# RULER — evaluations that know their optimum The object: an integer x in Z/N, N = Mc, read through the residue r = x mod M, held in its **channels**, the residues of x at the prime powers exactly dividing M; the fiber over r is the c integers r + jM, j its index, and c is the cofactor, with M > 1 and c > 1 throughout. An **eval** is a task with a prior, this evidence and a score (0-1 accuracy, or log-loss in nats). Its ceiling is the Bayes-optimal expected score, its floor the best score with the evidence severed, and a cell (N, M) is dead when the two coincide; a **gauge** is the closed form naming the dead cells. The tasks ask one question of size or order: the sign bit [2x ≥ N], the threshold [x ≥ t], the cyclic orientation of three points, the comparison of two, and the magnitude class ⌊kx/N⌋. The prior is uniform or the tilt θ^x, whose ratio between neighbours in a fiber is q = θ^M. The exact Bayes error then makes the family a **ruler**: an estimator of the Bayes error is scored on its bias against the truth and on the coverage of its own stated 95% interval, the fraction of trials in which that interval holds the truth. The verifiers are ceiling.py, estimators.py, statements.py, audit.py, selfreport.py, lira.py and sets.py. The thesis: each eval read here at uniform ships its 0-1 ceiling as a closed form, a solver of a CRT combination or two and a comparison (a division, for the magnitude class) that attains it, and a gauge; which channels pay is the task's, not the ring's. The threshold family's dead cells at uniform, the even-c ones, are a resonance of that one prior: every rational tilt off uniform gives each a live threshold, and only the seam, where the threshold sits on a fiber boundary (RULER.md#one-threshold-range-one-tent-the-threshold-family), stays dead at every tilt, the sign bit at even c among it, and the k-ary parent trades the isolated resonant tilts for one live arc of tilts, ending at resonance roots once M > k and c > k with k ∤ c. Turned on the textbook Bayes-error estimators, the ruler finds intervals that fail in ways more data does not repair, or for a stretch makes worse, and that a truth-free check sees only in part; the failures are the estimators', the ring supplying only their exact placement, and the two distribution-free brackets are one bound, correct and certifying about one in c of the family's own live thresholds. Read as a statement, estimate ± h or one side of it, an interval is protected only on the side its omitted term's known sign favours; under the normal approximation its coverage is a function of two ratios, and the one that decides, the bias's, is read from inside only through the fitted class's own functional. Turned on a published deletion audit, which estimates a mixture weight in the box [0, 1] and whose target the depth column (FORGET.md#two-witnesses-buy-the-robust-grade) places exactly, read against that exact target, the estimator is followed root by root: where its objective's vertex lies off the box, its wrong-side answers are the sign of one scalar, its band omits the noise of the samples it holds fixed, and a statement that reports the root set, its objective's roots in the box, keeps the sign law where the target is not identified. An auditor can measure the spread of their own answer from inside, but in the normal model with that scalar's spread σ known, no statement of its instability read off the scalar alone is right in the mean at every distance of the scalar's mean from 0. A second, per-example tool converges to its model's Gaussian limit rather than to the posterior, and its membership call survives only where the two share a side. ## The parity law of the sign bit Tier: theorem (squarefree N); observation (211 dead of the 602 cells over the 57 squarefree rings from two or more of the first six primes). Verifier: proof; ceiling.py::section_sign. The below-count of fiber r is ⌈c/2 − r/M⌉. At even c it is c/2 on every fiber, so the posterior is the prior; at odd c it is (c+1)/2 for r < M/2 and (c−1)/2 otherwise, and every fiber scores (c+1)/2c. So the 0-1 ceiling is (c+1)/2c at odd c and 1/2 at even c, and the log-loss is H₂((c+1)/2c) and log 2, H₂ the binary entropy in nats. At even squarefree N, c is even iff 2 does not divide M: those cells, and only those, are dead (211 of 602). At odd N, c is odd and no cell is dead. The achiever reconstructs r from the channels read and compares 2r with M. A lone channel p is the same law at M = p: at even squarefree N only channel 2 lifts the score, to 1/2 + 1/N, and at odd N every channel scores 1/2 + p/2N, so a channel set is priced against its own subsets (N = 2310 read at {2, 3, 5} scores 39/77, its best lone channel 578/1155). This is the hiding lemma of SIZE.md#the-hiding-lemma read as a score: a posterior that leans nowhere is a dead cell. ## One threshold range, one tent: the threshold family Tier: theorem; observation (the floor sets' sizes, over the 40,914 (cell, t) slices with N ≤ 500 a product of two or more distinct primes among 2, 3, 5, 7, 11, 13). Verifier: proof; ceiling.py::section_threshold. Write t = QM + s, 0 ≤ s < M. Fiber r holds Q + [r < s] points below t, so two fiber types exist. The 0-1 ceiling is the floor max(t, N − t)/N except when c = 2m + 1 is odd and t lies strictly inside the middle range (mM, (m+1)M), where it is (c+1)/2c whatever t is. The lift there is the tent min(s, M − s)/N, highest at the midpoint: the sign bit is the family's extremal member. The achiever guesses below at fiber r iff Q + [r < s] > c/2. The two scores disagree about deadness. The log-loss is (s/M)H₂((Q+1)/c) + ((M−s)/M)H₂(Q/c), at the prior's entropy iff M | t, so its floor set is the thin **seam** {M | t}, the thresholds on a fiber boundary; the 0-1 floor set is everything outside the range, where the posterior, off the seam, moves but never crosses 1/2 (the log-loss floor set 4,599 slices, inside the 0-1 floor set's 38,217). ## The gauge is the task's: orientation and comparison Tier: theorem. Verifier: proof; ceiling.py::section_orientation. Orientation. A uniform ordered triple of distinct points, all three read mod M, is positive iff (y − x) mod N < (z − x) mod N. With the differences reduced mod M to (α, β), a fiber with α ≠ β, both nonzero, leans (c+1)/2c toward the residue order (the differences α + jM and β + j′M compare by j against j′ and at a tie by α against β; at α = 0, j runs from 1 and half the pairs of blocks order each way); α = β, or a zero among them, is balanced. The ceiling is 1/2 + c(M−1)(M−2)/(2(N−1)(N−2)), the floor 1/2, and no fiber's posterior reaches 0 or 1. The one dead cell shape is M = 2: channel 2, without which the sign bit is dead at even squarefree N, is orientation's only blind channel, and a lone channel is worth more the larger its prime. The achiever subtracts the residues channel by channel, CRT-combines the two differences and compares them. Comparison. [x < y] on a uniform ordered pair of distinct points, both read mod M, to residues r_x and r_y: r_x < r_y leans (c+1)/2c, r_x = r_y is balanced, and the ceiling is 1/2 + (M−1)/(2(N−1)), with no condition on c, over a floor of exactly 1/2. No cell is dead, M = 2 included. The achiever CRT-combines both residues and compares them. The **skew unit** (property) is the score (c+1)/2c. In the sign, threshold, orientation and comparison tasks every fiber of a live cell whose posterior is not exactly 1/2 is a staircase count on a c-point ladder and scores exactly the skew unit. The four tasks carry three gauge shapes: the threshold range (sign and threshold; at the midpoint threshold its live condition is c odd), the inverted gauge (orientation, dead only at M = 2) and the constant gauge (comparison, never dead). The inversion is the prime 2's: at even squarefree N the sign bit is dead exactly when 2 does not divide M, since c is then even, and orientation exactly when M = 2. ## The threshold-range law under a tilted prior Tier: theorem. Verifier: proof; ceiling.py::section_tilt. Under the tilt a fiber of below-count J lies below t with probability Pb(J) = (1 − q^J)/(1 − q^c), and leans below iff J > Q*, Q* = log_q((1 + q^c)/2), which lies strictly in (0, c) whenever q ≠ 1 and rises with q: at q = e^(−u) < 1 it is c·log(2/(1 + e^(−cu)))/(cu), the numerator concave and 0 at 0, and Q*(1/q) = c − Q*(q). So a threshold is live iff s ≠ 0 and Q < Q* < Q + 1: at every tilt off uniform the live thresholds form one range of M − 1 thresholds at index ⌊Q*⌋, empty exactly when Q* is an integer. The seam {M | t}, where every fiber has one type, has posterior equal to prior at every tilt and is the whole log-loss floor set at every tilt. An even-c cell, dead at every threshold at uniform, has a live threshold at every rational tilt off uniform (the resonance criterion below), and the map x ↦ N − 1 − x sends the ceiling at (θ, t) to the ceiling at (1/θ, N − t). A single fixed threshold is live only on a range of tilts, Q* being monotone; the family loses every live threshold only at its cell's irrational resonant tilts and, at even c, at uniform, and its one live range slides toward index 0 as q → 0 and toward index c − 1 as q → ∞. The sign bit's own row is the k = 2 row of the magnitude arc: at odd c = 2m + 1 ≥ 3 it is alive exactly while q lies in (ρ_m, 1/ρ_m), ρ_m the root in (0, 1) of q^(2m+1) − 2q^m + 1, the values of q putting Q* in (m, m + 1), an interval in q free of M, dead outside, and at even c its threshold N/2 sits on the seam, dead at every tilt. ## The resonance criterion Tier: criterion. Verifier: proof; ceiling.py::section_resonance. Off uniform, q ≠ 1, the range of a cell with cofactor c is empty iff q is a root of f(q) = q^c − 2q^J + 1 for some 0 < J < c; at uniform it is empty iff c is even, the parity law. Proof. At q ≠ 1, Q* = J is 2q^J = 1 + q^c, and Q* lies in (0, c). f is monic with constant term 1, so its only rational roots are ±1: no rational tilt off uniform is ever resonant. Its signs +, −, + allow two positive roots, one of them q = 1 with f′(1) = c − 2J. From f(0) = 1 and f → ∞, at 2J < c the other root lies in (0, 1) and at 2J > c in (1, ∞), one algebraic irrational per pair; q ↦ 1/q exchanges J and c − J. At 2J = c, f = (q^J − 1)², so q = 1 is a double root and the only one: the uniform prior is where the 2J = c