# AMRAL × Lebesgue Universal Covering — Round 09
## Adaptive Atlas Refinement and Certificate-Cost Minimization

**Document ID:** AMRAL-LUC-FC-R09  
**Version:** v0.1  
**Date:** 2026-09-18  
**Research status:** Round 09 / Adaptive certificate optimization  
**Research mode:** Human-Directed + Semi-Autonomous AI Mathematical Research  
**Research initiation and methodology source:** Neo.K  
**AI collaborating researcher and primary executor:** Aletheia / ChatGPT, GPT-5.6 Sol  
**Parent methodology:** Relational Constraint–Handoff Methodology (RCHM)  
**Prerequisite documents:** AMRAL-LUC-FC-R00 v0.2; R01–R08 v0.1

---

# 0. Round summary and verdict

Round 08 already established the complete nested certificate grammar:

$$
\text{base atlas}
\to
\text{cell-local witness lift}
\to
\text{independent verifier}.
$$

Round 09 does not add any new proof domain. Instead, for certificate cost, it establishes a set of **adaptive strategies that do not affect soundness in any way**.

This round completes the following main results.

---

## Conclusion A: The base hull cell has an explicit Hausdorff radius

For a finite placed-family configuration cell:

$$
C,
$$

if each placed body's Hausdorff displacement relative to the cell-center placement is respectively at most:

$$
\delta_i(C),
$$

then the entire convex hull satisfies:

$$
\boxed{
d_H
\left(
H(q),
H(q_C)
\right)
\le
\tau(C)
:=
\max_i\delta_i(C)
}
$$

for all:

$$
q\in C.
$$

This converts the size of the five-dimensional placement box directly into geometric hull uncertainty.

For:

$$
D+B_3+B_5,
$$

we have:

$$
\delta_3
=
\sqrt{
h_{x3}^2+h_{y3}^2
},
$$

$$
\delta_5
=
\sqrt{
h_{x5}^2+h_{y5}^2
}
+
2R_5\sin\frac{h_{\phi5}}2,
$$

so:

$$
\boxed{
\tau(C)=\max\{\delta_3,\delta_5\}.
}
$$

---

## Conclusion B: Round 07's area-transfer error can be tightened

Round 07 used the safe but loose perimeter bound:

$$
\operatorname{Per}(U)\le4\pi T.
$$

This round uses the fact that every relevant hull contains:

$$
B_r(0),
\qquad
r=\frac12,
$$

to obtain:

$$
\operatorname{Area}(U)
\ge
\frac r2
\operatorname{Per}(U).
$$

Hence, whenever:

$$
\operatorname{Area}(U)<T,
$$

we have:

$$
\boxed{
\operatorname{Per}(U)
<
\frac{2T}{r}.
}
$$

At:

$$
r=\frac12
$$

this gives:

$$
\boxed{
\operatorname{Per}(U)<4T.
}
$$

So the representative-transfer error can be rewritten as:

$$
\boxed{
\Omega_{T,r}(\tau)
=
\frac{2T}{r}\tau
+
\pi\tau^2.
}
$$

For this problem:

$$
\boxed{
\Omega_T(\tau)
=
4T\tau+\pi\tau^2.
}
$$

This is:

`R07-STRENGTHENING-001`

not a correctness correction.

---

## Conclusion C: The representative-area cell closure theorem

Let:

$$
H_C
$$

be the cell-center full hull representative.

If there is a rigorous lower bound:

$$
\underline A(H_C)
$$

with:

$$
\boxed{
\underline A(H_C)
\ge
T+\Omega_T(\tau(C)),
}
$$

then the entire configuration cell satisfies:

$$
\boxed{
\operatorname{Area}(H(q))\ge T
\qquad
\forall q\in C.
}
$$

Hence many boxes can skip the erosion core entirely and prune directly using:

$$
\boxed{
\text{center representative}
+
\text{Hausdorff uncertainty}.
}
$$

---

## Conclusion D: The margin-to-resolution compiler

If the representative rigorous margin:

$$
m
=
\underline A(H_C)-T
>
0,
$$

then cell closure only requires:

$$
\Omega_T(\tau)\le m.
$$

For:

$$
r=\frac12,
$$

the maximum admissible radius can be solved explicitly:

$$
\boxed{
\tau_\star(m,T)
=
\frac{
\sqrt{
4T^2+\pi m
}
-
2T
}{
\pi
}.
}
$$

Hence every positive representative margin translates directly into:

> how fine the base box must be refined for closure to hold.

