1RB0RE_1LC0RA_1LD0LB_1LA1LF_1LE1RA_0LC--- never halts
- Machine:
1RB0RE_1LC0RA_1LD0LB_1LA1LF_1LE1RA_0LC---, a BB(6) holdout. It is on every holdout list from 2024-06-14 through 2026-10-07 (line 622 of the 778-machine list). - Class: "Skelet 1-like Class 2" (spreadsheet Results row 1289 and the Classifications tab, class only, with no notes or claims; nothing in our Discord exports). It is the mirrored, state-renamed twin of our kill #3,
1RB0LD_1RC0RA_1RD1RF_1LA0LE_1RE1LD_0RB---(row 1278). - Result:
Lemma nonhalt : ~ halts tm c0.in Coq 8.20.1 against unmodified busycoq (commit0940bb9c).Print Assumptions nonhaltis closed (no axioms). It compiles in 40.0 s with-native-compiler yes, and coqchk re-checks it in 14.6 s (check_kill.txt).
Mechanism
- #3's table, mirrored, with a different start. Flip every direction (L↔R) and rename the states by φ: A→D, B→A, C→B, D→C, E→E, F→F, and you get #3's table, all 12 entries. φ is the only one of the 720 × 2 maps that does this. The twin starts in A, and φ(A) = D, so it is #3's machine run from
0^inf D> 0^inf: a different orbit of the same macro map. - #3's counter, mirrored. The twin's tape is the mirror image of #3's
K(h; P; T) = 0^inf <A (10)^h 11001 [p_1,q_1,d_1] .. [p_L,q_L,d_L] T 0^inf, with pairs(10)^p (01)^q s(d), so it grows to the right where #3's grows to the left. Each macro step decrements a binary number. Merges, an End tail, a re-pairing overflow and the rare X1/X2 detour all occur, under #3's 9 macro rules. - Only the start changes. From
(D, tape0)the run reaches K(2; (11,12,0); −) at step 1,021. At step 78,911 it reaches the E-start K(2; (95,68,0),(34,33,1),(10,15,1),(1,25,1); −), which is #3's base except for the far q (25 instead of 24). #3's invariant holds there with the same constants.
How the proof works
- #3's whole development, kept as
tm3. It has 14essweeps, the 9 macro rulesK h P T -->+ K h' P' T', the per-step invariantJ(a ghost phase of the two-cycle period plus per-pair bounds on p + Rt and q − Rt) andJstep. This provesnonhalt_D : ~ halts tm3 (D;; tape0), starting from a newinit(78,911 steps bysolve_initD) andJ_base. - The conjugacy uses busycoq's own lemmas.
perm_twin : Perm (flip tm) tm3 phiis checked entry by entry. ThenFlip.flip_halts(the blank tape is symmetric, soflip_conf c0=A;; tape0) andPermute.perm_halts(φ(A) = D) turnhalts tm c0intohalts tm3 (D;; tape0), whichnonhalt_Drefutes.
Template and what changed
The file extends kill #3's .v (1,303 lines; in #3's own package). A script generated it, asserting each edit, and no hand edits were made afterwards. The changes:
- tm is the twin, and #3's table becomes tm3, with the local -->*/-->+ bound to it;
- solve_initD is busycoq's solve_init started in D (esx from a non-c0 start falls back to symbolic es, which did not finish);
- init0, and base with far pair (1,25);
- #3's nonhalt becomes nonhalt_D;
- the conjugacy (phi, perm_twin, nonhalt, about 15 lines).
Every sweep, macro rule, J, Jstep and J_base is #3's text, unchanged, and the file compiled on the first attempt. In all, 22 of #3's lines were replaced and 82 lines added. The file is 1,363 lines and imports only Individual62 from busycoq. It is a direct proof on the holdout string. The second TM_from_str literal (#3's string, as tm3) is the only warning from our checker.
Files
1RB0RE_1LC0RA_1LD0LB_1LA1LF_1LE1RA_0LC---.v: the proof.proof.md: the informal delta proof.parent-proof.md: #3's informal proof, whichproof.mdcites by section.check_kill.txt: our verifier's full output.discord.md: the draft announcement, not posted.
AI assistance: the informal proof and the Coq formalization were produced with the help of Claude (Anthropic). They were checked by our checker, check_kill: compile against pinned busycoq, theorem statement, forbidden tokens, Print Assumptions, coqchk.