Prove one crossing path keeps the proper-terminal rank-bill maximum and four break it (no κ)

research/proper-terminal-helper-rank-bill/PROOF.md

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+# Proof: proper helper terminals and the rank bill
+
+**Proof. No new κ.** The selected conditional exponent stays
+`363005817/100000000000`. No word is emitted and no cut is declared legal.
+
+## Hypotheses
+
+Fix a real exponent `0 < t < 1` and integers `0 < T < D`. A helper boundary
+runs from the zero subspace to a proper terminal of dimension `T`, strictly
+inside the ambient `D`-dimensional space. The helper is cut at an intermediate
+dimension `x`, using `m ≥ 1` contact-free transport paths. Path `i` has integer
+endpoints `0 ≤ a_i ≤ b_i ≤ D` and old charge `(b_i - a_i)^t`. A path with
+`b_i > T` leaves the boundary.
+
+Write `h(u) = u^t` for `u > 0` and `h(0) = 0`. On the closed feasible interval
+`[A, B]`, with `A = max a_i` and `B = min(T, min b_i)`, every leg length is
+nonnegative and the rank bill is
+
+```text
+G(x) = h(x) + h(T-x) - h(T)
+       - Σ_i ( h(x-a_i) + h(b_i-x) - h(b_i-a_i) ).
+```
+
+This is the #7283 formula with the helper dimension `D` replaced by the
+terminal `T`. A zero leg is omitted from a physical word; the formula keeps
+`h(0) = 0`. The value `G` is the rank-bill factor only. It is not an ordinary
+cost, a legality test or a supplier invoice.
+
+The interval endpoints are feasible dimensions. `A` is the largest path start,
+and `B` is the terminal when at least one path crosses it, otherwise the
+smallest path end. Joins or meets can only shrink `[A, B]`, and the argument
+below applies to every closed subinterval on which all legs are nonnegative.
+
+## Lemma 1 (concavity and its derivative)
+
+For `0 < u < v`,
+
+```text
+h'(u) = t u^(t-1) > t v^(t-1) = h'(v).
+```
+
+Both sides are positive because `0 < t < 1`, and `u^(t-1) > v^(t-1)`
+because `u < v` and `t - 1 < 0`. Hence `h` is strictly concave and `h'` is
+strictly decreasing on `(0, ∞)`. The verifier checks the equivalent integer
+form: for `t = p/q` in lowest terms, `s = q - p > 0`, and `u ≤ v` implies
+`u^s ≤ v^s`.
+
+The same strict monotonicity gives the increment form used below. For
+`0 ≤ y < z` and `x > 0`,
+
+```text
+h(z) - h(z-x) < h(y+x) - h(y)
+   if and only if z-x > y,
+```
+
+because both increments are `∫_0^x h'(base + s) ds` and `h'` is strictly
+decreasing. We use the case `y = T-x`, `z = b` with `b > T`.
+
+## Lemma 2 (subadditivity)
+
+For `u, v ≥ 0`,
+
+```text
+h(u+v) ≤ h(u) + h(v).
+```
+
+For positive `u, v` this is the strict concavity inequality
+`h(u) ≥ h(u+v) + h'(u+v)(u - (u+v))`, together with the symmetric bound for
+`h(v)`; the boundary cases follow from `h(0) = 0`. Equality holds only when
+`u = 0` or `v = 0`.
+
+## Lemma 3 (the terminal is not a positive bill)
+
+Suppose every path has `a_i ≤ T ≤ b_i`. Then
+
+```text
+G(T) = - Σ_i ( h(T-a_i) + h(b_i-T) - h(b_i-a_i) ) ≤ 0.
+```
+
+The helper terms vanish because `h(T) - h(T) = 0` and `h(0) = 0`. Each summand
+is nonnegative by Lemma 2, applied to `u = T - a_i` and `v = b_i - T` whose sum
+is `b_i - a_i`. A summand is zero exactly when `T` equals `a_i` or `b_i`.
