Incoprea research · part 2 of 5 · reversals and constraints · 3 November 2026

Reversals, model freedom, and the constraint protocol

Computation, one synthetic demonstration, and one short proof. A selectivity reversal was manufactured from random coefficients at four levels of model freedom: 0% of draws at one or two free coefficients, 24% at four. Three constraint tests follow — one clean refutation, one synthetic power check, one construction failure. Part C proves a parity constraint on crossings (0 violations in 3,000 trials) and thereby refutes the earlier two-crossing conjecture.

Part A · free-parameter base rates and the crossing bound

A "reversal" is any claim that a ranking flips: gas A separates better than B at low pressure and worse at high; membrane X is selective one way and then the other; a diode rectifies and then inverts. These claims are common and they are easy to produce. This page asks how easy.

The answer: the probability of manufacturing a reversal rises steeply with the number of free coefficients in the model, and a reversal that survives when the coefficients are constrained is a different and much stronger thing than one that does not.

1. Base rate versus freedom

A three-channel permeance model was run 400 times at each level of freedom. At level k, exactly k of the four surface coefficients (affinity and capacity for each of two gases) were drawn from decade-wide ranges; the others were fixed.

Free coefficientsDraws showing an α = 1 crossing
10.0%
20.0%
313.2%
424.0%
1 free coefficient(s) 0.01x 2 free coefficient(s) 0.01x 3 free coefficient(s) 13.25x 4 free coefficient(s) 24.00x
Frequency of an apparent selectivity reversal against the number of free coefficients. With one or two free coefficients the reversal does not appear; by four it appears in roughly one draw in four.

With one or two free coefficients the reversal essentially never appears. With three it appears in 13.2% of draws; with four, in 24.0%. Nothing physical changed between those rows. The only difference is how much freedom the modeller allowed.

Consequence: a published reversal should be accompanied by a count of free versus measured coefficients. Without that count, the reader cannot distinguish a mechanism from a fitting freedom.

2. Is there a bound on how many crossings are possible?

If each saturating channel can contribute at most one sign change to the difference of two sums, then with S saturating channels the number of crossings should be at most S − 1. The test: 600 random channel sets with 2–5 channels and 0–5 saturating channels.

Result: 0 violations in 600 trials. With zero saturating channels the maximum number of crossings observed was 0 — as the bound requires, since a sum of monotone pressure-independent terms cannot cross.

The bound is not a deep theorem, but it is a useful screening rule: if a reported reversal has no saturating channel in the model, the reversal is not in the model's physics — it is in its parameters.

3. The membrane case, constrained

The same model with one shared surface coefficient and fixed isotherm values:

ksurfcrossingsα at 0.1 barα at 100 bar
0.0502.7322.751
0.202.7582.825
0.802.8623.021
3.203.2753.298

0 crossings in 4 constrained cases. The selectivity stays between 2.73 and 3.30 across a 64-fold change in the shared coefficient. The surface term increases the same gas's advantage at low pressure and lets it decay at high pressure, which narrows the gap but does not close it.

4. The protocol this suggests

  1. Count the free coefficients in the model that produces the reversal.
  2. Count the ones that were independently measured.
  3. If free > measured, re-run with one shared coefficient and measured isotherms.
  4. If the reversal survives, report the pressure at which it occurs and the coefficient that sets it.
  5. If it does not, report that the reversal was an artefact of freedom. That is a publishable negative.

5. New hypotheses

#HypothesisTest
P2-H1Published reversals in membrane, thermal-diode and ionic-diode literature will correlate with free-coefficient count.Code 40 papers on both axes; test the association.
P2-H2Reversals in systems with no saturating channel are always parameter artefacts.Screen reported reversals for a saturating term.
P2-H3A "constrained replication" requirement would remove a large fraction of published reversals.Re-run three published models with shared coefficients.

6. Limits


Part B · three constraint tests

Three further constraint tests. Each asks the same question in a different field: does the claimed effect survive when the parameters are tied to something measurable rather than left free?

1. Response-time collapse

If tracking error depends only on R = τ/T, one curve should describe every system. At R = 1 the error spans 0.702 to 1.919 — a factor 2.74. Against the first-order curve the worst departure is 1.26 in absolute error amplitude.

Systemerror at R = 1
1st order0.7016
2nd zeta=0.31.9193
2nd zeta=0.71.2223
2nd zeta=1.01.1145
2nd zeta=2.01.0298

Verdict: not supported. R is the dimensionless frequency; it was never going to carry order and damping. The hypothesis was a restatement, and the restatement does not extend.

2. Enzyme competition, constrained

The competitive-kinetics crossover is already fixed by kcat/Km; the only testable content is a departure. A synthetic two-site system was generated (clearly labelled synthetic), then fitted with a one-site model in two ways:

FitRMSEMeaning
one-site, Km free0.020the model absorbs the second site into a shifted Km
one-site, Km constrained to measured value0.180the departure becomes visible

The constrained model is 9.0× worse than the free one. That is the test working as designed: when the free fit is much better than the constrained fit, the system is not described by single-site competitive kinetics, and the size of the gap measures how far off it is.

But this is a synthetic demonstration, not evidence about any real enzyme. Its only content is that the test has power when the departure exists. Whether real enzymes show it is untested and requires the measured constants.

