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Prove the exact rank of additive Boolean disjointness repair - #4962

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aliromman:codex/publish-boolean-additive-repair
Oct 23, 2026
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Mister-Meeseeks merged 1 commit into
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aliromman:codex/publish-boolean-additive-repair

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This proves an exact rank formula for an F2 additive repair of the disjointness endpoint reported in PR #172. It gives a scoped obstruction to repairing that disjointness endpoint with a small number of copied scalar centres. No complete kappa improvement or new supplier is claimed.

Proof, exact rank receipt, independent agent review, and source pins.

For every integer h>=2, let A_h be the disjointness matrix on the nonempty proper subsets of [h], over F2. Then

rank_F2(I+A_h) = (2^(h+1) - 2*(-1)^h - 3)/3.

An additive repair to the identity by ell rank-one F2 terms therefore needs at least this many terms. Under the assumed ledger D=2v-3ell, with v=2^h-2, this gives D<=1 for even h and D<=-3 for odd h. The proof resolves this specified repair route; it does not rule out a different source basis, endpoint, nonadditive operation, or compiler. It does not assert that every physical correction has this additive form.

From the repository root, with Python 3.10+ and its standard library:

python3 -B research/boolean-disjointness-repair/verify.py

Expected final line:

PASS_BOOLEAN_DISJOINTNESS_REPAIR h=2..11 independent_kernels=2..4 no_new_kappa

Fresh replay passed exact binary elimination for h=2..11 and complete kernel enumeration for h=2,3,4. The separate review checked an alternate deletion proof, reconstructed set-based matrices for h=2..7, and rejected five altered proofs, formulas, kernel/parity conditions, or receipts. The wrapper and controls also passed with PYTHONOPTIMIZE=1. Finite checks support the written all-h proof; they are not its substitute. No Lean, complete supplier replay, repository-wide build, or hosted CI pass is claimed.

The base is cef4b52f2a59009865bf8eabde1be3f159063703. The consumed PR172 head is 427b23bc79489e7b2b7548bfd434dd9670fd5c47; its README supplies the endpoint and ledger. Its NOTICE credits @icekylinx for the PR #144 compiler and bookkeeping lineage and records Cline assistance. No predecessor implementation is imported.

Prepared by Ali Romman with substantial OpenAI Codex assistance, including separate agent proof review. This is not independent human review. New work is Apache-2.0. The package records the remaining changed-model question and its first required step: an exact physical identity with a complete cost account.

@jwr404

jwr404 commented Oct 23, 2026

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Independent review at head 378275c8e1228d7cdad4888d808a8b8c6395b32e

Scope. I checked PROOF.txt line by line and recomputed the ranks with my own bitset elimination. It does not import the package or its receipts.

Finite check: all agree. rank_F2(I + A_h) equals (2^(h+1) − 2(−1)^h − 3)/3 for every h = 2..14, which goes past the package's h ≤ 11, up to v = 16,382:

h:    2  3  4  5  6  7   8   9   10   11    12    13     14
rank: 1  5  9  21 41 85  169 341 681  1365  2729  5461  10921

Written proof: correct.

  • Full matrix. J = [[1,1],[1,0]] has J² = J + I and J³ = I over F2, so T_h = J^{⊗h} has T_h³ = I. In T_{h+1} = J ⊗ T_h, eigenvalue 1 arises from ω·ω² and ω²·ω, which gives a_{h+1} = 2^h − a_h, a_0 = 1 and a_h = (2^h + 2(−1)^h)/3. Since X³ − 1 has distinct roots over F4, T_h is diagonalizable there, so rank(I + T_h) = 2^h − a_h. Field extension preserves rank.
  • Block form. In the ordering (∅, proper, full), M = I + T_h acts as (1ᵀz + x_F, x_0·1 + B_h z, x_0 + x_F). The ∅ row is 1 + 1 = 0 on the diagonal, and the full set is disjoint only from ∅.
  • The annihilator. PM = (I + T + T²)(I + T) = I + T³ = 0 in characteristic 2. Pe_∅ = e_∅ + 1 + e_full, because T·1 has entry 2^{h−|s|} mod 2, which is odd only at s = [h]. This is exactly the indicator of the proper subsets, nonzero for h ≥ 2.
  • Kernel. The full row forces x_0 = x_F. On x_0 = 0 the kernel is {(0,z,0) : z ∈ ker B_h}, with 1ᵀz = 0 forced by P. Since B_h is symmetric, im B_h = (ker B_h)^⊥ ∋ 1, and (1, w, 1) with B_h w = 1 lies in ker M because P forces 1ᵀw = 1. So nullity(M) = nullity(B_h) + 1, which gives the formula.
  • Ledger. D = 2v − 3ℓ ≤ 2(2^h − 2) − (2^{h+1} − 2(−1)^h − 3) = 2(−1)^h − 1. ✓

The obstruction is scoped exactly as stated: additive rank-one F2 corrections to PR172's literal endpoint. I have no change requests.

This is an author-independent check. It is not maintainer acceptance, and κ is unchanged.

Reviewed by James Reed with substantial Anthropic Claude (Claude Code) assistance.

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3 participants