+Therefore `G(T) ≤ 0`, with equality exactly when `T` is an endpoint of every
+path.
+
+## Lemma 4 (no path starts inside the boundary)
+
+Suppose every `a_i = 0`, at least one path crosses (`b_i > T`), and
+`0 < x < T`. Then `G(x) < 0`.
+
+Proof. Substitute `a_i = 0`:
+
+```text
+G(x) = - (h(T) - h(T-x)) + Σ_i ( h(b_i) - h(b_i-x) ) - (m-1) h(x).
+```
+
+For a crossing path, `b_i > T`, so `b_i - x > T - x`. Lemma 1 with
+`y = T-x`, `z = b_i` gives
+
+```text
+h(b_i) - h(b_i-x) < h(T) - h(T-x).
+```
+
+Summing over the `m` paths,
+
+```text
+Σ_i ( h(b_i) - h(b_i-x) ) < m ( h(T) - h(T-x) ),
+```
+
+so
+
+```text
+G(x) < (m-1) ( h(T) - h(T-x) ) - (m-1) h(x)
+     = -(m-1) ( h(x) + h(T) - h(T-x) ).
+```
+
+For `m ≥ 2` the right side is negative because `h(x) > 0` and
+`h(T) > h(T-x)`. For `m = 1` the strict increment inequality alone gives
+`G(x) < 0` without the last line. At `x = 0` every term vanishes, so
+`G(0) = 0` and the maximum is attained only at the left endpoint.
+
+## Theorem 1 (one crossing path)
+
+Suppose `m = 1` and `b > T`. Then `G` is strictly decreasing on `[a, T]`, so
+its maximum on the feasible interval is `G(a)`.
+
+Proof. For `a < x < T` the legs are positive and
+
+```text
+G'(x) = h'(x) - h'(T-x) - h'(x-a) + h'(b-x)
+      = [h'(x) - h'(x-a)] + [h'(b-x) - h'(T-x)].
+```
+
+Since `a ≥ 0`, we have `x - a ≤ x`, and Lemma 1 gives
+`h'(x-a) ≥ h'(x)`, with equality only if `a = 0`. Since `b > T`, we have
+`b - x > T - x`, and Lemma 1 gives `h'(b-x) < h'(T-x)`. The second bracket is
+strictly negative, so `G'(x) < 0` on `(a, T)`. A differentiable function with
+a strictly negative derivative on an open interval is strictly decreasing
+there; `G` is continuous on the closed interval because `h(0) = 0`. Hence the
+maximum is `G(a)`. When `a = 0`, Lemma 4 gives `G(x) < 0` on `(0, T)` directly.
+
+Consequently one path that leaves a proper helper boundary can never justify a
+new intermediate dimension: the candidate is the path's own lower endpoint,
+which is already a boundary dimension. This is the first step requested in
+#7283's Open next question 1.
+
+## Proposition (the endpoint maximum fails for four crossing paths)
+
+Take `t = 1/2`, `D = 26`, `T = 24`, and the four paths
+
+```text
+(3, 25), (17, 26), (0, 25), (6, 25).
+```
+
+Every path starts in `[0, T]` and crosses the terminal, and every endpoint is
+in `[0, D]`. The feasible interval is `[A, T] = [17, 24]`. At
+`x0 = 3819/160` the exact integer square-root certificate gives
+
+```text
+G(17) - G(x0) < 0,      G(24) - G(x0) < 0.
+```
+
+The decimal displays are `G(17) = -3.747477...`, `G(24) = -3.734846...` and
+`G(x0) = -3.554274...`. Thus `x0` strictly beats both endpoints. The endpoint
+maximum of #7283 therefore does not extend to four boundary-crossing paths.
+
+The witness has a negative interior value, so it does not exhibit a profitable
+rank bill. It shows that the two-endpoint comparison can understate the
+maximum. The sign question is left as an open next question.
