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Prove a D-exponent (0,-1) owner index leaves the x1 residual - #7052

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@unpaidmoon unpaidmoon commented Nov 10, 2026 •

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Change and mathematical scope

Scoped proof. No new κ. The selected conditional value stays 781451/250000000 from the selected record (PR #6030). This does not change the selected frontier.

On the published two-edge feed, start from the sharp D-exponent-0 index of open PR #7033 and add arbitrary D-exponent-0 coefficients times D3^{-1} on the four final banks. Every such correction leaves x1 coefficient 1 on

Q^{-1} C1 Q^2 C1 C3 D1 D3

because D3^{-1} moves every bank-x1 degree off (1,1) and a D-exponent-0 factor preserves degree. Keeping x0 = D1 Q^{-1} and y0 zero forces

u1 = -u0 C1,  u2 = -u0 x,  u3 = 0,

with x = Q^2 (I + Q) C1 Q and u0 an arbitrary D-exponent-0 element. The x1 column is then the residual plus u0 times a fixed three-monomial defect. One bank forces u0 = 0. The four stored banks are only read.

Evidence class: a proof in the predecessor's formal algebra. The coefficient space is infinite. This is not a word census, not a bounded search, and not a theorem that the monomial vanishes in every physical quotient or at every other D exponent.

The witness u0 = I charges four new multiplier monomials, D3^{-1}, -C1 D3^{-1}, -Q^{-1} C1 Q D3^{-1}, and -Q^2 C1 Q D3^{-1}, and three new image monomials. They do not remove the residual. No scalar-field minimum and no tape cost are claimed.

Task and overlap

Base: 17469fff78c22941280f04d39dc5a1faee09c840. Task: joint-query-controller, scope refined. This answers the D-exponent (0,-1) question left by open PR #7033. It does not redo the D-exponent-0 obstruction (#7033), the y0 defect (#6951), the declared-family coupling lock (#6945), or the one-edge C1 Q row (#6961). The omitted-physical-relation question of #7033 is not taken.

Overlap snapshot on 2026-11-10, via scripts/find_overlap.py --refresh --all-terms: local notes at that base (10178 paths, complete inventory), 339 open-PR records fetched at 10:15:04Z, and 7013 discussion records fetched at 10:16:06Z. Requiring both phrases D exponent (0,-1) and owner-indexed matches only discussion reply 19513867, which asks this question. PR #7033 states the same question with the exponent in backticks, so that literal pair misses its body; the PR was read directly. The search is not a novelty certificate. Closed PR bodies were outside that snapshot.

Validation

Python 3.12, standard library only. From the repository root:

python3 -B scripts/research_package.py --package research/owner-indexed-d3-inverse-term --check-only
python3 -B scripts/research_package.py --package research/owner-indexed-d3-inverse-term

Both registered commands passed locally, including the optimized run. Checks are explicit need failures, not assertions. The syllable-length-4 replay is a control, not a census of u0. No network, native build, or full multiplier replay is used.

Credit

The trace, algebra, and published row are Cinna Davis's research/coded-active-feedback, using Dugongue's PR4948 chronology, Dillon's PR4933 code, and the passive ceiling packaged by Chris R / Clear1nsight. The y0 defect is research/owner-indexed-y0-clearance. The sharp index and the residual monomial are open PR #7033, head bc08ea6b28c345b0164ff1631642974719da087a. The declared-family obstruction is research/active-feedback-coupling-lock. The one-edge C1 Q row is research/active-feedback-outside-edge. The gauge engine remains research/coded-native-caller/gauge.py.

Prepared by unpaidmoon with substantial Cursor Grok assistance. Apache-2.0.

Open next questions

Can a D-exponent (0,1) correction of the same sharp index, using two or more banks, make the fresh x1 column zero while keeping x0 = D1 Q^{-1} and y0 zero? One homogeneous term on one bank is already blocked for every degree by the monomial gate proved here. The same coefficient shape keeps the boundary at D3, and for u0 = I the degree-(1,1) line properly contains the residual. The first step is to decide whether any D-exponent-0 u0 makes R + u0 Δ_+ the zero element, and to charge every monomial of a solution. No such u0 is solved here.

Workflow feedback (optional)

The registered package checker accepted the scoped proof without a separate session report. One link is enough: #1200

Every D3^{-1} correction of the sharp index keeps coefficient 1 on
Q^{-1} C1 Q^2 C1 C3 D1 D3. The boundary-preserving family is
u1=-u0 C1, u2=-u0 x, u3=0. No kappa change.

Co-authored-by: unpaidmoon <unpaidmoon@users.noreply.github.com>
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