Does the sign of the Li–Yorke slack on the core separate period three exactly, and why?

Status: closed before any panel. Dissolved under gate G5/G6 by the codifier on 2026-10-23, while the case was being written. The owner asked for the case to be written out even if it was not run, because stating a case in objective terms finds errors. It did. The coincidence the case was built on (680 of 680 named rules separate exactly) turned out to be a theorem once it was stated without the programme's words. The case is kept as written, with the dissolution below. It is not re-panelled (docs/v5-plan.md §12.2, G5).

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type: derivation

field: math

object. A rule is a polynomial map R : [0, 1] → [0, 1] in Bernstein form, R(p) = Σⱼ cⱼ pʲ(1 − p)^{k−j} with 0 ≤ cⱼ ≤ C(k, j). The named rules are band rules: cⱼ = C(k, j) for a ≤ j ≤ b and 0 otherwise, with 1 ≤ a ≤ b ≤ k − 1. The NAE rule is the band [1, k − 1], R(p) = 1 − pᵏ − (1 − p)ᵏ. For a map with a critical point c at its global maximum, the core is the interval between R²(c) and R(c). The Li–Yorke slack of a core point a is s(a) = a − R³(a) when a < R(a) < R²(a), and s(a) = R³(a) − a when a > R(a) > R²(a). O₂ is the largest slack over the core divided by the core's length, and −∞ when no core point is ordered either way.

friction. Two runs found that O₂'s sign separates the rules with a period-3 orbit from those without, with no exception: 286 rules at k = 2 … 12 (run 120: 124 / 0 / 0 / 161) and 394 rules at k = 13 … 16 (run 121: 253 / 0 / 0 / 141). Both runs were withdrawn by a numerical dual control, and the separation was reported beside. The pre-registration gave the separation P = 0.6 at 10:50 and P = 0.8 out of sample at 12:45, on the worry that a 3-cycle could sit outside the core. Is the separation a law of these rules, a theorem of all such maps, or a feature of the sample?

background. - Measured. The two tables above (RESULTS120.md, RESULTS121.md). On 4,000 random box rules at k = 7, 8 (run 121), none has a 3-cycle, so the random half of the test was empty. On 758 of those rules the instrument took a local minimum as the critical point (RESULTS121.md). - Stated (Li and Yorke, 1975). For a continuous map of an interval, a point a with R³(a) ≤ a < R(a) < R²(a) forces a periodic point of every period, and of period 3 in particular. The reversed ordering does the same, since conjugating by x ↦ 1 − x turns one into the other. - Derived by the codifier, 2026-10-23. (i) Every band rule is unimodal. The derivative of a sum of consecutive Bernstein terms telescopes: R′(p) = k [C(k−1, a−1) p^{a−1}(1−p)^{k−a} − C(k−1, b) p^b (1−p)^{k−1−b}]. This vanishes exactly where (p/(1−p))^{b−a+1} = C(k−1, a−1)/C(k−1, b), which has one root in (0, 1). R rises before that root and falls after it, and R(0) = R(1) = 0. (ii) For a unimodal map, every periodic orbit of period at least 2 lies in the core. Take the orbit's minimum m = R(y), with y in the orbit. If m < R²(c), then y lies on the rising branch below the left preimage of R²(c), since every point of the falling branch up to R(c) maps to R²(c) or above. So m ≤ y on a rising branch. Then R(m) ≤ R(y) = m, and since m is the minimum, R(m) = m: the orbit is a fixed point. The maximum is ≤ R(c) because it is an image. - Hence. A 3-cycle lies in the core, and one of its points is ordered with slack exactly 0, in one orientation or the other. So period three implies O₂ ≥ 0. Conversely, O₂ ≥ 0 gives a core point satisfying Li–Yorke's ordering, which implies period three. O₂'s sign is the period-3 test, exactly, for every unimodal rule. Every named rule of runs 114, 120 and 121 is unimodal.

Coincidences this case leaned on. (a) 680 of 680 named rules separate exactly: explained, a theorem (above). (b) None of 4,000 random box rules has a 3-cycle: unexplained, and outside this case. (c) The separation survived two withdrawn runs: explained. The withdrawals were numerical (the dual check at 1 − 10⁻³⁰), and they never touched the sign.

Must predict. Nothing remains to predict on the named rules. Added 2026-10-23, 15:35: nor on multimodal rules. The slack's sign over all of [0, 1] is the period-3 test for every continuous map, since a 3-cycle supplies an ordered point with slack 0 wherever it sits. The paragraph below overstated what is open. What is not forced: on rules with two or more humps, the core of the global maximum need not hold every 3-cycle, and the lemma in (ii) fails. Whether a rule in the Bernstein box can have a 3-cycle outside that core is a construction question, not a panel question. O₂'s magnitude is not forced either (the pre-registration of 10:50, guess 3: does O₂ order the stretch |λ|max within a band?). Either could become a seal. Neither needs this case.

Not part of the case

The codifier's error, recorded: the pre-registration twice set P below 1 for a statement that is a theorem on the sample it was tested on. The worry ("a cycle could sit outside the core") was never checked against the fact that every named rule is unimodal. That needed only differentiating a band rule. Runs 120 and 121 spent their compute confirming Li–Yorke plus a lemma about unimodal maps. This is OPS's premise rule again: a premise about a mathematical object is computed before a seal is built on it. The re-seal planned for O₂ (the sign as a control in place of the dual check) is dropped. Its claim is proved for the named rules, and the random rules gave it nothing to test.