSFC-R26-v0.1`  
**Version:** v0.1  
**Date:** 2026-09-24  
**Author:** Neo.K  
**AI Research Collaborator:** Aletheia / OpenAI GPT-5.6 Sol  
**Series:** Worm Sequence Freedom Closure  
**Prerequisite:** `WSFC-R25-v0.1`  
**Status:** `CONJECTURE (6) RESEARCH-PROOF CLOSURE / COMPUTER-ASSISTED KERNEL / FORMAL AUTHORITY OPEN`  
**Canonical Source:** UTF-8 Markdown

---

# Abstract

Round 25 corrects:

$$
\boxed{
b_n=t_nd_n^\star
}
$$

and proves:

$$
d_{n+1}^\star<d_n^\star.
$$

But strictness of the unit-length endpoint distance still requires:

$$
\mathcal E
=
V V_{dq}
+
dV_qV_{dd}
>0.
$$

Round 26 fully decomposes this kernel.

Let:

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

First:

$$
-g_q
=
2(z\sec^2z-\tan z),
$$

$$
-g_{sq}
=
\frac{2z\sec^2z\tan z}{q}.
$$

This round proves:

$$
\boxed{
\frac{-g_{sq}}{-g_q}
\ge
\frac3s.
}
$$

Equivalently:

$$
z\sec^2z-\tan z
\le
\frac23z^2\sec^2z\tan z.
$$

Let the right side minus the left side be $R(z)$.

Its derivative is:

$$
R'(z)
=
-
\frac{
2z[
2z\cos^2z-3z+\sin z\cos z
]
}{
3\cos^4z
}.
$$

And:

$$
2z\cos^2z-3z+\sin z\cos z
\le
2z-3z+z=0.
$$

Hence:

$$
R'(z)\ge0,
\qquad
R(0)=0.
$$

The ratio lemma holds.

R25 establishes:

$$
V_{dq}=(-g_{sq})(-s').
$$

Thus:

$$
V_{dq}
\ge
\frac{
3(-g_q)(-s')
}{s}.
$$

On the other hand, the breadth-one optimizer has a top point of height $1$.

If the endpoint distance is $d$ and the top point's horizontal coordinate is $x$, then:

$$
V
\ge
\sqrt{1+x^2}
+
\sqrt{1+(d-x)^2}
\ge
2\sqrt{1+\frac{d^2}{4}}.
$$

The physical support cell satisfies:

$$
d<\frac2{\sqrt3}.
$$

Therefore:

$$
2\sqrt{1+d^2/4}>2d.
$$

Hence:

$$
\boxed{
b=\frac dV<\frac12.
}
$$

Now:

$$
\mathcal E
=
V V_{dq}
-
d(-V_q)V_{dd}.
$$

The ratio lemma gives:

$$
\mathcal E
\ge
(-V_q)
\left[
\frac{3V(-s')}{s}
-dV_{dd}
\right].
$$

Since:

$$
d<V/2,
$$

it suffices to prove:

$$
\boxed{
\mathcal K
=
6(-s')
-sV_{dd}
>0.
}
$$

Round 26 constructs a computer-assisted interval certificate for $\mathcal K$.

---

# 1. Continuous Count Domain

Write the stage interpolation as:

$$
(q_L,q_R)
=
(q,q+\lambda),
$$

$$
q\ge1,
\qquad
0\le\lambda\le1.
$$

Let:

$$
r=1/q\in[0,1],
$$

Then:

$$
1/q_R
=
\frac{r}{1+\lambda r}.
$$

Thus the infinite support-count family is compactified to:

$$
(r,\lambda)\in[0,1]^2.
$$

---

# 2. Endpoint Domain

The comparative statics of R25 make $d^\star$ strictly decrease as the support count increases.

The continuum broadworm's dual endpoint is approximately:

$$
1.043590>1.02.
$$

While for $n=4$:

$$
d_4^\star<1.062.
$$

thus all finite-stage and continuous-interpolation optimizers lie in:

$$
\boxed{
1.02<d^\star<1.08.
}
$$

---

# 3. Physical Apex Cell

Let:

$$
\sigma=\sqrt{d^2-1},
\qquad
\rho=d-\sigma.
$$

Round 24:

$$
\sigma<x<\rho.
$$

Parametrize:

$$
x
=
\sigma
+
u(\rho-\sigma),
\qquad
u\in[0,1].
$$

Thus the full interval-box domain is:

$$
[1.02,1.08]\times[0,1]^3.
$$

---

# 4. Stationarity Pruning

Only the profiled apex is needed:

$$
\mathcal L_x=0.
$$

If an interval box satisfies:

$$
0\notin[\mathcal L_x],
$$

it cannot contain the optimizer branch and is discarded directly.

For the remaining boxes, verify both:

$$
[\mathcal K_L]>0,
\qquad
[\mathcal K_R]>0.
$$

---

# 5. Interval Audit

Divide $\lambda$ into $16$ closed slices.