---

## Conclusion E: Establishing the Hausdorff-radius optimal one-step split

For each bisectable coordinate:

$$
a
$$

of cell:

$$
C,
$$

let:

$$
\tau_a
$$

be the common Hausdorff-radius upper bound of the two children obtained by bisecting that coordinate.

Choose:

$$
\boxed{
a^\star
=
\arg\min_a\tau_a.
}
$$

Then, among all single-coordinate bisections, this split is one-step optimal for this round's certified hull-radius bound.

For:

$$
D+B_3+B_5
$$

rooted at:

$$
T=0.8350,
$$

this round computes:

$$
\tau_{\rm root}
\approx
0.6046803712.
$$

After one split on each axis:

- $x_3$: does not change the current maximum radius;
- $y_3$: does not change the current maximum radius;
- $\phi_5$:

  $$
  \tau\approx0.4442456056;
  $$

- $x_5$:

  $$
  \tau\approx0.5460899294;
  $$

- $y_5$:

  $$
  \tau\approx0.5460899294.
  $$

So the first cut:

$$
\boxed{
\phi_5
}
$$

is the exact radius-bound-optimal split.

This agrees with the Mishra code's use of:

$$
R_5h_{\phi5}
$$

as the effective split weight, but this round gives the exact nonlinear radius criterion.

---

## Conclusion F: The official split weight is a first-order approximation of the new criterion

For small:

$$
h_\phi,
$$

we have:

$$
2R\sin\frac{h_\phi}{2}
=
Rh_\phi
+
O(h_\phi^3).
$$

So the official effective weight:

$$
Rh_\phi
$$

is exactly the first-order approximation of the motion-radius rotational contribution.

Round 09 upgrades it to:

$$
\boxed{
\text{the exact split-gain scheduler}.
}
$$

---

## Conclusion G: Establishing a three-tier low-cost prune cascade

For each cell, try, in order of increasing cost:

### Tier 0 — a-priori geometry

For example, the farthest-corner area bound.

### Tier 1 — representative-transfer prune

Use:

$$
\underline A(H_C)
-
\Omega_T(\tau(C)).
$$

If:

$$
\ge T,
$$

close directly.

### Tier 2 — common-core erosion

Use Round 08's:

$$
C_i
\subseteq
K_i(q)
$$

common core, and compute:

$$
\operatorname{Area}
\operatorname{conv}
(D,C_3,C_5,\ldots).
$$

### Tier 3 — split

Only bisect if none of the first three tiers can close the cell.

This does not change correctness — it only reduces how often the expensive core / lift evaluation is used.

---

## Conclusion H: Establishing cross-cell common-core dominance reuse

Let:

$$
G(C)
$$

be the base cell's verified common-core hull.

If:

$$
\boxed{
G(C_a)
\supseteq
G(C_b),
}
$$

then for any identical witness common core:

$$
W,
$$

we have:

$$
\operatorname{conv}
(
G(C_a)\cup W
)
\supseteq
\operatorname{conv}
(
G(C_b)\cup W
).
$$

So:

> if some entire witness lift tree can close $C_b$, the same tree, under the same witness placement partition, also closes $C_a$.

This only requires the additional verification:

$$
G(C_b)\subseteq G(C_a).
$$

This provides a sound condition for certificate reuse / hash-consing.

---

## Conclusion I: Witness dominance can descend to a single-lift functional

If:

$$
K\preceq L
$$

denotes that some congruent copy of $K$ is contained in $L$, then for any base hull:

$$
H,
$$

we have:

$$
\boxed{
J_L(H)\ge J_K(H).
}
$$

So a stronger witness:

$$
L
$$

is never weaker, in mathematical closing power, than:

$$
K.
$$

Hence the witness pool can first undergo geometric dominance pruning.

