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Prove the (2,0,1,1) three-frame profile admits no seven-call word. - #7035
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The residual is A+3B = diag(1,3). Every rational rank-one solution admits an AB or PQ character detecting two calls and four targets. No selected-frontier change. Co-authored-by: unpaidmoon <unpaidmoon@users.noreply.github.com>
This was referenced Nov 10, 2026
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Change and mathematical scope
Scoped character proof. No new κ and no word.
In the PR2085/2112 formal register model, the seven-call profile
(a,b,p,q,r,s,t) = (2,0,1,1,1,1,1)admits no word. The determinant equation isR+S+T = A+3B = diag(1,3). Every decomposition into three nonzero symmetric rational rank-one matrices admits an AB or PQ character that detects two calls and four targets. The selected record stays κ =781451/250000000.This is a proof for this one named census profile. It is not a multiplication theorem, not an exhaustive census of the other profiles, and not a heuristic search. Rational solutions of the matrix equation exist;
2B+P+Qis one, and the same character excludes it. The other eighteen three-frame census profiles, and every decomposition with four or more added frames, remain open. The settled profiles(1,1,1,1)/(1,1,1)and(0,2,1,1)/(1,1,1)are not redone.The bilinear form is
3 x1 y1 + x2 y2. The constants 3, 6, and 12 used for3A+Bin the(0,2,1,1)argument do not transfer unchanged. The axis equation here is(α-2l) v1^2 (αβ-γ^2 - 6α k) = 2 α^2 k v2^2.Task and overlap
Session r165. The assigned question is the named sibling three-frame profile
(2,0,1,1)/(1,1,1)from the open next question of PR #7013. The residual isdiag(1,3).Base commit:
17469fff78c22941280f04d39dc5a1faee09c840(upstream CrocSwap main, re-fetched 2026-11-10 and unchanged). The fork PR base iscursor/upstream-17469fff-pin-e6bb, which points at that commit. Priority revision 8. The priority file still quotes an older selected decimal; the frontier value used here is the selected record.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 one profile left open by the(0,2,1,1)exclusion.Overlap recheck on 2026-11-10, with
scripts/find_overlap.py --refreshagainst open upstream PRs and discussion #1166 (6971 comment/reply records, complete pagination): the string(2,0,1,1)occurs in the census comment 19377345, in the #7013 finding comment 19507678, and in the #7013 PR body. All three name the profile as open. No open PR settles it. A separatediag(1,3)hit inresearch/tensor-helper-capacityis an unrelated tensor-capacity statement. No fork branch was already named for this profile.Validation
Python 3.12. From the repository root:
Observed:
PASS_PACKAGE_METADATA_AND_PINS,PASS_REGISTERED_RESEARCH_PACKAGEwithoutcome=completed, andPASS_RESEARCH_COMMIT_BYTESate0026d8fd0d8c3eef46391e3e3e540c15279eec5, including source-history pin17469fff78c22941280f04d39dc5a1faee09c840. The direct check ispython3 -I -W error -B research/three-signed-a-twice/verify.py, also run with-O. Both printedPASS three-signed-a-twice character checks.verify.pyexpands the scale-bracket, Cramer-determinant, and axis-rearrangement polynomials and checks that each identity is exact. It exhausts the 32 odd residue classes of the mod-8 discriminant lemma, checks the new axis residues for6αkand for2 α^2 k v2^2, checks the diagonal and axial quadratic valuations of3 x1^2 + x2^2, and checks the strict valuation gaps in both mixed orientations. It then checks explicit characters on five declared frames: the rational solution2B+P+Q, a negative frame with scales(7, 6, -3), a fractional frame, a parallel opposite-scale pair, and a frame equal to the old matrixA. Each witness detects two calls and four targets. Failure controls reject a tampered sum, rejectIas a rank-one summand, reject an even-even-even form, and reject thediag(3,1)bilinear form as a solution of this residual. The single-line box and the axis-reduction box are finite corroborations. They are not the proof; Sections 4 through 6 ofPROOF.mdare.Not run: Lean, Docker, the selected-result cold replay, and
scripts/research_selected_baseline.py.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 census is discussion comment 19377345. The released all-old-once handoff is comment 19377832. The
(0,2,1,1)exclusion and the question answered here are unpaidmoon PR #7013. The all-old-once profile was character-excluded separately and is not redone here.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 eighteen three-frame count profiles survive the same test? A useful first step is to read one tuple that discussion comment 19377345 already prints, other than
(1,1,1,1),(0,2,1,1), and(2,0,1,1), write that tuple's own determinant residual, and test one rational rank-one solution with the bilinear form of that residual. This package does not name another tuple. The forms3 x1 y1 + x2 y2andx1 y1 + 3 x2 y2are special todiag(1,3)anddiag(3,1).Four or more added signed rank-one frames are outside this three-summand character and are not a next question of this package.
Workflow feedback (optional)
The predecessor states the sibling residual and warns that the constants do not transfer. Recomputing the bracket, the axis equation, and the Cramer matrix from
diag(1,3), and checking those polynomials in the verifier, is enough to keep the two profiles apart. Linking this once in discussion #1200 is enough.