Within each slice, perform adaptive subdivision in $(d,u,r)$.

Results:

```text
processed boxes             28244
discarded by stationarity    6545
certified leaf boxes         7585
unresolved terminal boxes       0
maximum depth                  17
```

The smallest certified lower bound globally is:

$$
\boxed{
\mathcal K
\ge
0.0006339178281571378
>0.
}
$$

Authority:

```text
COMPUTER-ASSISTED INTERVAL CERTIFICATE
REPLAYABLE
NOT PROOF-ASSISTANT VERIFIED
```

---

# 6. Elasticity Sign

From:

$$
\mathcal K>0
$$

we obtain:

$$
6(-s')>sV_{dd}.
$$

Together with:

$$
d<V/2
$$

and the ratio lemma:

$$
\boxed{
\mathcal E>0.
}
$$

Hence:

$$
\boxed{
b_q<0.
}
$$

---

# 7. Stage-to-Stage

even:

$$
(q,q)
\rightarrow
(q,q+1).
$$

odd:

$$
(q,q+1)
\rightarrow
(q+1,q+1).
$$

The interval certificate covers the entire continuous imbalance:

$$
|q_L-q_R|\le1.
$$

Thus both kinds of step are strict:

$$
\boxed{
b_{n+1}<b_n
}
$$

for all $n\ge4$.

---

# 8. Exact Base

For $n=2$:

$$
b_2=\frac12.
$$

For the three-segment optimum, let:

$$
c_3=\cos\theta_3.
$$

The first-order cubic is:

$$
4c^3+12c^2+3c-4=0.
$$

It is strictly increasing for $c\ge0$, and:

$$
F_3(11/26)
=
-\frac{615}{2197}<0.
$$

Hence:

$$
c_3>\frac{11}{26}.
$$

The source endpoint formula is:

$$
b_3
=
\frac{1+2c_3}{3+2c_3}
>
\frac{12}{25}
=
0.48.
$$

And $b_2>b_3$.

---

# 9. Exact $b_4<0.47$

For $n=4$, the unique dual endpoint $d_4$ satisfies:

$$
35d^6+249d^4-680d^2+400=0.
$$

This polynomial is increasing on the relevant branch.

$$
Q_4(53/50)<0,
$$

$$
Q_4(531/500)>0.
$$

Hence:

$$
d_4<531/500.
$$

Moreover, the path must pass through the top point $(x,1)$:

$$
L_4
\ge
\sqrt{1+x^2}
+
\sqrt{1+(d-x)^2}
\ge
2\sqrt{1+d^2/4}.
$$

Therefore:

$$
b_4
<
\frac{d_4}{\sqrt{4+d_4^2}}
<
0.47.
$$

After squaring, the margin in the final inequality is:

$$
\boxed{
\frac{12241849}{250000}>0.
}
$$

Thus:

$$
\boxed{
b_2>b_3>0.48>0.47>b_4.
}
$$

---

# 10. Conjecture (6)

Base:

$$
b_2>b_3>b_4.
$$

Kernel certificate:

$$
b_{n+1}<b_n
\qquad
n\ge4.
$$

Round 22:

$$
b_n\rightarrow b_0.
$$

Hence:

$$
\boxed{
b_n
\text{ strictly decreases to }
b_0.
}
$$

This is precisely the 2007 conjecture (6).

---

# 11. Authority

The most precise current label is:

```text
CONJECTURE (6)

RESEARCH-PROOF-LEVEL
  PROVISIONALLY CLOSED

COMPUTER-ASSISTED COMPONENT
  K>0 interval certificate

FORMAL PROOF ASSISTANT
  OPEN

INDEPENDENT REPLAY / REVIEW
  OPEN
```

No peer-reviewed resolution or literature priority is claimed.

---

# 12. Next Steps

The major sequence conjectures in the 2007 list are now substantially consolidated.

The most reasonable next round is not immediately expanding the proof machinery again, but:

$$
\boxed{
\textbf{Conjecture Ledger / Authority Audit}
}
$$

reclassifying conjectures (1)–(7) one by one:

```text
source theorem
research proof
computer-assisted
formal open
remaining exact debt
```

This avoids conflating conditional claims with independent closures after more than twenty rounds.

---

# Conclusion

Round 26 closes the last issue left by Round 25:

$$
\mathcal E>0.
$$

Analytic reduction:

$$
\mathcal E>0
\Leftarrow
\mathcal K>0,
$$

The full continuous-interpolation interval certificate gives:

$$
\boxed{
\mathcal K>0.
}
$$

Therefore:

$$
\boxed{
b_{n+1}<b_n.
}
$$

Together with exact base cases and R22 convergence:

$$
\boxed{
b_n
\text{ strictly decreases to the broadworm endpoint distance}.
}
$$

===END===