+
+## What this does not prove
+
+The proper-terminal bill is a rank-bill factor. This package does not prove a
+legal cut, an arbitrary-input identity, a native realization, an ordinary
+cost, ownership, restoration or a paid-profile improvement. It does not change
+the selected exponent. The two-path and three-path cases are not proved; only
+the stated finite grids are censused. The sign of an interior maximum with
+nonpositive endpoints is not proved.
+
+## Dependencies
+
+The rank-bill formula, its zero-to-`D` endpoint theorem and the proper-terminal
+question are unpaidmoon's #7283. That package credits CubeJerry's shared-native
+helper-split exploration and the two-chain word from DreamingOfClouds in #6902.
+No part of #7283 is replayed or re-audited here. This package was produced by Nick Petalas.
+Apache-2.0.

research/proper-terminal-helper-rank-bill/README.md

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+# A proper helper terminal and the rank bill
+
+**Proof with an exact finite counterexample. No new κ.** The selected
+conditional exponent stays `363005817/100000000000`. No word is emitted, no cut
+is declared legal, and no ordinary or paid bill is priced.
+
+## Result
+
+Fix a real exponent `0 < t < 1`, integers `0 < T < D`, and `m ≥ 1`
+contact-free paths `0 ≤ a_i ≤ b_i ≤ D`. The helper boundary now runs from `0`
+to the proper terminal `T`, so the rank bill of a cut at dimension `x` is
+
+```text
+G(x) = x^t + (T-x)^t - T^t
+       - Σ_i ( (x-a_i)^t + (b_i-x)^t - (b_i-a_i)^t ),
+```
+
+with `0^t = 0`. This is the formula of
+[#7283](../helper-split-endpoint-rank-bill/PROOF.md) with `D` replaced by the
+terminal `T`. A path with `b_i > T` leaves the boundary. The feasible interval
+is `[A, B]` with `A = max a_i` and `B = min(T, min b_i)`.
+
+**Theorem 1 (one crossing path).** If `m = 1` and `b > T`, then `G` is strictly
+decreasing on `[a, T]`. Its maximum is `G(a)`. If moreover `a = 0`, then
+`G(0) = 0` and `G(x) < 0` for every `x ∈ (0, T)`.
+
+So the answer to the first step of the #7283 open question is: for one path that
+leaves the boundary, the endpoint maximum survives, and the only candidate
+dimension is the path's own lower endpoint.
+
+**Proposition 2 (the terminal is never a positive bill).** For every `m` and
+every path with `a_i ≤ T ≤ b_i`, `G(T) ≤ 0`, with equality exactly when `T` is
+an endpoint of every path.
+
+**Proposition 3 (no path starts inside the boundary).** If every `a_i = 0` and
+at least one path crosses the terminal (`b_i > T`), then `G(x) < 0` for every
+`x ∈ (0, T)` and every `m ≥ 1`. No number of boundary-crossing paths makes a
+proper interior cut profitable in this first-order sense.
+
+**Proposition 4 (the endpoint maximum fails for four crossing paths).** At
+`t = 1/2`, `D = 26`, `T = 24` and paths
+
+```text
+(3, 25), (17, 26), (0, 25), (6, 25),
+```
+
+the interior dimension `x0 = 3819/160 = 23.86875` satisfies
+
+```text
+G(17) = -3.747477... < G(x0) = -3.554274...
+G(24) = -3.734846... < G(x0).
+```
+
+The comparisons `G(17) < G(x0)` and `G(24) < G(x0)` are certified with exact
+integer square-root arithmetic, not floating point. At this witness both
+endpoint values are lower than the interior value, so the endpoint maximum of
+#7283 does not survive a proper terminal once four paths cross the terminal.
+
+The counterexample does not make the interior bill positive: it shows that the
+endpoint comparison alone can understate the maximum, not that a gain exists.
+Whether a positive proper-terminal bill can appear away from `A` is left open;
+see Open next questions.
+
+## Evidence classes
+
+1. **Written proof.** Lemmas 1-3 and Theorem 1 in [PROOF.md](PROOF.md), for
+   every real `0 < t < 1`, every integer `0 < T < D` and every `m ≥ 1`.