family collapses, which is the parity law's dead half. At (J, c) = (1, 3), f = (q − 1)(q² + q − 1), and the root is the reciprocal golden ratio. ∎ ## One live arc: the magnitude parent Tier: theorem (every N, squarefree or not); observation (the masses read at two tilted cells). Verifier: proof; ceiling.py::section_magnitude, sets.py::section_family. Write c = ak + v, 0 ≤ v < k. At an atom, the fiber over r, the classes of Y = ⌊kx/N⌋ are contiguous blocks of fiber indices of lengths a and a + 1, exactly v of them heavy (length a + 1), at {⌊k(ℓ + r/M)/v⌋ : ℓ < v}; the first heavy class is y₁(r) = ⌊kr/(vM)⌋. At uniform every atom's best class carries ⌈c/k⌉/c, so the ceiling is ⌈c/k⌉/c over a floor of ⌈N/k⌉/N, and guessing y₁(r) attains it (at k | c every class ties). A cell is dead at uniform iff k | c or M(k − v) < k, saturation, where class 0 is heavy at every atom; it needs k > M, and at k = 2 the gauge is the parity law's. Under the tilt a block starting at lo of length w carries mass q^lo(1 − q^w). Since q's powers are monotone, the first of equal-length blocks wins at q < 1 and a heavy block beats every later light one; so off uniform an atom's best class is 0 or y₁(r) at q < 1, the last heavy or k − 1 at q > 1 (at uniform all heavy classes tie). At q < 1 an atom with y₁ ≥ 1 overrules the floor's guess, class 0, iff g(q) = 1 − q^a − q^L + q^(L+a+1) < 0, L = a·y₁(r): g has one root in (0, 1), at a ≥ 1, and the atom overrules above it; at a = 0 its class 0 is empty and it overrules at every q < 1. The root rises with L, since g < 0 iff q^L > (1 − q^a)/(1 − q^(a+1)), a ratio falling from 1 to a/(a + 1) on (0, 1) while q^L rises from 0 to 1, so they cross once. The partition by ⌊kx/N⌋ is not symmetric under x ↦ N − 1 − x, so q > 1 is read on its own, where a cell is dead exactly when one class is best at every atom. An atom whose top class is heavy keeps class k − 1; one whose top class is light prefers its last heavy class y_last on (1, 1/q′), q′ the root of g at L = a(k − 1 − y_last), and k − 1 above it, or y_last at every q > 1 when a = 0. The top class is light at the atom r = 0 whenever k ∤ c, and if it is light at every atom, which needs Mv < k, the atoms r = 0 and r = M − 1 differ in y_last = ⌊k(v − 1 + r/M)/v⌋. So at a tilt q > 1 the cell is alive iff q lies in (1, 1/q′) for some light-topped atom: it is alive on (1, q_up), q_up the largest of those 1/q′ and infinite when a = 0. So every cell is of one of three kinds: dead at every tilt iff k | c; alive at every tilt when a = 0 and it is unsaturated; otherwise alive on exactly one open arc (q_lo, q_up), q_lo the least root in (0, 1) of g over the atoms with y₁ ≥ 1, or 1 if there is none, each end algebraic, 1 or infinite. A saturated cell has q_lo = 1: dead at uniform and below, revived by a small tilt upward, and dead again past q_up when a ≥ 1. When M > k, a ≥ 1 and k ∤ c the arc is (ρ_a, 1/ρ_a), ρ_a the root in (0, 1) of q^(2a+1) − 2q^a + 1, the f of RULER.md#the-resonance-criterion at J = a and cofactor 2a + 1, a function of ⌊c/k⌋ alone: a phase r/M in [v/k, 2v/k) gives y₁ = 1 and one in [1 − 2v/k, 1 − v/k) gives y_last = k − 2, each meeting [0, 1) in length at least 1/k > 1/M, so L = a is reached at each end. ρ₁ is the reciprocal golden ratio. Read set-valued, the class keeps what the fiber index loses. The index j reads q^j at every atom, the θ^r cancelling, while the class blocks move with the atom's phase r/M. At uniform every atom carries the same multiset of masses, so a set's size is blind to the atom and only its labels move; at k | c the blocks do not move at all, so every atom carries one law at every tilt; at a = 0, c classes hold one point each and the rest are empty, and every atom carries one multiset at every tilt; at the two tilted cells with k ∤ c that sets.py::section_family reads, the masses themselves depend on the atom. SETS.md scores set rules on this class. ## The knife: a consistent point inside an inconsistent interval Tier: observation (the ratio derived asymptotically; 40 trials at n = 500 to 32,000, on the ring and on bare Bernoulli bins). Verifier: estimators.py::section_knife, estimators.py::section_transient. Drawing x uniform and reading (x mod M, [x ≥ t]) draws a bin r uniform and a label Bernoulli(η_r), η_r = (c − J(r))/c with J(r) the below-count of fiber r, its number of points below t, so any failure an estimator shows on the ring it shows on bare bins with those η_r (property); what the ring adds is placement, η_r = 1/2 exactly at a knife slice (c even, t = N/2), with the truth computed in exact rational arithmetic. There resubstitution on n points, (1/n)Σ_r min(n_r0, n_r1) with n_r0 and n_r1 the label counts in bin r, falls short by the mean absolute deviation of a fair binomial, (1/2)√(2M/πn), while its stated half-bar is 1.96/(2√n): the point estimate is consistent, but the ratio of deficit to half-bar is √(2M/π)/1.96, free of n (1.5766 at M = 15, printed −1.46 to −1.68 as a signed ratio over 40 trials a point), and the interval covers at 0 to 3% at every n on the ring and bare alike. The failure is a band at 1/2: on bare bins at n = 8000 a label rate one sixty-fourth