If:

- $L$ dominates $K$;
- and $L$'s certificate cost is no higher than $K$'s;

then:

$$
K
$$

can be removed from the priority pool.

---

## Conclusion J: A positive margin gives a finite depth upper bound

If a rigorous margin:

$$
m>0
$$

already holds for some representative / witness,

then as soon as refinement brings:

$$
\tau(C)
\le
\tau_\star(m,T),
$$

representative closure is guaranteed to succeed.

If every round's full cycle bisects every relevant parameter half-width once, then there exists a constant:

$$
S_0
$$

such that:

$$
\tau_r
\le
2^{-r}S_0.
$$

Hence sufficiently many full refinement cycles:

$$
\boxed{
r
\ge
\left\lceil
\log_2
\frac{S_0}{\tau_\star(m,T)}
\right\rceil.
}
$$

This is an extremely conservative worst-case finite-depth bound.

In practice, adaptive splitting plus early pruning will do far better than this.

---

This round's overall verdict:

$$
\boxed{
\text{ADAPTIVE CERTIFICATE-COST LAYER: CLOSED}
}
$$

$$
\boxed{
\text{SOUNDNESS-PRESERVING REUSE RULES: CLOSED}
}
$$

$$
\boxed{
\text{HEAVY }0.8350\text{ CERTIFICATE: STILL COMPUTE-DEFERRED}
}
$$

---

# 1. Convex hull is Hausdorff non-expansive

## Theorem 1.1

For compact sets:

$$
X,Y,
$$

if:

$$
d_H(X,Y)\le\delta,
$$

then:

$$
\boxed{
d_H
(
\operatorname{conv}X,
\operatorname{conv}Y
)
\le\delta.
}
$$

### Proof

$$
X\subseteq Y+\delta B.
$$

Hence:

$$
\operatorname{conv}X
\subseteq
\operatorname{conv}(Y+\delta B).
$$

Since:

$$
Y+\delta B
$$

is already a subset of:

$$
\operatorname{conv}Y+\delta B,
$$

and:

$$
\operatorname{conv}Y+\delta B
$$

is convex, we have:

$$
\operatorname{conv}(Y+\delta B)
\subseteq
\operatorname{conv}Y+\delta B.
$$

The reverse direction follows symmetrically.

Q.E.D.

---

# 2. Family cell Hausdorff-radius theorem

For placed bodies:

$$
K_i(q)
$$

and cell-center:

$$
K_i(q_C),
$$

if:

$$
d_H
(
K_i(q),
K_i(q_C)
)
\le
\delta_i
$$

for all:

$$
q\in C,
$$

then the union:

$$
X(q)
=
\bigcup_iK_i(q)
$$

satisfies:

$$
d_H
(
X(q),
X(q_C)
)
\le
\max_i\delta_i.
$$

Then, by convex-hull non-expansiveness:

$$
\boxed{
d_H
(
H(q),
H(q_C)
)
\le
\max_i\delta_i.
}
$$

---

# 3. F3 explicit cell radius

Base:

$$
D+B_3+B_5.
$$

Disk fixed:

$$
\delta_D=0.
$$

$B_3$ orientation fixed:

$$
\boxed{
\delta_3
=
\sqrt{
h_{x3}^2+h_{y3}^2
}.
}
$$

$B_5$:

$$
\boxed{
\delta_5
=
\sqrt{
h_{x5}^2+h_{y5}^2
}
+
2R_5
\sin
\frac{h_{\phi5}}2.
}
$$

So:

$$
\boxed{
\tau(C)
=
\max\{
\delta_3,\delta_5
\}.
}
$$

---

# 4. Full base + lift joint radius

If a witness:

$$
K
$$

placement box is further added, with motion radius:

$$
\delta_K
$$

computed by Round 08,

then the full hull relative to the joint box-center satisfies:

$$
\boxed{
d_H
(
H_{\rm full}(q,g),
H_{\rm full}(q_C,g_C)
)
\le
s
:=
\max
\{
\tau(C),\delta_K
\}.
}
$$

Hence the representative-area prune can be applied directly to the joint base/lift box.

