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Machines / R3

`1RB0RF_1RC0RE_1LD1RA_1LA0LD_1RA---_0RA0LB`

R3Claimed elsewhereBell eats counterbbchallenge ↗Explorer

Upstream result

Claimed elsewhere

The blank-start machine never halts: BECv10.v.TM28 ↗. Recorded commit time: 2026-10-13T13:42:06Z. First public push time is not established.

Absent from the current 663 holdout list. Earlier local observations remain below.

Overview

In one sentence

Gap bounces in (101)^a 00 (101)^b, counter counts; at b=0 the low digits pick eat E or bell B0/B1

Upstream: The blank-start machine never halts: BECv10.v.TM28 ↗. Recorded commit time: 2026-10-13T13:42:06Z. First public push time is not established.
Status: Externally Solvedcrossing · updated 2026-10-11Not blocking · Stuck at the crossing stage. Answer Q1–Q3 whenever; a “no” redirects the work.
idle since Oct 11 08:19 · status.json 2026-10-11
Class: Bell eats counterours: Bell eats counter (sheet) · ours adds: FractalBEC-like 101/00 bouncer plus a 3-cell binary counter with transparent J blocks; the bell shifts the counter one cell · med confidence
On the list: not on BB6_holdouts_663.txt (line 202 of the previous list) · historical 740 line 202
R3eqv ×1sim 1e15

Your calls on file 6

priority call (answered): R3's cleanup is now exact and safety reduces to one per-epoch inequality (2^P <= N, via L*), the same crossing arithmetic R2 is stuck on: fund one joint crossing-lemma stage, start Rocq on the exact rules, or park?
parked: Parked 2026-10-11 (Dennis: go with recommendations): Park R3 with R2 (stop rule).
funded: option: crossing · date: 2026-10-10 · auto: true
2026-10-10 · crossing: R3's cleanup is now exact and safety reduces to one per-epoch inequality (2^P <= N, via L*), the same crossing arithmetic R2 is stuck on: fund one joint crossing-lemma stage, start Rocq on the exact rules, or park? (auto-applied: Claude's recommendation, under Dennis's standing instruction 'do your recommendations and continuously orchestrate' (2026-10-10, while he was away). Override any time.)
2026-10-10 · prep: Should R3 get a proof-prep stage on the cleanup lemma (x/z/w marks left by the bell vanish before the next b=0), done together with R2's size invariant, or be parked? (auto-applied: Claude's recommendation, under Dennis's standing instruction 'do your recommendations and continuously orchestrate' (2026-10-10, while he was at lunch). Override any time.)
decision.json · answered 2026-10-11: park: The crossing stage hit its stop rule (Lemma S proven, the Big-Y lemma open, with 3 instances in 1e7 epochs): park R3 with R2, fund one Big-Y stage, or start Rocq with Big-Y as the one hypothesis? Decisions →

Mechanism

How it works: 3 notes

Hub K(a,b,G) = 0^inf (101)^a 00 (101)^b A> 0 enc(G) 0^inf (enc 0=000, 1=010, J=100). R1: each bounce moves the 00 gap one block right and adds 1 to the counter G (J skipped).
At b=0 the low digits decide: 0 1^r J^k = partial eat E (J's become k-1 fresh blocks); 0 1^r 0 / 1... = bell B0/B1: sweep to the left end, counter shifted one cell (σ).
σ leaves x/z/w marks that the carry cleans within 2^q <= 2^floor(log2 N) increments (19,773/19,773 macro epochs); lockstep from blank holds to 1e9 (all hubs) and 3e10 (b=0 hubs).

Figures for Checkpoint A

Fig 1 · bbchallenge view, first 10,000 steps
Time runs down. Follow the diagonal bands and note where each one starts and ends.

Fig 2 · Hub rows: gap and counter, four epochsQ1
Left: where the cyan gap sits. Right: the counter word. Compare the rows just before and after each cyan bar.

Fig 3 · Close-up: bounces, partial eat, bellQ2
Count the bounces; watch where the right end jumps left and where the head runs to the far left.

Checkpoint A · 3 open

check: does the figure show what I claim? A “no” stops proof work built on a wrong reading. decision: your call.

