[JosieElliston]

Change and mathematical scope

This PR adds an independent adversarial audit of #6114 (Matarener, head 3183c6553e), in
research/p104-controller-income-audit/. It makes no new κ claim and leaves the supplier and the selected
record unchanged.

#6114 proves a theorem about the literal p16 Laurent caller alphabet of
research/p104-productive-controller/caller.py. From a constant invertible boundary, a word with s paid odd
shears twists at most s − 1 outputs. So controller income per port is at most thick − 1, and the retained
controllers attain it (17 = 18 − 1).

• The theorem holds. I re-derived it. A separate exact engine, which does not import caller.py or the
  Controller income is capped at full calls minus one: even-part rank bound in the p16 Laurent alphabet (scoped negative result; no new kappa) #6114 code, reproduces every number on all three retained controllers: 18 thick + 15 thin shears, 32 twisted
  outputs, even-rank trace 32, 32, 31, …, 0, and 194 lifts. The engine runs a symbolic Laurent replay to the
  declared diagonal. It also tracks the even rank after every event under Q² → 13/7, 19/11 with random nonzero
  stage values; Controller income is capped at full calls minus one: even-part rank bound in the p16 Laurent alphabet (scoped negative result; no new kappa) #6114 uses D_s → 1 with 7/3 and 11/5.
• Two gaps in the written proof, now closed (PROOF.md A1–A2). (1) The final inequality uses zero rows of
  L·M rather than M. A left-kernel argument repairs it without assuming L is invertible. (2) The invoice
  step "every output stage was paid once as a lift" was asserted. It follows from the determinant, stage by
  stage, and the verifier checks it on all 33 stages plus the central exponent.
• The scope claim needs correcting (A4). Controller income is capped at full calls minus one: even-part rank bound in the p16 Laurent alphabet (scoped negative result; no new kappa) #6114 says a paid primitive touching several banks or ports is
  "the only way" to escape the bound. For a paid job A = A_e + Q·A_o, Sylvester's inequality bounds the
  even-rank drop by (n − rank A_e) + rank A_o. Before any odd content exists, the bound is n − rank A_e.
  • Any job that writes one row, y ← u·y + Σ f_j·x_j with u nonzero and even, obeys Controller income is capped at full calls minus one: even-part rank bound in the p16 Laurent alphabet (scoped negative result; no new kappa) #6114's bound
    exactly, including the − 1. This holds for any number of source banks and any Laurent coefficients f_j.
  • Three escapes exist, each with an exact control. A paid one-bank odd translation gives 1 twist from 1 job.
    A Laurent entry boundary gives 2 twists from 2 shears. A two-row job gives 2 twists from 1 job.
  • None of these is in the literal alphabet, so Controller income is capped at full calls minus one: even-part rank bound in the p16 Laurent alphabet (scoped negative result; no new kappa) #6114's 17 = 18 − 1 conclusion stands.

Not audited: the surplus formula M = v·income − n_thick·c with c = 56 and M = 760 on PR5907, and the moment
side.

Task and overlap

The task came from the adversarial-audit track. The audited result belongs to the joint-query-controller card;
TASK.json records the scope change and the reason for it. Base: 721c03822c421b7f7fcbb2598993b947c4d88419.
When this PR was opened, #6114 had no comments or reviews, and the open PRs and the overlap cache showed no other
audit of it. #6097 is being audited separately (session 47) and does not overlap this work.

Validation

python3.12 -B research/p104-controller-income-audit/verify.py
python3.12 -B scripts/research_package.py --package research/p104-controller-income-audit

The verifier uses only the standard library and takes about 0.6 s. It passes on Python 3.10–3.14, both normal
and -O. RESULT.json is compared in full. Random words with one-destination jobs (n = 2, 4, 6; 3,909 paid
jobs) never exceed the bound. In 41 of them the even-rank deficit n − rank E equals s − 1; this counts only
the rank deficit, not constructed outputs. The test supports the A4 proof; it does not replace it. An independent
read-only OpenAI Codex review found no counterexample to A1, A2 or A4 and confirmed that the engine's semantics
match caller.py. Its two wording fixes are applied: the condition that u is even, and the label of the
random statistic.
No selected-result replay was run, because nothing in the selected construction changes.

Open next questions

1. Can a physical paid odd translation Q^{±1}, or a supplier boundary that carries odd content, be realized at
   the P104 caller? Either would lift the cap to thick per port: up to 104 rank units on PR5907, before the
   primitive's own charge.
2. Before any search, compute the twist capacity (n − rank A_e, rank A_o) of each priced multi-row primitive.

Credit

The theorem is by Matarener (#6114). The controllers are by Cinna Davis (p16/a7 lattice) and Ali Romman (PR5398
replay, all-forward wrapper). The binary predecessor is Cinna Davis's research/productive-controller-basis-search.
The audit is by Josie Elliston, prepared with Claude Code (Opus 5.5). Apache-2.0.

🤖 Generated with Claude Code