---

# 5. Inradius-perimeter inequality

If convex:

$$
U
$$

contains:

$$
B_r(0),
$$

then:

$$
h_U(u)\ge r.
$$

Minkowski's area formula:

$$
\operatorname{Area}(U)
=
\frac12
\int_{S^1}
h_U\,dS_U.
$$

And:

$$
S_U(S^1)
=
\operatorname{Per}(U).
$$

Hence:

$$
\boxed{
\operatorname{Area}(U)
\ge
\frac r2
\operatorname{Per}(U).
}
$$

So:

$$
\boxed{
\operatorname{Per}(U)
\le
\frac{
2\operatorname{Area}(U)
}{
r
}.
}
$$

At:

$$
r=\frac12,
$$

this gives:

$$
\boxed{
\operatorname{Per}(U)
\le
4\operatorname{Area}(U).
}
$$

---

# 6. Sharpened representative-transfer theorem

## Theorem 6.1

Let convex:

$$
A,A_0
$$

both contain:

$$
B_r(0),
$$

with:

$$
d_H(A,A_0)\le\tau.
$$

If:

$$
\boxed{
\operatorname{Area}(A_0)
\ge
T
+
\frac{2T}{r}\tau
+
\pi\tau^2,
}
$$

then:

$$
\boxed{
\operatorname{Area}(A)\ge T.
}
$$

### Proof

Suppose, for contradiction, that:

$$
\operatorname{Area}(A)<T.
$$

Since:

$$
A_0
\subseteq
A+\tau B,
$$

the Steiner formula gives:

$$
\operatorname{Area}(A_0)
\le
\operatorname{Area}(A)
+
\tau\operatorname{Per}(A)
+
\pi\tau^2.
$$

By the inradius-perimeter inequality:

$$
\operatorname{Per}(A)
<
\frac{2T}{r}.
$$

So:

$$
\operatorname{Area}(A_0)
<
T
+
\frac{2T}{r}\tau
+
\pi\tau^2,
$$

a contradiction.

Q.E.D.

---

# 7. Round 07 strengthening

Round 07 used:

$$
4\pi T\tau+\pi\tau^2.
$$

This comes from:

$$
U\subset B_{2T}
$$

and:

$$
\operatorname{Per}(U)\le4\pi T.
$$

This round uses the inradius to directly improve this to:

$$
\boxed{
4T\tau+\pi\tau^2.
}
$$

Both are valid.

This round only improves efficiency — it does not change Round 07's theorem status.

---

# 8. Margin-to-resolution theorem

Let the representative rigorous area lower bound be:

$$
\underline A_0.
$$

Define:

$$
m
=
\underline A_0-T.
$$

If:

$$
m>0,
$$

closing the cell only requires:

$$
\frac{2T}{r}\tau+\pi\tau^2
\le
m.
$$

The positive root:

$$
\boxed{
\tau_\star(m,T,r)
=
\frac{
\sqrt{
T^2/r^2+\pi m
}
-
T/r
}{
\pi
}.
}
$$

At:

$$
r=\frac12,
$$

$$
\boxed{
\tau_\star(m,T)
=
\frac{
\sqrt{
4T^2+\pi m
}
-
2T
}{
\pi
}.
}
$$

---

# 9. Illustrative resolution numbers

The following is only cost planning, not a lower bound proof.

Take:

$$
T=0.835.
$$

If the representative margin:

$$
m=0.001494901
$$

equals the search headroom of the official full four-body search ceiling relative to the milestone,

then:

$$
\boxed{
\tau_\star
\approx
4.47387\times10^{-4}.
}
$$

The old Round 07 bound would require approximately:

$$
1.42462\times10^{-4}.
$$

So the sharpened bound relaxes the required Hausdorff resolution by about:

$$
3.14
$$

times.

Note:

$$
0.001494901
$$

is `SEARCH-ONLY` headroom, not a rigorous representative margin.

This number is used only to quantify the certificate scale.