3. Fog collection: a model that failed to discriminate

The claim was that the optimum spacing-to-droplet ratio moves when collection and drainage are both scored. In the illustrative model used here, interception rises with spacing while the drainage fraction falls, and the product is maximised at the same spacing as interception alone — a shift factor of 1.00, i.e. no shift.

This test is uninformative as constructed, and the failure is mine, not the hypothesis's. The drainage term was monotone in the wrong direction, so no interior optimum could form. A useful version needs a drainage model with a genuine maximum (shedding that improves with spacing up to a point, then degrades), which was not built. The hypothesis remains planned, not refuted.

4. The constraint protocol, stated once

  1. Write the model with every parameter either measured or explicitly free.
  2. Count both.
  3. Run the free version — expect the effect (that is the base rate, not evidence).
  4. Run the constrained version.
  5. Report both. The gap between them is the finding.

Applied three times on this page, the protocol produced one clean refutation, one synthetic demonstration of test power, and one honest construction failure. That mix is what a working protocol looks like.

5. New hypotheses

#HypothesisTest
P3-H1Two-parameter collapse (R, ζ) is minimal for second-order systems; no one-parameter version exists for order ≥ 2.Fit surfaces; held-out prediction across unseen ζ and order.
P3-H2Measured enzyme pairs will show constrained-fit residuals systematically larger than free-fit residuals, and the gap will track known allostery.Use BRENDA/SABIO-RK constants; compare to allostery annotations.
P3-H3A drainage model with a non-monotone shedding term produces an interior fog optimum that monotone drainage cannot.Build the non-monotone term; map the optimum.

6. Limits


Part C · a parity theorem for crossings, and the refutation of the two-crossing conjecture

1. The setup

Let the selectivity be α(P) = N(P)/D(P), where each side is a sum of Langmuir-type channels

channel(P) = K + A · bP/(1 + bP)      K ≥ 0, A ≥ 0, b > 0

Every channel is monotone increasing and bounded, and so is every sum. The two limits are immediate:

α(0)   = K_N / K_D
α(∞)   = (K_N + A_N) / (K_D + A_D)

Both are finite and positive. They are generally different, because the amplitudes A enter one limit and not the other. This is the fact the earlier conjecture missed.

2. The parity theorem

If α is continuous on (0, ∞) and α(P) = 1 at finitely many points, then the number of crossings is odd when α(0) and α(∞) lie on opposite sides of 1, and even when they lie on the same side.

Proof. α is continuous and never zero or infinite (both K terms are positive in the cases of interest). Let g = α − 1, also continuous. Each crossing is a sign change of g. A continuous function whose endpoint signs agree must change sign an even number of times; if the endpoints disagree, an odd number. Since crossings are exactly the sign changes, the count has the parity of the endpoints. ∎

The theorem is elementary — it is the intermediate value argument applied to a ratio of monotone sums. Its value is not depth but use: it converts a numerical conjecture into a constraint that any reported reversal must satisfy, and it tells you which crossing counts are even possible.

3. The test

3,000 random channel sets, 1–4 channels per side, coefficients drawn over 1.5–2.5 decades, crossings counted on a 4,000-point log grid from P = 10⁻³ to 10³.

Crossings01234+
Count2,3105939520

Parity violations: 0. Every case's crossing count was consistent with its own two end signs. That is the theorem being exercised 3,000 times without a counterexample.

4. The conjecture is dead

Two cases produced three crossings, and both have the signature the theorem requires — opposite limits:

casechannels (N, D)α(0)α(∞)crossings
13, 10.5351.0493
22, 10.4031.1593

Both start below 1 and finish above it. The curve must therefore cross at least once; it happens to cross three times, which requires two returns — a shape the earlier "at most two" claim excluded.

Rate: 2 in 3,000, or 0.07%. Rare, but real, and reproducible from the seed.

5. What replaces the conjecture

Old claimNew statement
Form"At most two crossings"Parity fixed by the two limits; no upper bound established
StatusConjecture, no counterexample foundTheorem for parity; open for the upper bound
Screening use"A third crossing indicates a model outside the class""A crossing count inconsistent with the end signs indicates an error or a model outside the class"

The new rule is weaker but usable, and it cannot be wrong in the way the old one was. A reported reversal with three crossings and both limits on the same side of 1 is now a definite error — either in the model or in the analysis.

6. New hypotheses

#HypothesisDisposition
N-23An upper bound of 4 crossings exists for 4-channel-per-side Langmuir sums, and the bound grows more slowly than the channel count.test now — extend the sweep to S = 6 with more trials
N-24Three-crossing cases require at least one side to have more channels than the other; equal-channel sums cannot produce three.test now — stratify by channel count
N-25Published reversals with an even number of crossings and equal limits are parameter artefacts; odd-crossing reversals with unequal limits are the physically possible class.note — needs a corpus of reversals
N-26The parity constraint applies to any ratio of monotone bounded sums, so it transfers to thermal rectification and to diode rectification without modification.test now — restate the proof for the general class

7. Limits

If one sentence survives: the number of selectivity crossings has the parity of the two limits of α, this is provable and held in 3,000 trials, and it kills my own earlier claim of a maximum of two.


Incoprea Research · batch 1 · 3 November 2026. Computational model results and one real-catalogue re-analysis. No new physical experiment was performed. No novelty claimed; no systematic literature search. Code, data and JSON accompany this page set.