Q1
check Hub rows: does each block of rows read as one epoch (the cyan gap walks right through the grey blocks while the white/grey digits between the orange J columns count up), with the low digits eaten at each cyan bar? see Fig 2
Q2
check Close-up from K(5,2,0JJ1): do you see two bounces, then the partial eat E (the right end jumps left, the two J blocks become one new block), one more bounce, then the bell B1 running to the far left? see Fig 3
Q3
decision Confidence is med: everything to 3e10 fits R1/E/B0/B1, but the cleanup lemma (x/z/w gone before the next b=0) is only observed. Should we fund a proof-prep stage on it, shared with R2's size invariant, or park R3?

More figures (2)Fig 4–5: Records to 100 million steps · Blaze, 10 million steps

Rules (6/6 validated)exact tape rewrites the proof will formalize, each re-checked by simulation

rule	from ⟶ to	cases	steps	checked
init	0^∞ A> 0^∞ ⟶ K(0, 2, ε) = 0^∞ (101)² A> 0 0^∞	—	12	✓1/1
R1 bounce	K(a, b+1, G) ⟶ K(a+1, b, inc G)	a >= 0, b >= 0, G in {0,1}{0,1,J}*; inc = +1 on the digit groups (LSB first), J skipped, a new top digit appended after the last group; a = 0 is the bell sweep to the left end. Orbit: 189,846/189,846 to 1e9	—	✓1800/1800
R1x bounce on marked words	K(a, b+1, G), G with x/z/w groups ⟶ K(a+1, b, T(G))	T = the counter's own carry transducer, the same for every a, b (x -> z on a carry; z then x-runs -> 0 J... on the next). Orbit: 2,196/2,196 to 1e9	—	✓900/900
E partial eat	K(a, 0, 0 1^r J^k G') ⟶ K(a+2+r, k-1, G')	r >= 0, k >= 1, G' empty or starting with a digit. Orbit: 229/229 to 1e9	—	✓1048/1048
B0 bell (low digit 0)	K(a, 0, 0 1^r 0 G'') ⟶ K(0, a+2+r, σ(0 G''))	r >= 0, also G = 0 1^r with nothing after; σ = shift every 1-cell of enc one cell left. Orbit: 36/36 to 1e9	—	✓1244/1244
B1 bell (low digit 1)	K(a, 0, 1 G') ⟶ K(0, a+1, σ(inc(1 G')))	any standard G'. Orbit: 222/222 to 1e9	—	✓2208/2208

Compressed historythe tape as 0^∞ (101)^a 00 (101)^b A> 0 enc(G) 0^∞ at 40 moments: where the rules are read off

0^∞ (101)^a 00 (101)^b A> 0 enc(G) 0^∞

step	`a`	`b`	`G`
2,858,672	188	0	0JJJJJJJJJJJJJJJJJJJJJ011J1101
2,861,162	210	0	010J0111
2,861,177	0	213	x00JJJ
2,867,625	3	210	1JJJJJ1
2,876,115	7	206	1JJJJJ11
2,892,621	15	198	1JJJJJ111
2,923,719	31	182	1JJJJJ1111
2,978,241	63	150	1JJJJJ11111
3,056,571	127	86	1JJJJJ111111
3,097,347	213	0	1JJJJJ0101011
3,097,412	0	214	xxxxx0JJ0J0JJ
3,104,064	3	211	1JJJJJ1000JJJ
3,112,594	7	207	1JJJJJ1100JJJ
3,129,180	15	199	1JJJJJ1110JJJ
3,160,438	31	183	1JJJJJ1111JJJ
3,215,310	63	151	1JJJJJ1111JJJ1
3,294,340	127	87	1JJJJJ1111JJJ11
3,336,140	214	0	0JJJJJ1101JJJ011
3,336,316	220	0	1111JJJ011
3,336,379	0	221	000xxx0JJJ
3,343,009	3	218	110xxx0JJJ
3,351,719	7	214	111xxx0JJJ
3,368,665	15	206	111zxx0JJJ
3,400,751	31	190	1111JJJ1J0J
3,457,103	63	158	1111JJJ1J1J
3,539,103	127	94	1111JJJ1J1J1
3,585,305	221	0	1011JJJ1J0J11
3,585,320	0	222	0JJwxx0w0x0JJ
3,592,154	3	219	1JJJJJJz0x000J
3,601,080	7	215	1JJJJJJ1JJJJ1J
3,618,436	15	207	1JJJJJJ1JJJJ1J1
3,651,234	31	191	1JJJJJJ1JJJJ1J11
3,709,156	63	159	1JJJJJJ1JJJJ1J111
3,794,286	127	95	1JJJJJJ1JJJJ1J1111
3,845,182	222	0	0JJJJJJ1JJJJ1J11011
3,845,583	229	0	0JJJJ0J10111
3,845,712	234	0	1J01111
3,845,737	0	235	x0JJJJJ
3,852,835	3	232	1J10000J
3,862,125	7	228	1J11000J