---

# 10. Exact one-step split gain

Cell:

$$
C
$$

half-width:

$$
\mathbf h.
$$

For axis:

$$
a,
$$

after bisection the children's half-width becomes only:

$$
h_a
\mapsto
\frac{h_a}{2}.
$$

Define:

$$
\tau_a
$$

as the child's family hull-radius upper bound.

Since both children have the same half-width:

$$
\tau_a
$$

does not depend on the left/right child.

Define the split gain:

$$
\boxed{
G_a
=
\tau(C)-\tau_a.
}
$$

Choose:

$$
\boxed{
a^\star
=
\arg\max_aG_a
=
\arg\min_a\tau_a.
}
$$

This is optimal for this round's radius bound among all single-coordinate bisections.

---

# 11. Explicit F3 split formulas

Let:

$$
d_3
=
\sqrt{
h_{x3}^2+h_{y3}^2
},
$$

$$
d_5
=
\sqrt{
h_{x5}^2+h_{y5}^2
}
+
2R_5
\sin\frac{h_\phi}{2}.
$$

Originally:

$$
\tau=\max(d_3,d_5).
$$

If split:

$$
x_3,
$$

$$
d_3'
=
\sqrt{
(h_{x3}/2)^2+h_{y3}^2
}.
$$

If split:

$$
y_3,
$$

symmetric.

If split:

$$
x_5,
$$

$$
d_5'
=
\sqrt{
(h_{x5}/2)^2+h_{y5}^2
}
+
2R_5\sin\frac{h_\phi}{2}.
$$

If split:

$$
y_5,
$$

symmetric.

If split:

$$
\phi_5,
$$

$$
\boxed{
d_5'
=
\sqrt{
h_{x5}^2+h_{y5}^2
}
+
2R_5
\sin\frac{h_\phi}{4}.
}
$$

Each candidate axis directly recomputes:

$$
\tau_a.
$$

---

# 12. F3 root split sanity

At:

$$
T=0.835,
$$

the Round 07 root:

$$
h_{x3}=h_{y3}=0.194856180909,
$$

$$
h_{x5}=h_{y5}=0.197820670401,
$$

$$
h_{\phi5}=\frac{\pi}{5},
$$

and:

$$
R_5=0.525731112119.
$$

This gives:

$$
d_3
\approx
0.2755682538,
$$

$$
d_5
\approx
0.6046803712.
$$

So:

$$
\tau_{\rm root}
\approx
0.6046803712.
$$

Per split:

$$
\tau_{x3}
=
\tau_{y3}
\approx
0.6046803712,
$$

$$
\tau_{\phi5}
\approx
0.4442456056,
$$

$$
\tau_{x5}
=
\tau_{y5}
\approx
0.5460899294.
$$

So the first cut:

$$
\boxed{
\phi_5.
}
$$

---

# 13. Witness-box exact split scheduler

For a witness:

$$
K\subseteq B_R
$$

placement cell:

$$
(h_\phi,h_x,h_y),
$$

motion radius:

$$
\delta
=
\sqrt{h_x^2+h_y^2}
+
2R\sin\frac{h_\phi}{2}.
$$

After each of three kinds of split:

## $x$

$$
\delta_x
=
\sqrt{
(h_x/2)^2+h_y^2
}
+
2R\sin\frac{h_\phi}{2}.
$$

## $y$

Symmetric.

## $\phi$

$$
\boxed{
\delta_\phi
=
\sqrt{
h_x^2+h_y^2
}
+
2R\sin\frac{h_\phi}{4}.
}
$$

Choose the smallest:

$$
\delta_a.
$$

This replaces the naive surrogate:

$$
\max\{h_x,h_y,Rh_\phi\}.
$$

---

# 14. Why official weights were sensible

For small:

$$
h_\phi,
$$

Taylor expansion:

$$
2R\sin\frac{h_\phi}{2}
=
Rh_\phi
-
\frac{
R h_\phi^3
}{
24
}
+
O(h_\phi^5).
$$

So in the official code:

$$
R h_\phi
$$

is the correct first-order term for rotation uncertainty.