off 1/2 already covers at 0.90. The band's width in the scaled offset 2δ√(n/M), δ the label rate's distance from 1/2, is measured only, at n = 8000, M = 15. ## More data, a worse held-out interval Tier: observation (40 trials per point; the direction on the ring and on bare bins, the recovery on bare bins to n = 512,000). Verifier: estimators.py::section_transient. Held-out fits the per-bin majority on half the sample and scores it on the other half, so it measures a fitted rule, whose excess over the optimum is Σ_r P(r)|2η_r − 1|·P(the bin's majority flips). At |η_r − 1/2| = δ that decays exponentially in δ²n_r only once n_r passes 1/δ², while the stated bar shrinks as 1/√n from the start. So coverage falls as data grows, then recovers: at M = 105, η = 6/11 on bare bins it runs 0.88 at n = 500 to 0.17 at n = 32,000 and back to 0.95 at 128,000 over 40 trials a point (a second run of 400 reads 0.24 at 32,000, RULER.md#coverage-is-a-function-of-two-ratios), and the ring slice with those η_r falls alike, 0.72 at n = 500 to 0.12 at 8000 and 0.30 at 32,000. Without the exact truth the tightening interval reads as improving evidence. ## One bound at two statistics Tier: theorem. Verifier: proof; estimators.py::section_bound. Two distribution-free brackets are standard: Cover–Hart's [(1 − √(1 − 2R_NN))/2, R_NN] from the nearest-neighbour error, and the bracket [(1 − √u)/2, (1 − u)/2] of Berisha, Wisler, Hero and Spanias (IEEE Trans. Signal Process. 2016, Theorem 2) from a Friedman–Rafsky spanning-tree count, u = 1 − 2W/n with W the number of the tree's edges joining two labels, taken as 0 when negative. On a categorical feature with every bin populated (n ≫ M), ties among zero-distance neighbours broken blind to the label, R_NN tends to E[2η(1 − η)], and under a tree blind to the labels (the next block) u has population value E[(2η − 1)²]; (1 − u)/2 = E[2η(1 − η)] identically, so the two are one bound read through two statistics, and a second bracket buys a second estimator, never a second opinion. Since the Bayes error is R* = 1/2 − E|2η − 1|/2, the lower end's gap is the Jensen gap of |2η − 1|: zero iff |c − 2J(r)| is constant in r, iff M | t or t lies in the odd-c middle range. On that range, c = 2m + 1, the upper end is 2m(m + 1)/c² = 1/2 − 1/(2c²), so the bound is correct and excludes the no-evidence value min(t, N − t)/N only at |t − N/2| < M/(2c), about M/c of the M − 1 live thresholds: at N = 105, M = 15 the exact bracket is [3/7, 24/49] and holds the no-evidence value at 12 of the 14 thresholds inside it, so even exact knowledge of the statistic certifies the family's lift at two. ## The spanning tree is not identified Tier: property. Verifier: proof; estimators.py::section_mst. The Friedman–Rafsky count assumes a unique minimum spanning tree. On a categorical feature two points of one bin sit at distance 0, so any spanning tree of a bin's clique belongs to a minimum spanning tree, and every choice is genuine. A chain in label order puts one cross edge in each two-label bin, so W/n → 0 and the bracket collapses to [0, 0], the classes declared separable; a spanning tree of the bipartite graph between a bin's two label classes makes all n_r − 1 of its edges cross, so W/n tends to the mass of the two-label bins, which is 1 where every fiber holds both labels, as on the odd-c range; there u → −1 is clipped to 0 and the bracket sits at [1/2, 1/2]. Exchanging one edge of a tree moves W by at most 1, and such exchanges walk from one spanning tree to any other; every tree crosses at least once in each two-label bin, so where every fiber holds both labels the bracket's upper end takes every value from order M/n to 1/2 in steps of 1/n: the values the bracket states over the minimum spanning trees, its identified set, are every value a binary Bayes error can take, to within order M/n. A tree whose shape ignores the labels has the labels exchangeable on its n_r − 1 edges, expected cross count 2n_r0 n_r1/n_r, which is the population quantity; label-blindness is the repair, and randomness buys nothing beyond it. The edges between bins number at most M − 1, so they move W/n by at most M/n. ∎ At n = 8000 on the slice N = 105, M = 15, t = 45 the three within-bin trees give W/n = 0.0019, 0.998 and 0.4885 against 24/49 = 0.4898. ## The sign law of a stated interval Tier: property (proved, up to the binomial bar's normal approximation; measured at 8 settings, two worlds at four sample sizes, 400 trials each, and under non-identification in the audit's Gaussian worlds at mixture weights 1/5 and 4/5 with equal class variances, and in one where the second root has left the box, 200 trials each). Verifier: proof; statements.py::section_sign, audit.py::section_band. A statement "estimate ± h", or one side of it, counts only the error its own arithmetic admits; what it omits is its uncounted term. A **held-out bound** fits a rule on one half and states err + h, one-sided, on the other: the rule is fixed before the test half is read, so err is binomial about the rule's own risk and the bar is the right width for that, and every rule's risk is at least the Bayes error. So the bound covers