Hunt #13 (gate5_p6_a): property 6 is VACUOUS, and it is also not well posed

Verdict: VACUOUS. A rival satisfies property 6 in the very box chosen in advance to make that hardest.

Grade: measured (rung 1 of the ladder in AGENTS.md: one route, float grade, no enclosures). Nothing here is evidence for or against RH (docs/08). This is about what separates zeta from RH-violating look-alikes, and nothing more.

The property, and what was fixed before running

in a box strictly off the critical line, the completed function has no zeros

Operationalized as count_zeros_box(...) == 0 on the completed function, at dps = 20, over boxes fixed in MISSION.md and committed in 14dbb17, one commit before probe.py existed. The reasoning for the boxes is in that file; the short version is that box A was chosen adversarially. It is the one region in this repository where a rival is known in advance to have an off-line zero (the Davenport-Heilbronn zero 0.808517... + 85.699348...i pinned in zeta.epstein.OFFLINE_ZERO_RE / OFFLINE_ZERO_IM), so it is the box most favourable to a DISTINGUISHES verdict. Box B is the same shape at an unrelated ordinate, and exists to measure how far the verdict moves with the box.

MISSION.md also recorded a prediction before the run: VACUOUS. That is what happened.

The table

Counts are zeros of the completed function inside the closed rectangle, by the argument principle.

Box A (sigma in [0.70, 0.92], t in [85.55, 85.85]), the favourable box:

Box B (sigma in [0.70, 0.92], t in [70.10, 70.40]), the generic box:

Box B is settled too, and by its second row: the Davenport-Heilbronn function satisfies property 6 here. Its two Epstein cells did not finish, and cannot change that.

Controls, both run and reported before the verdict was read:

• Straddling-box control. xi over sigma in [0.20, 0.80], t in [13.60, 14.60] returned 1, the expected single zero at gamma_1 = 14.134725141734694. The contour machinery works.
• Known-answer control. The Davenport-Heilbronn count on box A returned 1, matching the zero pinned inside it. That column was known before the run, which is exactly why it is a control and not evidence.

Robustness of the one cell the verdict rests on, and it did not deliver. epstein_2_1_3 on box A was re-counted at halved and quartered forced-subdivision length (probe.py --confirm). Both re-counts hit the 420 s cap, on a box the same function completed in 29 s at the default step. So the decisive count of 0 was not independently reconfirmed at finer resolution, and this write-up does not claim it was.

What did hold is the routine's own guard, which is not nothing: the default step already forces every boundary segment under about 1/8 of a phase turn before the adaptive refinement runs, and count_zeros_box raises rather than rounds if the accumulated winding lands further than 1e-6 from an integer. The count of 0 passed that. Read the verdict at that strength, not higher.

Reading it

Box A settles the question in the direction that costs the property everything. battery's own rule is that a structure must be ungrantable to every counterexample: a claim zeta shares with any rival cannot be the load-bearing step, however many other rivals it excludes. The Epstein zeta of the non-principal discriminant -23 form has no zeros in box A. It satisfies property 6 there in precisely the sense zeta does. So property 6 is VACUOUS on box A, and the unfinished epstein_1_1_6 cell cannot change that: one survivor is enough, and a second survivor would only make the verdict more so.

That the favourable box came out vacuous is the point of having chosen it in advance. Any cheaper box has a strictly worse chance, so there is nothing to be gained by shopping for another one.

The larger finding: property 6 is not well posed as stated

Box A and box B give different answers for the same function. The Davenport-Heilbronn function fails property 6 in box A and satisfies it in box B. Property 6 is therefore not a property of a function at all. It is a property of a (function, box) pair, and the gate is being asked to read a one-bit verdict off a quantity that has a free parameter in it.

Resolve the free parameter either way and the property stops being useful:

• Quantify over all boxes. "For every rectangle strictly off the critical line, the completed function has no zeros" is, for zeta, exactly RH restricted to the off-line region. It cannot be verified for zeta, by anybody, so it cannot be the true half of a gate #3 comparison. A gate whose zeta column is unknowable decides nothing.
• Instantiate one box. Then it is vacuous, and measurably so, for a structural reason rather than a numerical accident: the off-line zeros of the three rivals sit at unrelated ordinates, so a box small enough to compute is overwhelmingly likely to miss at least one rival's zeros. A box that caught an off-line zero of all three at once would have to be large, and confirming zeta's column in a large box is the expensive direction. That is the same wall the previous attempt hit at 50 minutes on 0.6+80i to 0.9+90i, and it is not an artifact of that particular box.

So the answer to the question as put is VACUOUS, and the reason it is vacuous is not "we picked a bad box". It is that no affordable box can make it otherwise.