alast 7

blast 228

Glast 1J11000J

not plottable

Template comparisonFractalBEC.v Section: partly applies · the nearest solved proof, what carries over, and the transitions that differ

	0	1
A	1RB	0RF
B	1RC	0RE
C	1LD	1RAtpl 1RB
D	1LA	0LD
E	1RA	---
F	0RA	0LB

differs from the template

file: FractalBEC.v:206
module: Section FractalBEC, instance tm1 / nonhalt1, cert1 B D [1] [0] [0;1] [1;0;1] [0;0] (mirrored L<->R)
machine: 1RB0RF_1RC0RE_1LD1RB_1LA0LD_1RA---_0RA0LB (1 transition apart)
diff: C1: 1RA (ours) vs 1RB (tm1 mirrored)
note: Same bouncer: w1 = 101, w0 = 00, and the gap step of R1 is the template's RInc. The counter differs: 3-cell digits 000/010 with transparent J = 100 blocks instead of 2-cell digits (LInc not run on ours: the encoding does not match). No overflow Ov and no fractal recursion: the b=0 step is E/B0/B1 by the low digits, with the one-cell shift σ, and epoch lengths are irregular (2, 5, 6, 11, 18, 44, ...) instead of 2^i - 1. Same family as R2 (BECv1 tm108 twin: 'drop a suffix, shift one cell' eat, pending-size invariant).

Proof strategy & riskshow the formal proof will go · 3 risks to watch

Proof plan: K(a,b,G) with G an Inductive group list; R1 by a carry lemma (ind over digits, J transparent); E/B0/B1 as finite case lemmas with σ as a list shift. Invariant: G is standard at every b=0 hub. Open: the cleanup lemma on σ-words (x -> z on the first carry, z x^j -> 0 J^(j+1) on the second; time 2^(p+1) <= N). Start from FractalBEC.v for the bouncer and a BECv1-style digit list for the counter; share the size lemma with R2.

▲Cleanup lemma unproved: arbitrary x/z/w words halt (K(5,6,1z0101) halts in 19 steps), so the invariant must track where σ puts x/w.
▲The carry through z/w is a cell-level transducer with context (z followed by x-runs); formalizing it may take many small lemmas.
▲Same family as R2 and the #10/#30 sibling, which got stuck on a size invariant; that invariant may be hard here too.

Stats & provenance3·10¹⁰ steps simulated · 15,368 cells · the status in full · files

simulated: 3·10¹⁰ steps
tape: 15,368 cells
growth: span ~ t^(1/3): 677 cells at 3e6, 4,850 at 1e9 (3,147 nonzeros), 15,368 at 3e10; N = 1,603 at 1e9; macro model to 20,000 epochs (N = 192,042, about 1e15 steps by a 3N^2-per-epoch estimate)
status now: Parked 2026-10-11 (Dennis: go with recommendations): Park R3 with R2 (stop rule). Funded crossing stage (2026-10-10) ended STUCK by its stop rule (no inductive form found), so R3 parks with R2; confidence stays med. New exact integer model (word = digit value v + multiset M of j depths; matches the token model on 100,000/100,000 epochs from blank) runs 1e7 epochs to N = 1.44e8: every epoch safe. Proved on paper and checked (100,000 trajectory epochs, 59,887 random states): Lemma S, the bell after epoch e is safe whenever zb(Y_e) <= bitlen(N_e) - 1 + s_e (s = eat shift) or s = 0 and Y_e = 2^bitlen(N_e); in particular whenever Y_e < 2^(bitlen N_e + 1). This replaces L*, which fails at 6 epochs to 1e7 (7, 8, 14, 15, 350,273, 350,274; all Y = 2^m exactly). Open: the Big-Y lemma, Y_e >= 2^(bitlen N_e + 1) => zb(Y_e) <= bitlen(N_e) - 1 + s_e. Its hypothesis holds only at epochs 9, 16 and 350,275 to 1e7 (C-margins 3, 3, 15), each right after an exact-power pair; the candidate C0 is not inductive on random near-crossing states. The safety margin is not 2.3 bits: it falls to 1.000 bit at epoch 350,274 (N = 2^22 - 4,693), a structural case covered by Lemma S. Not a cryptid. (status.json, 2026-10-11)
observer's note: Med confidence: rules R1/E/B0/B1 cover every hub from blank (lockstep to 1e9, b=0 hubs to 3e10, macro to N=192,042); non-halting still needs the cleanup lemma for x/z/w words. Not a cryptid: the eat choice is a regular function of the low digits.
files: observe.json · notes.md · status.json