the Bayes error at least as often as it covers the rule's risk: the omitted term has a known sign and only adds margin, whatever the world. ∎ That the bar covers the rule's risk at its nominal is the normal approximation's, and fails near a risk of 0, where err = 0 states a bar of 0. A fit (a plug-in with its model-implied bar) has a signed term only in the mean, never trial by trial: the per-bin plug-in reads low in the mean in every world, since min is concave, while a pooled fit, the bins merged into groups, adds its pooling gap, the excess of the pooled limit over the Bayes error, against that pull, so its sign is the world's. Nothing forces a fit to fail either: at the knife the per-bin bar is too wide for the estimate's sd S, h/S = 3.25 derived, 3.09 to 3.26 printed, and its two-sided coverage of 0.02 to 0.03 is lost to bias alone. The protection is one-sided: the same held-out error stated two-sided for the Bayes error has the term against its lower end and fails (0.18 at n = 8000, on the plane of RULER.md#coverage-is-a-function-of-two-ratios). Conservative is not empty: on a world with no bin rate nearer 1/2 than 1/10 the bound excluded the no-evidence 1/2 in every trial at every n, and on it and at the knife the bound covered 0.970 to 0.988 against 0.975. The law survives a target that is not identified, where the omitted term is not a displacement but which root is the truth. On the audit below (RULER.md#the-audit-world-and-its-two-roots, the box being [0, 1]), report from the same resamples of the audited set the 95% quantile of the largest in-box root as an upper bound and the 5% quantile of the smallest as a lower one: the truth is one of the roots, so the root term only widens. At equal class variances and the audit's mixture weight κ = 1/5, the pair closes on [1/5, 4/5], the span of the identified set {1/5, 4/5}, and not on a point, its excess 0.194, 0.146, 0.047 below and 0.194, 0.147, 0.048 above at n = 50, 200, 1000; the bound on the second root's side covers in every trial and closes on 4/5, never on κ, the other tightens and covers 0.93 to 1. At κ = 4/5 the two trade roles, the slacks within 0.003 of their mirrors and the tightening bound covering 0.925 to 1. With the second root pushed out of the box the same statement contracts (upper slack 0.047 at n = 1000) and covers 0.87 to 0.96 against 0.95: the root term is gone in the population, and what is left omits the class samples' noise, as the band below does (whether that omission is the shortfall is not read). The pair is never wrong together by construction. The tool's own band, the quantiles of the argmin, chooses a mode and has no such term. ## Coverage is a function of two ratios Tier: observation (44 settings over five worlds, four sample sizes and five kinds of statement; the formula follows from construction under normality, and what is measured is that the approximation holds). Verifier: statements.py::section_plane. With b the estimate's mean deviation from the truth, S its sd and h the mean stated bar, a two-sided statement covers at Φ((h − b)/S) − Φ((−h − b)/S), Φ the standard normal distribution function, and a one-sided bound at Φ((b + h)/S). Every setting landed within 0.034 of that (mean 0.006), so at these sizes, n ≥ 500, mechanisms that read as distinct act only by where they put a statement on the plane (b/S, h/S). With h/S settled, whether data carries coverage to its value at b = 0 is the trajectory of b/h: to 0 or negligible, remedied (a well-specified plug-in, |b|/h 0.038 to 0.048, within noise of 0 at every n); constant, sample size irrelevant (the knife, √(2M/π)/1.96 = 1.5766, |b|/h printed 1.57 to 1.60 over 400 trials a point); to infinity, more data is worse. The third regime has two residents. A model class that cannot hold the truth, pooling bins whose group straddles 1/2, keeps b on its exact pooling gap 7/750 while h falls: |b|/h 0.22 to 1.81, coverage 0.94 to 0.06, and the coverage passes 1/2 between the printed |b|/h of 0.914 (0.580) and 1.813 (0.058), near 1 by the plane's formula. A shrinkage toward 1/2 at weight λ = n^(−1/4) is (1 − λ)·(plug-in) + λ/2 exactly, since the lesser of a rate and its complement is linear on each side of 1/2 and the pull never crosses, so b decays as n^(−1/4): divergent while still moving (|b|/h 1.24 to 3.79). Held-out's transient (RULER.md#more-data-a-worse-held-out-interval) sits on the same plane: in a second run of 400 trials beside that block's 40, on a world of 105 bins at η = 6/11, coverage 0.89, 0.51, 0.18, 0.24 at n = 500, 2000, 8000, 32,000, each within 0.026 of it. ## The kernel of the inside instruments Tier: property (the kernel and the shrinkage identity, proved; the knife factor derived asymptotically; the detection rates measured at 400 trials). Verifier: proof; statements.py::section_kernel. An auditor holding one sample estimates S by resampling, as Ŝ, the sd of the estimate over bootstrap resamples of that sample, so h/S can be read from inside; b/S decides, and it is the hidden coordinate. On one world a pooled and a per-bin fit printed h/Ŝ within 0.2 of each other at every n while their coverages parted to 0.06 