[unpaidmoon]

Change and mathematical scope

Support classification for one equation, plus an integral character obstruction. No new κ and no word.

Fix one old projector E in {A,B,P,Q}. Five nonzero symmetric rational rank-one matrices sum to 4I-E and meet all four targets as subset sums if and only if three of them are the complementary projectors and the other two are a rational rank-one splitting of 2I. The integral splittings are exactly {2A,2B} and {2P,2Q}. Each integral solution detects two calls and four targets under the inherited P-coefficient character, so it is not a word. One pairwise nonparallel splitting, at chart parameter t=2, is a support witness only.

The selected record stays κ = 781451/250000000. This package does not change the selected frontier and does not lower the eight-call minimum.

Evidence classes, kept apart:

1. Construction proof. For each projector E, the three complementary projectors plus any rank-one splitting of 2I meet support. The identity Q(2I-X)-Q(X)=4(2-tr X) classifies those splittings. Integral points are read off the discriminant 4-c^2.
2. Exhaustive finite census, hence the converse. Subset sizes 2, 3, and 4 give 230300 unordered 4-tuples. Twins: 22370. Rank 3: 2450, all inconsistent. Full-support kernels: 3840, all killed by the trace test 2,3,4 ≠ 7. One frozen frame: 27840, none on the cone. Three frozen frames: 53400, of which 240 are consistent and on the cone, in bijection with placements of the three complementary projectors. Invertible: 120400, zero cone hits.
3. Explicit witnesses, not a word. Integral witness A,P,Q,2A,2B with old call B. Nonparallel t=2 frames (-2/5,4/5,8/5) and (12/5,6/5,-8/5).

Task and overlap

Base commit: 17469fff78c22941280f04d39dc5a1faee09c840 (upstream CrocSwap main, “Checkpoint completed November 10 queue review and remaining obligations”).

Task card: bit-new-register-model, priority revision 8. Scope changed. The card asks for a larger or nonabelian native alphabet. This package stays in the abelian symmetric 2-by-2 model and answers open question 2 of unpaidmoon PR #7024 (discussion comment 19514165): can five added signed rank-one frames plus one old call meet support at length 6? First step was the equation R+S+T+U+V=4I-E.

Concrete object: one fixed projector plus five free rank-one frames. Domain: the 230300 unordered 4-tuples of subset masks of sizes 2, 3, and 4, for each of the four projectors and each target assignment.

Not redone: the five-call 4I theorem of #7024, the four-call theorem of merged #6952, six frames with no old call, and four frames plus old calls. A second old call is not added. No open upstream PR states this equation. Discussion #2091 fixes one fifth address in a different alphabet and is not contradicted, because no word is exhibited.

Validation

Environment: Python 3.12. Registered replay:

python3 scripts/research_package.py --package research/five-signed-one-old --check-only
python3.12 scripts/research_package.py --package research/five-signed-one-old

Both registered commands passed (PASS_REGISTERED_RESEARCH_PACKAGE, outcome completed), including the python -O rerun. Direct check: python3 -I -W error -B research/five-signed-one-old/verify.py (about 90 seconds). Checks use require, so -O still executes them.

The census is an exhaustive finite classification, not a heuristic search. The splitting identity and the construction are proofs inside the inherited model. The determinant identity, subset-sum necessity, and character inequality are inherited from research/bit-two-signed-frame-floor and are not reproved. Predecessor replays, Lean, Docker, and the selected-result cold replay were not run.

Failure controls: I, zero, and A+P are not rank one; shifting the total by A stops recovery of the t=2 witness. Positive control: the invertible solver recovers the t=2 witness from its own subset sums.

Credit

The register model is Cinna Davis's PR2085/2112 formal model. The determinant, subset-sum necessity, and character inequality are the PR2910 package as recorded in apoursh's two-signed-frame floor, PR6190. The three-frame all-old-once character is research/three-signed-old-once. The four-call support proof is research/four-signed-once-parity, merged as #6952. The five-call census and this open question are #7024, discussion comment 19514165. None of those theorems is redone.

Prepared by unpaidmoon with substantial Cursor Grok assistance. No independent human or formal-kernel review. New work is Apache-2.0.

Open next questions

Can one rational splitting of 2I make this length-6 support solution into a word? A word would lower the eight-call minimum. The integral pairs already fail the P-coefficient test. The smallest follow-up is the same count on the displayed t=2 frames (-2/5,4/5,8/5) and (12/5,6/5,-8/5), with old call B and frozen frames A, P, and Q. Passing that count would still not be a word: support is necessary, not sufficient. Six added frames with no old call, and four added frames plus an old call, are different equations.

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