Round 09's split scheduler is its nonlinear exact-bound generalization.

---

# 15. Representative prune for full base + witness box

Joint box-center full hull:

$$
H_0.
$$

Joint uncertainty:

$$
s
=
\max\{
\tau_{\rm base},
\delta_K
\}.
$$

If the verifier can give:

$$
\underline A(H_0)
$$

with:

$$
\boxed{
\underline A(H_0)
-
4Ts
-
\pi s^2
\ge
T,
}
$$

then the entire joint box can be pruned.

This prune:

- does not require erosion;
- does not require a witness core;
- only needs the representative inner area lower bound and the motion radius.

Hence it should be placed before the common-core step.

---

# 16. Adaptive prune cascade

Recommended order:

```text
P0  a-priori single-point / translation bound

P1  representative-area transfer
        if A_center^- - Omega_T(s) >= T:
            prune

P2  common-core lower hull
        if A_core^- >= T:
            prune

P3  stronger witness-specific lower bound
        optional

P4  split by exact radius-gain scheduler
```

The search emitter can add more heuristics.

The independent verifier only needs to verify the actual proof mode used by the certificate leaf.

---

# 17. Proof-mode tags

A certificate leaf can explicitly record:

- `APR`
- `REP`
- `CORE`
- `SUPPORT-CORE`
- `OTHER-VERIFIED`

The verifier uses a fixed, independent rule for each mode.

This lets later statistics on

> which proof mode actually eliminates the most nodes

be read directly off the certificate.

---

# 18. Common-core dominance preorder

For base cell:

$$
C,
$$

let:

$$
G(C)
$$

be the verified base common-core hull.

Define:

$$
\boxed{
C_a\succeq_{\rm core}C_b
}
$$

if:

$$
G(C_a)\supseteq G(C_b).
$$

---

# 19. Core-dominance reuse theorem

## Theorem 19.1

If:

$$
G(C_a)\supseteq G(C_b),
$$

then for any witness common core:

$$
W,
$$

we have:

$$
\boxed{
\operatorname{Area}
\operatorname{conv}
(
G(C_a)\cup W
)
\ge
\operatorname{Area}
\operatorname{conv}
(
G(C_b)\cup W
).
}
$$

Hence:

> any lift certificate that closes $C_b$ using the same witness root / split tree and the same witness-core construction can be reused by $C_a$.

Provided that:

- the base-core containment has already been proved independently by the verifier;
- the witness tree and arithmetic policy are exactly identical.

---

# 20. Core inclusion check

Because the base common-core hull is convex:

$$
G(C_b)\subseteq G(C_a)
$$

is equivalent to:

$$
\boxed{
h_{G(C_b)}(\theta)
\le
h_{G(C_a)}(\theta)
\quad
\forall\theta.
}
$$

The Mishra geometry kernel's body representation is piecewise:

$$
c_x\cos\theta+c_y\sin\theta+r.
$$

Hence the support difference over merged cells is still a single sinusoid plus a constant.

Containment can be checked in closed form, without dense sampling.

---

# 21. Cross-cell lift-tree cache

Recommended cache key:

```text
witness_id
witness_root_domain_hash
witness_split_rule_hash
base_core_geometry_hash
threshold
arithmetic_policy_hash
```

If a new cell:

$$
C_a
$$

core-dominates the cache origin:

$$
C_b,
$$

the entire witness tree need not be rerun.

It is only necessary to:

1. verify:

   $$
   G(C_a)\supseteq G(C_b);
   $$

2. verify the referenced tree hash;
3. reuse.

---

# 22. Single-lift witness dominance

Recall:

$$
K\preceq L
$$

if:

$$
g_0K\subseteq L
$$

for some fixed rigid motion:

$$
g_0.
$$

## Theorem 22.1

For any base hull:

$$
H,
$$

$$
\boxed{
J_L(H)\ge J_K(H).
}
$$

### Proof

Take any placement:

$$
gL.
$$

It contains:

$$
gg_0K.
$$

So:

$$
\operatorname{Area}
\operatorname{conv}(H\cup gL)
\ge
\operatorname{Area}
\operatorname{conv}(H\cup gg_0K).
$$

The right-hand side is some legitimate placement of:

$$
K
$$

placement.

