1RB0RC 1RC0LC 0RD1LE 1LE1RF 0LB--- 0RA1LA: Difference between revisions
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m add category BB(6) |
not holdout |
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{{machine|1RB0RC_1RC0LC_0RD1LE_1LE1RF_0LB---_0RA1LA}} | {{machine|1RB0RC_1RC0LC_0RD1LE_1LE1RF_0LB---_0RA1LA}} | ||
{{TM|1RB0RC_1RC0LC_0RD1LE_1LE1RF_0LB---_0RA1LA}} is a | {{TM|1RB0RC_1RC0LC_0RD1LE_1LE1RF_0LB---_0RA1LA}} is a non-halting [[BB(6)]] TM proven infinite in Rocq ([https://github.com/ccz181078/busycoq/blob/0a61549704e6305f752a4ad1f0faa5cf41dc0e61/verify/SBCv3.v#L8-L201 link]) by @mxdys on 6 Dec 2024. | ||
Analysis by @mxdys: | Analysis by @mxdys: | ||
Latest revision as of 04:47, 28 September 2025
1RB0RC_1RC0LC_0RD1LE_1LE1RF_0LB---_0RA1LA (bbch) is a non-halting BB(6) TM proven infinite in Rocq (link) by @mxdys on 6 Dec 2024.
Analysis by @mxdys:
f(n0, ()) := n0 f(n0, (n,ls)) := 2*(f(n0, ls) - n) start: (2, (), 1) (k, ls, n) --> (k+1, (n,ls), 2*(f(2^k, ls)-(n+2))), f(2^k, ls)-(n+2) >= 0
Follow up by User:sligocki:
Let (k, ls, n) --> (k+1, [n, ls], n')
Then:
n'+2 = 2 (f(2^k, ls) - n) - 2
f(2^{k+1}, [n, ls]) = 2 (f(2^{k+1}, ls) - n)
So:
f(2^{k+1}, [n, ls]) - (n'+2) = 2 (f(2^{k+1}, ls) - f(2^k, ls)) + 2
Which indicates that the condition should always hold. But this is not detectible nor provable by any existing deciders.