[unpaidmoon]

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, every owner-indexed D-exponent-0 left map of the four final banks that keeps x0 = D1 Q^{-1} has x1 coefficient

(const - m0 C1) D1 + (1 - m0 x) Q^{-1} C3 D1 D3

with x = Q^2 (I + Q) C1 Q and const = Q^{-1}(C1 Q C1 - I). The second summand is never zero. The Q-letter parity character sends I + Q to 0 and Q^{-1} C1 to -1, so m0 x = 1 is impossible. Keeping y0 zero fixes m1 = const - m0 C1 and does not cancel the summand. Banks 0, 1, and 2 have zero y1, and m3 = 0, so y1 stays zero. The four stored banks are only read.

Evidence class: a proof in the predecessor's formal algebra. The multiplier 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.

The unique index that also clears the D1 line is m0 = const C1, m1 = 0, m2 = Q^{-1} C1 Q^2 C1 Q. Its fresh image keeps x0, y0, and y1 as required and leaves the single monomial Q^{-1} C1 Q^2 C1 C3 D1 D3. That image removes the three published keys and is still not the autonomous row [A, 0, 0, 0].

Task and overlap

Base: 17469fff78c22941280f04d39dc5a1faee09c840. Task: joint-query-controller, scope refined. This answers the x1-column question left by merged PR #6951 and research/owner-indexed-y0-clearance. It does not redo the y0 defect (#6951), the declared-family coupling lock (#6945), or the one-edge C1 Q row (#6961).

Overlap snapshot on 2026-11-10, via scripts/find_overlap.py: local notes at that base (10178 paths, complete inventory), 317 open-PR records fetched at 08:49:48Z, and 6971 discussion records fetched at 08:50:49Z. Requiring both phrases three-term x1 and owner-indexed matches the predecessor package and discussion reply 19507118, which ask this question, and no open PR. Closed PR bodies were outside that snapshot. The search is not a novelty certificate.

Validation

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

python3 -B scripts/research_package.py --package research/owner-indexed-x1-obstruction --check-only
python3 -B scripts/research_package.py --package research/owner-indexed-x1-obstruction

Both registered commands passed locally, including the optimized run. Checks are explicit need failures, not assertions. The run prints PASS_OWNER_INDEXED_X1_OBSTRUCTION and new_kappa None.

Not run: multipliers with a nonzero D exponent, owner-mixing maps, any feedback operator other than the published pair, quotients by further physical relations, and any tape or minimum field cost.

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 and constant-index charge are research/owner-indexed-y0-clearance. 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.

Open next questions

The package records two follow-ups.

Can one owner-indexed multiplier whose D exponent is not zero cancel Q^{-1} C1 Q^2 C1 C3 D1 D3 while keeping x0 = D1 Q^{-1}, keeping y0 zero, and leaving the four banks stored? The first step is to add one term of D exponent (0,-1) to the sharp index above and recompute the four columns. This proof separates the D1 and D1 D3 lines only for a D-exponent-0 factor.

Does one physical relation omitted by this formal algebra send that residual word to zero without collapsing Q^{-1} C1? The first step is to name one source relation outside Q^4 = C1^2 = C3^2 = I and the inherited D conjugation rule, then reduce only that word. No such relation is imposed here. A quotient that kills the residual would still not be an autonomous caller.

Workflow feedback (optional)

The x1 column is an identity once the four banks are expanded, so a bounded word search is the wrong instrument. The parity character finishes the infinite family in one step. Link once: #1200