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Character-exclude the all-old-once three signed frames - #6942
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Mister-Meeseeks merged 2 commits intoNov 10, 2026
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Every rational decomposition of 2I into three nonzero symmetric rank-one matrices admits an AB or PQ character detecting two calls and four targets, so the seven-call profile (1,1,1,1,1,1,1) has no word. The selected kappa is unchanged. Co-authored-by: unpaidmoon <unpaidmoon@users.noreply.github.com>
Discussion 1166 still ends at comment 19481299, so the all-old-once handoff remains unclaimed. Co-authored-by: unpaidmoon <unpaidmoon@users.noreply.github.com>
This was referenced Nov 10, 2026
Prove five once-each rank-one frames miss the five-call endpoint.
unpaidmoon/integer-mult-bounds#199
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Change and mathematical scope
Scoped character proof. No new κ and no word.
Every rational decomposition
2I = R+S+Tinto three nonzero symmetric rank-one matrices admits a first-order character that detects two calls and four targets. In the PR2085/2112 model, the seven-call profile(a,b,p,q,r,s,t) = (1,1,1,1,1,1,1)therefore admits no word. The selected record stays κ =781451/250000000.The character is an integer form of shape
(odd,odd,even)or(even,even,odd)on the coefficient vector(a,b,c). The mod-2 image of the integral relation lattice misses(1,1), so at least one shape extends off that lattice. The extension detects either{A,B}or{P,Q}and all four targetsA+P,A+Q,B+P,B+Q. The inherited inequality "detected calls ≥ detected targets" is used as recorded by PR2910 and is not reproved.This is a complete argument for this one profile. It is not a multiplication theorem, not a search failure, and not a certificate that changes the selected frontier. The other twenty three-frame census profiles, and every decomposition with four or more added frames, remain open.
Task and overlap
Session r108 of the parallel donation. Ordered eligible list after excluding claimed, delivered, and open unpaidmoon work, following the r105 order and rechecked against discussion #1166 through comment 19481299 on 2026-11-09T13:58:36Z:
p104-side-correction,coverage-law-change).bit-helper-contract).supplier-boundary-audit).uniform-transfer-audit) — taken earlier by r105 / Record the scalar prime support of the selected PR6030 word #6928.replay-runtime).bit-new-register-model.Base commit:
7c49fe32c55de02c2c873655c804b16e4c3fef0a(upstream CrocSwap main). The fork PR base iscursor/upstream-7c49fe32-pin-a03a, which points at that commit. Priority revision 8.Task card:
bit-new-register-model,scope_status: changed. The card asks for a larger or nonabelian native alphabet. This package stays in the abelian symmetric 2-by-2 model and answers the three-frame question left open byresearch/bit-two-signed-frame-floor(PR #6190), for the single all-old-once profile. That floor's stop rule was two added frames. Open PR #6799 is a different seven-call question. No unpaidmoon branch answers comment 19377832.Validation
Python 3.12. From the repository root:
Observed:
PASS_PACKAGE_METADATA_AND_PINS,PASS_REGISTERED_RESEARCH_PACKAGEwithoutcome=completed, andPASS_RESEARCH_COMMIT_BYTESat84f70e14432a975bcdaeae37fcfecbf3890bda67. The direct check ispython3 -I -W error -B research/three-signed-old-once/verify.py, also run with-O. Both printedPASS three-signed-old-once character checks.verify.pyexpands the cleared scale-formula polynomial and checks that it is identically zero, exhausts the 32 odd residue classes of the mod-8 discriminant lemma, and checks explicit characters on nine declared frames. Each witness detects two calls and four targets. Failure controls reject a tampered sum, rejectIas a rank-one summand, and reject an even-even-even form. The finite fixture list is not the proof; Theorems 5 and 6 inPROOF.mdare.Not run: Lean, Docker, the selected-result cold replay, the full predecessor replay of PR #6190, and any search of the other twenty profiles.
Credit
The register model is Cinna Davis's PR2085/2112 formal model. The determinant and first-order inequality are the PR2910 package as recorded in apoursh's two-signed-frame floor, PR #6190. The concrete handoff is discussion comment 19377832, following the census in comment 19377345 and the claim in comment 19377792.
Prepared by unpaidmoon with substantial Cursor Grok assistance. No independent human or formal-kernel review. New work is Apache-2.0.
Open next questions
Which of the other twenty three-frame count profiles from the seven-call census survive the first-order character test? Only
(1,1,1,1)with three added labels used once is settled here. A useful first step is to take one other profile, write its determinant equation, and test the same AB/PQ obstruction on one rational rank-one solution. No reduction of those profiles toR+S+T = 2Iis supplied.Can four or more added signed rank-one frames lower the eight-call minimum? The two-summand sign reduction and this three-summand character both stop at the number of added frames they assume. A useful first step is four primitive directions and the shortest determinant equation that uses each new label once.
Workflow feedback (optional)
The handoff's exclusion criterion,
A−Poutside the lattice generated byR,S, andA−Bmodulo2Λ, is the right obstruction, and the character proof is the global form of that test. A raw valuation screen on the uncorrected scales is not a proof: some triples need an even correction before either family is integral. Linking this once in discussion #1200 is enough.