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

# WSFC Round 25
## Endpoint-Distance Strictness Reframed: Dual Comparative Statics and the Unit-Length Elasticity Gap

**English Title:** *Endpoint-Distance Strictness Reframed: Dual Comparative Statics and the Unit-Length Elasticity Gap*  
**Document ID:** `WSFC-R25-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-R24-v0.1`  
**Status:** `DUAL ENDPOINT STRICTNESS CLOSED / UNIT-ENDPOINT STRICTNESS SHARPLY REDUCED`  
**Canonical Source:** UTF-8 Markdown

---

# Abstract

The 2007 conjecture (6) is:

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

Round 22 already addressed:

$$
b_n\rightarrow b_0.
$$

Round 25 first corrects a normalization-semantic bug.

Our reduced functional uses breadth-one normalization:

$$
L_n^\star
=
\text{minimum breadth-one length},
$$

$$
d_n^\star
=
\text{breadth-one endpoint distance},
$$

$$
t_n
=
1/L_n^\star.
$$

The 2007 quantity $b_n$, however, is the endpoint distance of a unit-length arc.

Therefore:

$$
\boxed{
b_n
=
\frac{d_n^\star}{L_n^\star}
=
t_nd_n^\star.
}
$$

For example:

$$
d_4^\star\approx1.061526,
$$

$$
t_4\approx0.438208,
$$

Thus:

$$
b_4\approx0.465169,
$$

corresponding to $0.4652$ in the source table.

Thus Round 24's final direct identification of $b_n$ with $d_n^\star$ was a shorthand error; the uniqueness theorem itself is unaffected.

---

# 1. Reduced Profile

Fix the support counts:

$$
q_L,q_R.
$$

Define:

$$
\mathcal L(d,x)
=
2\sqrt{d^2-1}
+d
+
g(q_L,s_L)
+
g(q_R,s_R),
$$

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

$$
s_L
=
C(d)-2\arctan(d-x),
$$

$$
s_R
=
C(d)-2\arctan x,
$$

$$
C(d)=\arcsin\frac1d.
$$

Round 24 gives, for fixed $d$, a unique $x^\star(d)$, and the profiled objective:

$$
V(d;q_L,q_R)
=
\mathcal L(d,x^\star(d))
$$

is strictly convex in $d$.

---

# 2. Support Spans Strictly Decrease with $d$

Let:

$$
c=C'(d)<0.
$$

Left responsibility:

$$
a_L=d-x,
$$

$$
p=A'(a_L)<0,
\qquad
p_2=A''(a_L)>0.
$$

Right:

$$
a_R=x,
$$

$$
r=A'(a_R)<0,
\qquad
r_2=A''(a_R)>0,
$$

where:

$$
A(a)=-2\arctan a.
$$

Also let:

$$
\phi_i=g_s>0,
\qquad
\psi_i=g_{ss}>0.
$$

We have:

$$
\mathcal L_{xx}
=
\phi_Lp_2+\psi_Lp^2+\phi_Rr_2+\psi_Rr^2>0.
$$

And:

$$
\mathcal L_{dx}
=
-\phi_Lp_2
-\psi_Lp(c+p)
+
\psi_Rcr.
$$

For the right span:

$$
s_R'(d)
=
\frac{
c\mathcal L_{xx}
-r\mathcal L_{dx}
}{
\mathcal L_{xx}
}.
$$

The numerator has the exact decomposition:

$$
\boxed{
\psi_Lp[c(p+r)+rp]
+
\phi_Lp_2(c+r)
+
c\phi_Rr_2.
}
$$

Because:

$$
c,p,r<0
$$

while:

$$
c(p+r)+rp>0,
$$

every term is strictly negative.