and 0.97. Two instruments reach for b. With Ê the estimate of R* and Ê* its value on a resample, the bootstrap term, mean Ê* − Ê at the same n, estimates the bias against the estimator's own estimand, the class's functional on the population; the drift, Ê on the sample minus Ê on an exact half run at n/2, reads b(n) − b(n/2). So the gap between a class's limit and the truth is invisible to the bootstrap by construction, the drift differences a settled term away, and a term that has stopped moving is in the kernel of both. ∎ Both reach b only through the class's own functional, which is why the nested repair escapes by fitting a richer class. At the shrinkage the bootstrap is blind by identity, the λ/2 parts cancelling at equal n (its term is (1 − λ) times the plug-in's to 2·10⁻¹⁶), while the drift reads −(2^(1/4) − 1)λ(1/2 − R*), which is −0.189|b|, and lands within 0.09 h of that at the measured b at every n. At the knife the bootstrap sees the term and reads only √2 − 1 of it: a resampled bin rate carries twice the fitted one's variance where |·| is not differentiable (0.402 to 0.420 printed). The drift's mean reads the same fraction with the opposite sign, b falling as 1/√n there, but a single drift has sd near S and its mean is not measured. The failure is anti-monotone: at thresholds holding alarms to 0.10 on the settings that cover at 0.90 or more, the bootstrap caught every knife trial and 0.27, 0.01, 0, 0 of the pooled fit's on the world of pooling gap 29/150, not the one above, as its |b|/h grew 4.4 to 35, since severity there is bought by the term converging. And the readable coordinate is biased where the functional is not smooth: at the knife Ŝ overstates S, Ŝ/S 1.18 to 1.24. ## The nested repair reads the term, not the model Tier: observation (three pooled worlds at four sample sizes, 400 trials; Δ̂ ≥ 0 and its limit, proved). Verifier: statements.py::section_repair. Where a richer class nests the model, read the two fits' difference in the functional, Δ̂ = pooled − per-bin: min is concave, so Δ̂ ≥ 0, its limit is the pooling gap, and it vanishes where every bin of a group sits on one side of 1/2. On a world pooled wrongly about 14 of its 15 bins and exactly about the Bayes error (gap 0, coverage 0.95 to 0.96), a goodness-of-fit statistic fired hard, standardized means 8 to 548, while Δ̂/h alarmed on 0.06 then no trials; Δ̂ caught every trial of a world of pooling gap 29/150 and 0.998 of one of gap 7/750 at n = 32,000. It reads the term's size, never its lethality: at n = 500 it alarms on 0.44 of the small gap's trials while that statement still covers at 0.94, h alone deciding when the term kills. The kernel narrows and does not empty: error outside every class the auditor can nest stays uncounted. ## The audit world and its two roots Tier: property (proved; the depth-cell roots and the tangency identity checked in exact rational arithmetic, the Gaussian second root 1 − κ + γ/4 in floating point at κ = 1/5 and the four γ in {0, 1/4, 1/2, 4/5} where it lies in the box, and found outside the box at γ = 1 and 2). Verifier: proof; audit.py::section_controls, audit.py::section_tangency. A mixture-proportion deletion audit (Statistical Membership Inference, arXiv:2602.01150, Lemma 3.2 and Algorithm 1, read at source) models the audited features as F_u = κF_out + (1 − κ)F_in, F_in the member class and F_out the non-member class, and reports κ. For a scalar feature, with d = μ_out − μ_in, μ_out and μ_in the class means, B = d², A = var_u − var_in and G = var_out − var_in + B, it returns the minimizer of (A − pG + p²B)² over the candidates 0, 1 and the objective's critical points in [0, 1], the vertex G/(2B) and the roots of Bp² − Gp + A. It never reads the audited set's mean; its band is the {5%, 95%} quantiles of the estimate over resamples of the audited set. The world it is turned on is the depth column of FORGET.md: a history is a route of moves, each multiplying the state by 2 or by 3, and a **depth cell** is named by its counts of moves by 2 and by 3, the cell (12, 3) holding the routes to the state 2^12 3^3 in every arrangement, equally likely. The audited datum is the first move, the member class the routes opening with 2 and the non-member class those opening with 3. That depth cell is an exact mixture of the two classes at weight κ, the share of moves by 3 (1/5 at (12, 3)), whose distance from ½ is the size of the count leak of FORGET.md#readability-descends-spread-ascends-flatness-neither, and with the feature the number of moves by 2 in the three positions after the first, each class is hypergeometric: the audit's model holds exactly and every moment is rational. For any two-class mixture var_u = κ var_out + (1 − κ) var_in + κ(1 − κ)d², so A = κG − κ²B identically: κ is a root of the population objective in every world, and the other root is A/(Bκ). ∎ With F_in = 𝒩(0, 1) and F_out = 𝒩(2, 1 + γ) it is 1 − κ + γ/4: equal variances give the twin {κ, 1 − κ}, and a variance excess γ > 0 slides the second root up until it leaves the box past γ = 4κ. The box, not the data, identifies. The depth cells (12, 3) and (2, 8), κ = 1/5 and 4/5, have second roots −113/65 and 17/10: both are identified. For a {0, 1} feature A − Bκ² = κd(1 − 2μ_u), μ_u the audited set's mean, its rate, since var_out − var_in = d(1 − μ_out − μ_in): for d ≠ 0 the two roots meet at κ exactly when the audited set is balanced, a tangency. ∎ ## The flip is the sign of one scalar Tier: theorem (off the box); observation (inside it, the cases and their counts; the rates, 1,000 replicates per row). Verifier: proof; audit.py::section_level. Write L(p) = Gp − Bp², the level the objective needs at p, so the estimate is the p whose L(p) is nearest Â. When a replicate's own vertex lies off [0, 1], L is monotone on the box and the estimate is  clamped to L's range and inverted, so it lies past 1/2 exactly when  is past the half-box level L(1/2) = G/2 − B/4, the direction set by the side of the box the vertex sits on. ∎ The flip, an estimate nearer 1 − κ than κ, is therefore the sign of one scalar at the replicate's own coefficients, D̂ =  − (Ĝ/2 − B̂/4) = s_u² − s_in²/2 − s_out²/2 − d̂²/4, s_u², s_in² and s_out² the sample variances (divisor n − 1) of the audited, member and non-member samples and d̂ the difference of the two class sample means, the flip being D̂ < 0 at both depth cells, for a replicate whose vertex sits on the population's side, where D itself is positive; D̂ draws noise from all three samples, so the flip is not Â's event. At n = 200 and 1000 its rate is Φ(−D/sd(D̂)) within 4 standard errors, sd(D̂) by the delta method from the exact moments: 0.247 and 0.063 against 0.275 and 0.060 measured at (12, 3), n = 200 and 1000, and 0.137 and 0.007 against 0.132 and 0.003 at (2, 8); at n = 50 the normal reading of D̂ degrades (0.368 against 0.440 at (12, 3)). Holding Ĝ and B̂ at the population and reading sd(Â) alone gives 0.100 at (12, 3), n = 1000, 4.2 standard errors off. With the vertex in the box L is concave and the winning candidate decides. At the vertex the side is G > B; at an endpoint it is the endpoint; a lone in-box root keeps the level set on its branch; two in-box roots both zero the objective and the winner is floating-point residue. When the winner is the vertex or an endpoint, the side of 1/2 is fixed by the class samples alone: of the 565 replicates at the depth cells whose vertex lies in the box, 553 fall in the first three cases and 370 in the first two, so crossing into the box the estimator changes which sample decides. L(1/2) lies strictly inside L's range, so the flip and the event of no in-box root,  leaving that range, cut  against L̂ at two places: at n = 200 and 1000 a flip is a root drifting past 1/2 (0.92 to 1 of flips), not the silent fallback, which takes 0.384 of the flips at (12, 3), n = 50. The no-root rate is near the two normal tails of  past the range's edges, Ĝ and B̂ held at the population, the near margin 0.0160 and 0.0156 in units of the level: within 0.037 of measurement at n = 200 and 1000 (observation), the class samples' noise left out. A tangency puts an edge on A itself, out of the tails' scope. ## The audit's band counts one sample of three Tier: observation (two depth cells and three Gaussian worlds, κ = 1/5 and 4/5 at γ = 0 and κ = 1/5 at γ = 2, three sizes, 200 replicates of 200 resamples). Verifier: audit.py::section_band. The band resamples the audited set and holds the class samples fixed, so it omits their noise, which enters D̂ at weights 1/2, 1/2 and 1/4 and has no known sign. At the identified depth cells it covers 0.69 to 0.78 against its nominal 0.90 at every size, while it narrows with data (0.71 to 0.37 wide at (12, 3)). The same quantiles with all three samples resampled cover 0.87 to 0.95, above the band at all six rows: the omitted term is chiefly the held samples' noise, and no root choice is involved, the cells having one root in the box. The one-sided bound on the side of the box end nearest the truth fails there as well against 0.95, 0.83 to 0.92 (0.98 at (12, 3), n = 50), since the sign law covers the root set's term and never the counted spread's. At equal class variances the band covers 0.83 to 0.89, choosing a mode where the bound pair reports the set. ## With σ known, no self-report read off D̂ is unbiased at every distance Tier: theorem (in the normal model with σ known; the laws measured at two depth cells, three sizes, 1,000 replicate pairs each). Verifier: proof; selfreport.py::section_theorem, selfreport.py::section_statements. Off the box, with both vertices on one side of it, two audits of one world land on opposite sides of 1/2 exactly when their D̂s differ in sign (0 exceptions on 4,819 such pairs), an event the one unknown direction cannot touch, so an auditor can state their answer's instability from inside: Φ(−|D̂|/(ζσ̂)) at a scale ζ, or the pair form 2p̂(1 − p̂) with p̂ = Φ(−|D̂|/σ̂). The spread is solved: σ̂ from the replicate's own moments sits within 5% of the exact delta-method sd at n = 200 and 1000. The form is not. With ν = D/σ, signed, the **pair rate** is g(ν) = 2Φ(ν)Φ(−ν), even in ν, and a statement's mean is its folded law E[Φ(−|ν + ε|/ζ)], ε standard normal (within 0.010 of measurement at n = 200 and 1000). Without the fold this is Φ(−|ν|/√(1 + ζ²)), so noise in D̂ flattens every tail exponent and every ζ > 0 diverges against g. On the grid ν ∈ [0, 6] of step 0.01, each form tested is right in the mean at one distance |ν| = ν* (1.51, 1.13, 1.07, 2.18 for ζ = 1, ζ = √2, the pair form, and the pair form sharpened twofold, its argument doubled, each the grid point at or below the crossing), under-warns below it and over-warns above it; past ν = 6 no crossing was searched; the divergence above says only that each form over-warns far enough out. No form escapes. Take X = D̂/σ ~ 𝒩(ν, 1) with σ known; g, being even, does not see ν's sign, and no bounded function T of D̂ alone has E_ν T(X) = g(ν) for every ν. A bounded T's mean Γ(ν) = E_ν T(X) is entire in ν with |Γ(iξ)| ≤ sup|T|·e^{ξ²/2}, ξ real. But writing the pair as U = (X₁ + X₂)/√2, V = (X₁ − X₂)/√2, the signs differ iff |V| > |U|, so g(ν) = E 2Φ(−|U|) with U ~ 𝒩(√2ν, 1), and g(iξ) = e^{ξ²}ψ̂(√2ξ) for ψ(u) = 2Φ(−|u|)φ(u), φ the standard normal density, ψ̂(ω) = ∫ψ(u)e^{iωu}du its Fourier transform (a transform here, not an estimate). The kink of ψ at 0 gives ψ̂(ω) ~ 2/(πω²), so g(iξ) ~ e^{ξ²}/(πξ²), beyond any bound of the form C·e^{ξ²/2}. ∎ Given the auditor's own D̂, the truth takes the two values Φ(−|ν|) and 1 − Φ(−|ν|) by whether D̂ has D's sign (measured 0.136 and 0.892 against the predicted 0.137 and 0.863 at (2, 8), n = 200). A reliability diagram of any such statement therefore slopes, and the weights that would flatten it are functions of ν. ## The per-example audit states its Gaussian limit Tier: property (proved; the convergence measured at (12, 3), (2, 8) and the two-point world and bounded at (5, 5), the side census over every depth cell with 2 to 30 moves by 2 and by 3). Verifier: proof; lira.py::section_tool, lira.py::section_census. U-LiRA (arXiv:2403.01218, Algorithm 1) fits a Gaussian of mean μ₁ and variance s₁² to an example's scores under K shadow models that saw and then unlearned it and one of mean μ₀ and variance s₀² to its scores under K that never saw it, and states at the example's score o p = 𝒩(o; μ₁, s₁²)/(𝒩(o; μ₁, s₁²) + 𝒩(o; μ₀, s₀²)), 𝒩(o; μ, s²) the normal density at o, calling a member when p > 1/2. Its estimand is the equal-prior posterior τ, exact at a depth cell with the feature of the audit world (RULER.md#the-audit-world-and-its-two-roots), the number of moves by 2 in the three positions after the first. The fits converge to the class moments, so p converges to the Gaussian limit p_lim, and the calibration error converges to Σ P_u|p_lim − τ| over the scores, P_u a score's mass in the depth cell, a property of the model class that no shadow count removes; at a finite count it can sit below that limit (0.080 against 0.088 on the two-point world below at K = 32). At (12, 3), τ = (1, 11/15, 5/9, 3/7) against p_lim = (0.982, 0.840, 0.554, 0.414): the certain score is stated short of certainty and the tail score past its truth. Two support points {0, 3}, the member's masses 9/10 and 1/10 and the non-member's mirrored, give p_lim = 1/(1 + e^{−40/9}) = 0.988 against 9/10, because the Gaussian prices the far support point by a squared distance. At K = 512 the stated mean at every score of mass above 0.01 sits within 0.02 of p_lim, and the error within 0.015 of its limit, which runs from 0.0031 at (5, 5) to 0.088 on the two-point world; at (5, 5) that band is wider than the limit (0.0132 measured), so there the run bounds the error rather than showing it converge. The share of mass stated more extremely than τ reaches the limit's share only where p_lim clears τ by more than the noise. The call is a separate matter. Balanced accuracy is maximized by taking τ's side of 1/2, so the limit call reaches the ceiling iff p_lim and τ sit on the same side at every score with mass, τ = 1/2 counting as either side, and each exception costs |P_in − P_out|/2, P_in and P_out the two classes' masses at that score. ∎ The source names the Gaussian fit as a risk to accuracy; at (5, 5), (12, 3), (2, 8) and on the two-point world the call reaches the ceiling at K = 512, and the cost lands on the stated probability. The census finds 90 of 841 cells with an exception, the worst costing 11/1120 at (5, 12), where τ = 0.4898 and p_lim = 0.5019: in the census's cells no exception cost more than 11/1120, so it is common and cheap there. Off the depth column a far support point can move p_lim across 1/2 at a score far from a tie: member masses 0.45, 0.1, 0.45 at 0, 1, 10 against non-member 0.2, 0.8 at 1, 5 put τ = 1/3 at score 1 and p_lim near 0.65, an exception costing 1/20. ## Open fronts Where do orientation's and comparison's dead cells sit under the tilt? The tilted prior is read here only for the sign bit, the threshold and the magnitude class. Which cells does a prior beyond the tilt kill or revive, a two-block prior being the simplest? What gauge does a quantile task carry? Outside the normal model, or with σ estimated rather than known, is there a bounded self-report whose mean is the pair rate at every distance? And can an eval be built whose ceiling is known exactly while its floor is a functional of a process, held at a scale its user cannot enumerate?