What a well posed version would be

Replace the box with a fixed region, chosen so that zeta's column is a theorem rather than a computation:

the completed function has no zeros in Re s > 1

• Zeta's column is settled and provable. zeta(s) != 0 for Re s >= 1 (Hadamard and de la Vallée Poussin, 1896), and xi(s) = (1/2)s(s-1)pi^{-s/2} Gamma(s/2) zeta(s) has no extra zeros there, since Gamma never vanishes and s(s-1) vanishes only at s = 0, 1. No computation, no height limit, no box.
• The rivals' column is the classical Davenport-Heilbronn theorem (1936): both the Davenport-Heilbronn function and the Epstein zeta of a positive definite binary form of class number greater than one have infinitely many zeros with Re s > 1. This is quoted from the literature and was not recomputed here (see "what I could not settle").
• It is box-free, so it has no free parameter for a verdict to hide in.

This version DISTINGUISHES. But it should be adopted with its own caveat attached, because it distinguishes for a reason the module already names: zeta's nonvanishing in Re s > 1 follows immediately from the Euler product, absolutely convergent there. So the well posed version is a restatement of "has an Euler product", which is already property 4 or 5 territory. It is a correct gate input and it is not an independent one. A gate #3 property earns its keep by isolating a structure other than the Euler product; this one does not, and saying so is more useful than banking a DISTINGUISHES.

What I could not settle

• Three of the eight cells did not finish inside their 420 s caps (epstein_1_1_6 on box A, both Epstein cells on box B) and are recorded as gaps. None of them affects either verdict, since one survivor settles a box. The single-point cost probe puts the two Epstein functions at about 1 s per evaluation against 0.0099 s for completed_dh and 0.0009 s for xi, so the Epstein contour work is roughly 100x the Davenport-Heilbronn contour work and 1000x the zeta contour work. That ratio, not the box, is what makes this question expensive.
• The decisive cell was not reconfirmed at finer resolution, as above. A count that cannot be re-run at half the step inside seven minutes is a count with one route and one resolution behind it.
• The rivals' half of the proposed well posed version was not verified here. Box C (sigma in [1.02, 1.60], t in [0.50, 50.50]) was fixed in advance for exactly that purpose and gated on a measured cost projection. The gate said no, on numbers: the measured per-evaluation cost is 0.0009 s for xi, 0.0099 s for completed_dh, 0.9895 s for epstein_2_1_3 and 1.0154 s for epstein_1_1_6, and box C forces about 747 boundary segments, projecting 2.7 s and 29.6 s for the first two against 2957.7 s and 3035.1 s for the two Epstein functions. So box C was not attempted. That is the preregistered cost rule doing its job rather than a result. Finding an explicit zero with Re s > 1 for each of the three rivals would upgrade the proposal from "quoted theorem" to "quoted theorem plus a witness in this tree", which is what this repository normally requires of a claim it leans on.
• Nothing here bounds where the rivals' off-line zeros actually are. The claim "an affordable box cannot catch one of each" is an argument from their being at unrelated ordinates plus the two boxes measured, not a proven density statement.

Loose threads

• The battery's property-6 column is not literally the same claim across the four functions. zeta.epstein.dh_interface wires its count_zeros_box to the uncompleted dh_f, while zeta_interface uses xi and epstein_interface uses epstein_completed. In every box used here the gamma factor of completed_dh is finite and non-vanishing, so the two agree and nothing above is affected. In general they need not: the gamma factor has poles at s = -1, -3, -5, ..., and a box reaching into Re s < 0 would count different things in different columns. This probe used completed_dh explicitly to keep the four columns identical. Worth an issue.
• Where the Epstein contour cost actually goes. The two forms cost the same per evaluation (0.9895 s and 1.0154 s at 1.3 + 25i), yet (2,1,3) finished box A in 29 s while (1,1,6) did not finish in 420 s, and halving the step for (2,1,3) on the same box turned 29 s into a cap hit. Since the forced subdivision count is identical for both, the blowup is in the adaptive refinement of _arg_variation, which means the phase of epstein_completed is turning much faster along those edges than the step heuristic assumes. That heuristic is derived from the phase-turn rate of the Davenport-Heilbronn function and is simply the wrong scale for an Epstein zeta. A step matched to the Epstein density would probably make all three unfinished cells cheap, and it is the single change most likely to make this question affordable.
• Where the Davenport-Heilbronn zeros with Re s > 1 actually are. Spira 1994 computed off-line zeros of the Davenport-Heilbronn function; this tree pins exactly one, at Re = 0.8085, which is not in Re s > 1. A witness in Re s > 1 for each rival would let the proposed well posed version stand on something this repository measured rather than quoted.
• The gate itself has a shape problem worth naming. Three of the six battery properties came back vacuous and this one is vacuous and ill-posed. A gate input whose zeta column is unknowable is not a gate input, and it may be worth making that a stated admission rule for the battery rather than something each hunt rediscovers.