+2. **Exact finite certificate.** Proposition 4 is one exact instance at
+   `t = 1/2`. The verifier compares sums of integer square roots with a
+   certified truncation whose error is strictly less than one unit per term.
+3. **Exhaustive finite censuses**, only on the domains named in
+   [VERIFY.json](VERIFY.json): the one-path decrease on a rational grid for
+   `D ≤ 10`; the zero-start negativity for `D ≤ 8` and `m ≤ 3`; the terminal
+   sign for `D ≤ 8` and `m ≤ 2`; and the absence of an interior win for two or
+   three crossing paths on the stated grids (`D ≤ 8` for two paths, `D ≤ 7` for
+   three). A census is not a proof for larger parameters and is not used as
+   one.
+4. **Control rejections.** The verifier rejects a flipped endpoint comparison,
+   a one-path decrease claim applied to the four-path witness, and a positive
+   terminal bill.
+
+A failed or absent search is not used anywhere in this package.
+
+## Scope and connection
+
+Card `p104-side-correction`, priority revision 8, base commit
+`e54523517d0f03e58352e647c403a5af3e525a57`, selected record unchanged.
+Scope status: **changed**. The card asks whether a legal causal split improves
+the complete P104 invoice. This package answers one boundary case of the
+rank-bill factor that #7283 explicitly left open. It prices only the rank-bill
+term `G`. Legality, ownership, ordinary cost and any target/paid invoice remain
+where the selected record left them.
+
+## Reproduce
+
+From the repository root, CPython 3.12, standard library:
+
+```sh
+python3.12 -B research/proper-terminal-helper-rank-bill/verify.py
+python3.12 -O -B research/proper-terminal-helper-rank-bill/verify.py
+python3.12 -B scripts/research_package.py --package research/proper-terminal-helper-rank-bill
+```
+
+The direct verifier prints
+
+```text
+PASS_PROPER_TERMINAL_HELPER_RANK_BILL ...
+```
+
+Proof checks raise `SystemExit` rather than using `assert`, including under
+`python -O`.
+
+## Open next questions
+
+- **Can a positive proper-terminal bill appear away from the endpoints?** The
+  four-path witness has an interior maximum that is still negative. A search
+  over four million random integer configurations with `D ≤ 24`, `m ≤ 8` and
+  `t ∈ {1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, 4/5}` found no case with both
+  endpoint values nonpositive and an interior value above zero. That search is
+  floating-point evidence, not a proof and not part of the verifier. The
+  smallest next step is to fix `m = 4` and ask for the exact sign structure of
+  the unique interior critical point, or to prove the sign claim by a
+  maximum-principle argument on the tent decomposition in PROOF.md.
+- **Which crossing count is the threshold?** Theorem 1 proves one path. The
+  verifier finds no interior win for two paths with `D ≤ 8` or three paths with
+  `D ≤ 7` on the stated grids, and Proposition 4 gives four paths. Resolving
+  `m = 2` and `m = 3` for all `D` would make the threshold exact.
+- **Does a proper terminal help any actual helper?** Evaluate `G` at the two
+  endpoints for one pinned helper of the published 68-split word whose terminal
+  is proper, and compare with the zero-to-`D` certificate. A positive value
+  would still need legality, ownership and the paid profile.
+
+## Credit
+
+The rank-bill formula, the zero-to-`D` endpoint theorem, the one-path
+normalization and this open question are unpaidmoon's [#7283]. That package
+credits CubeJerry's shared-native helper-split exploration and the two-chain
+word from DreamingOfClouds in #6902. The present package adds the proper
+terminal case, the crossing-path theorem, the zero-start proposition and the
+four-path counterexample. Nothing from #7283 is replayed or re-audited.
+
+This package was produced by Nick Petalas. Apache-2.0.
+
+[#7283]: https://github.com/CrocSwap/integer-mult-bounds/pull/7283

research/proper-terminal-helper-rank-bill/SOURCES.json

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