So:

$$
\operatorname{Area}
\operatorname{conv}(H\cup gL)
\ge
J_K(H).
$$

Taking the minimum over:

$$
g
$$

we get:

$$
J_L(H)\ge J_K(H).
$$

Q.E.D.

---

# 23. Witness pool Pareto pruning

For witness:

$$
K
$$

record:

- geometric strength;
- estimated lift proof cost;
- symmetry period;
- core-provider cost.

If:

$$
K\preceq L
$$

and:

$$
\operatorname{Cost}(L)
\le
\operatorname{Cost}(K),
$$

then:

$$
K
$$

is dominated by:

$$
L
$$

on both axes — closure strength and proof cost — and can be removed from the priority pool.

If:

$$
L
$$

is stronger but more expensive, both stay on the Pareto frontier.

---

# 24. Weighted witness coverage

Given a certified incidence:

$$
C_{jK}\in\{0,1\},
$$

we can define a witness fixed cost:

$$
c_K.
$$

Solve:

$$
\min
\sum_K
c_Kx_K
$$

subject to:

$$
\sum_K
C_{jK}x_K
\ge1
\qquad
\forall j.
$$

This is a weighted set cover.

Correctness only requires coverage.

Optimization only determines certificate size.

A greedy method using:

$$
\frac{
\text{newly covered cells}
}{
\text{cost}
}
$$

can be used as an approximation.

---

# 25. Margin-to-depth upper bound

Let the cell's initial linear-motion surrogate be:

$$
S_0.
$$

For the family:

$$
S_0
=
\max_i
\left[
\sqrt{
h_{xi}^2+h_{yi}^2
}
+
R_i h_{\phi i}
\right].
$$

If every active parameter is bisected once within a full cycle, then after the cycle:

$$
\boxed{
\tau_r
\le
2^{-r}S_0.
}
$$

because the translation-norm half-width is entirely halved, while:

$$
2R\sin
\frac{
h_\phi/2^r
}{
2
}
\le
R\frac{h_\phi}{2^r}.
$$

If the representative rigorous margin:

$$
m>0,
$$

closure requires:

$$
\tau_r\le\tau_\star(m,T).
$$

So it suffices that:

$$
\boxed{
r
\ge
\left\lceil
\log_2
\frac{
S_0
}{
\tau_\star(m,T)
}
\right\rceil.
}
$$

This proves that positive-margin regions are guaranteed to close, in representative-transfer mode, at a finite depth.

---

# 26. What survives infinite refinement

Hence, if some nested branch:

- always has a uniform positive lower bound on the representative margin;
- but the search keeps splitting regardless,

this indicates a problem in the scheduler / implementation, not a mathematical necessity for infinite refinement.

The branches that can genuinely keep getting harder must approach:

$$
\boxed{
\text{zero-margin active frontier}.
}
$$

This again echoes Round 05's saturation picture.

---

# 27. Adaptive priority score

Recommended search-only priority:

$$
\boxed{
\operatorname{Score}(C)
=
\frac{
T-L(C)
}{
\max\{\tau(C),\tau_{\min}\}
}
\times
\operatorname{CostForecast}(C)
}
$$

or its reciprocal for priority ordering, where:

- $T-L(C)$: the unproved deficit;
- $\tau(C)$: geometric uncertainty;
- cost forecast: witness history / node count.

This is only a scheduling heuristic.

It must not enter the verifier's premises.

---

# 28. Split-vs-witness decision

For base cell:

$$
C,
$$

if some witness representative already has rigorous margin:

$$
m_K(C)>0,
$$

first compute:

$$
\tau_\star(m_K,T).
$$

If the current:

$$
\tau(C)
$$

is only slightly above:

$$
\tau_\star,
$$

prioritize splitting the base.

If:

$$
\tau(C)
\gg
\tau_\star
$$

and the witness core itself is strong, the conditional lift tree can be run directly.