Proof & checks

No proof work yet.

Files

machines/<tm>/: 10 entries · raw notes, status, scripts, images

name	size	changed
notes.md ↓	23 KB	2026-10-11 08:19
status.json ↓	1.7 KB	2026-10-11 08:19
observe.json ↓	13 KB	2026-10-10 13:24
deepen.json ↓	7.5 KB	2026-10-10 15:10
decision.json ↓	2.6 KB	2026-10-11 08:19
data/	4 items	2026-10-10 13:16
decisions/	2 items	2026-10-10 15:08
img/	5 items	2026-10-10 13:21
scripts/	11 items	2026-10-10 14:56
sim_1e9.jsonl ↓	149 B	2026-10-10 12:55

status.json in full

{
 "tm": "1RB0RF_1RC0RE_1LD1RA_1LA0LD_1RA---_0RA0LB",
 "status": "STUCK",
 "stage": "crossing",
 "updated": "2026-10-11",
 "confidence": "med",
 "cryptid_flag": false,
 "reason": "Parked 2026-10-11 (Dennis: go with recommendations): Park R3 with R2 (stop rule). Funded crossing stage (2026-10-10) ended STUCK by its stop rule (no inductive form found), so R3 parks with R2; confidence stays med. New exact integer model (word = digit value v + multiset M of j depths; matches the token model on 100,000/100,000 epochs from blank) runs 1e7 epochs to N = 1.44e8: every epoch safe. Proved on paper and checked (100,000 trajectory epochs, 59,887 random states): Lemma S, the bell after epoch e is safe whenever zb(Y_e) <= bitlen(N_e) - 1 + s_e (s = eat shift) or s = 0 and Y_e = 2^bitlen(N_e); in particular whenever Y_e < 2^(bitlen N_e + 1). This replaces L*, which fails at 6 epochs to 1e7 (7, 8, 14, 15, 350,273, 350,274; all Y = 2^m exactly). Open: the Big-Y lemma, Y_e >= 2^(bitlen N_e + 1) => zb(Y_e) <= bitlen(N_e) - 1 + s_e. Its hypothesis holds only at epochs 9, 16 and 350,275 to 1e7 (C-margins 3, 3, 15), each right after an exact-power pair; the candidate C0 is not inductive on random near-crossing states. The safety margin is not 2.3 bits: it falls to 1.000 bit at epoch 350,274 (N = 2^22 - 4,693), a structural case covered by Lemma S. Not a cryptid.",
 "funded": {
  "option": "crossing",
  "date": "2026-10-10",
  "auto": true
 },
 "parked": "Parked 2026-10-11 (Dennis: go with recommendations): Park R3 with R2 (stop rule).",
 "priority_call_answered": "R3's cleanup is now exact and safety reduces to one per-epoch inequality (2^P <= N, via L*), the same crossing arithmetic R2 is stuck on: fund one joint crossing-lemma stage, start Rocq on the exact rules, or park?"
}

observe.json contract (tools/README-card.md):

1 contract warning

rank: not an int

Rendered 2026-10-16 13:29 by tools/machinepage.py from status.json, observe.json (the card) and the dashboard records · rebuild: python3 -B tools/card.py 1RB0RF_1RC0RE_1LD1RA_1LA0LD_1RA---_0RA0LB