Therefore:

$$
\boxed{
s_R'(d)<0.
}
$$

Similarly on the left:

$$
\boxed{
s_L'(d)<0.
}
$$

---

# 3. Support-Count Cross Partial

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

Hence:

$$
\boxed{
g_{sq}
=
-
\frac{s}{q^2}
\sec^2\frac{s}{2q}
\tan\frac{s}{2q}
<0.
}
$$

The profile envelope gives:

$$
V_d=\mathcal L_d.
$$

Therefore:

$$
\boxed{
V_{dq_R}
=
g_{sq,R}s_R'(d)
>0.
}
$$

Similarly on the left:

$$
\boxed{
V_{dq_L}>0.
}
$$

---

# 4. Breadth-One Endpoint Strictness

The unique endpoint minimizer satisfies:

$$
V_d(d^\star,q)=0.
$$

Implicit differentiation gives:

$$
\boxed{
\frac{\partial d^\star}{\partial q}
=
-
\frac{
V_{dq}
}{
V_{dd}
}
<0.
}
$$

Thus increasing the support count on either side strictly decreases the breadth-one endpoint distance.

Round 24's allocation is:

even:

$$
(q,q),
$$

odd:

$$
(q,q+1)
$$

up to reflection.

Hence:

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

---

# 5. Why Conjecture (6) Cannot Yet Be Announced Directly

The source endpoint quantity is:

$$
b_n
=
t_nd_n^\star.
$$

But:

$$
t_{n+1}>t_n,
$$

while:

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

The two factors change in opposite directions.

Thus dual strictness alone is insufficient.

---

# 6. Exact Unit-Endpoint Elasticity Kernel

Continuously extend the support count on one side to $q$.

$$
b(q)
=
\frac{
d^\star(q)
}{
V(d^\star(q),q)
}.
$$

From:

$$
V_d=0,
$$

$$
d_q^\star
=
-
\frac{V_{dq}}{V_{dd}},
$$

and the envelope:

$$
V_q=g_q,
$$

we obtain:

$$
\boxed{
b_q
=
-
\frac{
V V_{dq}
+
dV_qV_{dd}
}{
V^2V_{dd}
}.
}
$$

Thus define:

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

Then:

$$
\boxed{
b_q<0
\iff
\mathcal E>0.
}
$$

This is the actual remaining scalar gate for strictness in conjecture (6).

---

# 7. Ratio Lemma

Let:

$$
z=\frac{s}{2q}.
$$

We have:

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

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

This round obtains:

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

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:

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

Therefore:

$$
R'(z)\ge0,
$$

with:

$$
R(0)=0.
$$

The ratio lemma holds.

---

# 8. Sufficient Kernel

Because:

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

the ratio lemma gives:

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

For genuine unit arcs with $n>2$:

$$
b_n<1/2.
$$

a sufficient condition is:

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

Define:

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

If one can prove:

$$
\mathcal K>0,
$$

then the strictness part of Conjecture (6) closes.

---

# 9. Numerical Red-Team

This round recomputes:

$$
n=2,\ldots,60.
$$

At every stage:

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

and:

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

both hold.

In particular:

$$
b_4\approx0.465169,
$$

$$
b_5\approx0.463026,
$$

$$
b_6\approx0.460562,
$$

reproducing the 2007 table.

This audit is not a general proof.

---

# 10. Current Status

```text
Conjecture (6)

convergence:
  CLOSED in R22

dual endpoint d_n* strictness:
  CLOSED at research-proof level
  conditional on G0/R24

unit endpoint b_n strictness:
  OPEN
  reduced to elasticity kernel E>0

numerical audit:
  strict for n=2..60
```

Thus this round does not promote Conjecture (6) to solved.

---

# 11. Correction to Round 24

Round 24's uniqueness theorem remains valid.

Only one shorthand statement is corrected:

```text
OLD:
  b_n = d_n*

CORRECT:
  b_n = t_n d_n*
```

Thus uniqueness already makes $b_n$ well-defined; strictness merely requires additional normalization analysis.

---

# 12. Next Steps

Round 26 should directly attack:

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

or the sufficient kernel:

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

There is now no need to revisit:

- optimizer uniqueness;
- allocation;
- support saturation;
- span monotonicity;
- dual-endpoint comparative statics.

What actually remains is a scalar curvature-versus-contraction inequality.

---

# Conclusion

Round 25 has two genuine results.

First, it identifies the semantic correction:

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

Second, it actually proves:

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

Meanwhile, strictness in the source conjecture (6) is compressed to:

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

Endpoint-distance strictness is therefore no longer a vague geometric problem; only the final normalization-elasticity gate remains.

===END===