Hence scheduling can be upgraded from:

> a fixed order

to:

$$
\boxed{
\text{margin-aware split/lift choice}.
}
$$

---

# 29. Numerical sanity at milestone $0.8350$

T:

$$
0.835.
$$

Round 07's official four-body search-only headroom:

$$
m_{\rm search}
=
0.001494901.
$$

New:

$$
\tau_\star
\approx
0.0004473869.
$$

The old Round 07 conservative-transfer requirement was approximately:

$$
0.0001424615.
$$

Ratio:

$$
\approx3.14.
$$

Once again, to stress:

$$
m_{\rm search}
$$

is not a rigorous margin.

This sanity check only illustrates the order-of-magnitude effect of the improved bound on certificate resolution.

---

# 30. COMPUTE-DEFERRED

## C09-1 — Add exact radius scheduler to base atlas emitter

Each node stores:

- $\delta_3$;
- $\delta_5$;
- $\tau$;
- per-axis child $\tau_a$;
- chosen axis.

---

## C09-2 — Add representative prune mode

Each node first computes:

$$
A_{\rm center}^{-}
-
4T\tau
-
\pi\tau^2.
$$

If:

$$
\ge T,
$$

emit a

`REP`

leaf.

---

## C09-3 — Add core-dominance cache

Build a support-hash index.

Attempt:

$$
G(C_a)\supseteq G(C_b).
$$

If it passes, reuse the lift tree.

---

## C09-4 — Proof-mode statistics

Output:

```text
APR leaves
REP leaves
CORE leaves
SUPPORT-CORE leaves
split nodes
reused lift trees
bytes saved
```

---

## C09-5 — Compare schedulers

A/B:

- official first-order weight;
- exact radius-gain scheduler.

Compare:

- retained base cells;
- total nodes;
- max depth;
- wall time;
- certificate bytes.

This is a performance experiment; it does not affect theorem status.

---

# 31. Round 10 assigned topic

## AMRAL-LUC-FC-R10
### Reference Atlas Emitter and Small-Scale Conditional-Lift Dry Run

If the heavy local result has not yet returned, Round 10 will go directly to a scaled-down reference implementation:

1. base atlas emitter;
2. exact radius scheduler;
3. `APR/REP/CORE` proof modes;
4. small-threshold dry run;
5. B7 conditional lift on retained cells;
6. independent replay skeleton;
7. certificate statistics.

The goal is not to pretend to a full proof of:

$$
0.8350,
$$

but to:

$$
\boxed{
\text{actually run Round 08–09's proof grammar through one finite, end-to-end sample}.
}
$$

If the local end has already reported back:

ingest the real:

$$
0.8350
$$

atlas directly.

---

# 32. Reproducibility checklist

## Family-cell Hausdorff radius

`PROVED`

## Sharpened area transfer

`PROVED`

## Margin-to-resolution formula

`PROVED`

## Exact one-step split scheduler

`PROVED FOR THE CERTIFIED RADIUS BOUND`

## Representative-area prune

`PROVED`

## Common-core dominance reuse

`PROVED`

## Single-lift witness dominance

`PROVED`

## Margin-to-depth finite bound

`PROVED`

## Scheduler performance gain

`COMPUTE-DEFERRED`

## $0.8350$ theorem

`NOT YET CERTIFIED`

---

# 33. Shortest handoff conclusion

Round 09 compresses certificate optimization into three core quantities:

$$
\boxed{
\tau(C)
=
\text{cell geometric uncertainty}
}
$$

$$
\boxed{
m(C)
=
\text{rigorous representative margin}
}
$$

and:

$$
\boxed{
\tau_\star(m,T)
=
\text{closure-required resolution}.
}
$$

From this:

- the split axis can be chosen exactly;
- a positive-margin cell has a finite closure depth;
- the representative prune can precede erosion;
- the common-core lift tree can be soundly reused across cells;
- the witness pool can use geometric dominance for Pareto pruning.

Hence the main question of heavy computation is no longer:

> how to brute-force split more boxes?

but has become:

$$
\boxed{
\text{how to concentrate computation only on the zero-margin active frontier}.
}
$$

This is Round 09's core progress.