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Convex Contact And Mixing

1 · Prerequisites

2 · Summary

This mathematics page develops forward sectors from A1–A7 and S1–S3, presents conditional results on their convex geometry, fixed-work orientation, Lipschitz boundary graph, nesting, and comparison completeness for finite scaled copies of one simple type. Thermal strips glue the local reference coordinates. The contact calibrator and arbitrary-family entropy results are explicitly conditional on the stated thermal assumptions, including join coherence C1 and universal thermal range T5. The later finite-chain cost and no-sinks offset results concern only the stated weak monotonicity across admitted transitions; no no-gap equality is claimed. These are mathematical deductions under declared hypotheses, not empirical laws.

All carriers are drafts. Their local deductions are reviewed separately from publication or workflow acceptance; the declared page prerequisites must also be resolved before page readiness.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

Forward-sector hypotheses for a simple state domain

Definition

Let P be a set of primitive type labels. A formal type Γ records a finitely supported amount function νΓ:P→R≥0, not identically zero. Primitive generators have their unit amount function. Positive scaling, separated product, and thermal-join constructors retain distinct formal tags and obey

νsΓ=sνΓ,νΓ⊗Γ′=νΓ+νΓ′(s>0).

Separated and joined types are distinct constructors even when their amount functions agree. Permutation, reassociation, and scaled-splitting conventions preserve the amount function. Define the positive bookkeeping weight wΓ:=∑p∈PνΓ(p), a finite sum over support. Thus wsΓ=swΓ and every product or join has weight wΓ+wΓ′. The optional empty unit has zero amount function and weight zero; it is represented only by the empty compound, not as a simple chart. These are formal amounts, not masses in kilograms or an empirical conservation law. The functions and pointwise operations use The vector space FX of all functions X→F with pointwise operations, and Fn as the case X=n={0,1,…,n−1}; their finite-support convention and finite sums are those of Tagged scalable compounds and Compound mass is well-defined and additive.

For Euclidean vectors use a⋅b=∑jajbj and ∥a∥=(a⋅a)1/2. Write B(a,r)={b:∥b−a∥<r}. Open means each point has such a ball contained in the set; convex means (1−t)a+tb belongs to the set for all its a,b and 0≤t≤1. In a subspace or subset, use intersections of these balls for the relative topology. A map P:D→Rn is locally Lipschitz if each a∈D has a relative neighborhood V and finite L≥0 with ∥P(b)−P(c)∥≤L∥b−c∥ for all b,c∈V. A set D is arcwise connected here if each two distinct points are joined by a continuous injective map γ:[0,1]→D with those endpoints; singleton and empty sets satisfy this condition vacuously. Continuity uses the indicated Euclidean relative topology. Let I index the declared family of simple types Γi. For each simple type Γ in that family, let nΓ≥1 and let Γ⊂RnΓ+1 be nonempty, open and convex, with coordinates x=(u,v). The notation for a type and its chart is shared, but each state has a distinct raw tag (Γ,x). Scaled charts are distinct tagged copies sΓ={sx:x∈Γ} with coordinate transport x↦sx; transported raw tags remain distinct from the original raw tags. Let e(Γ,x) be the unit point mass from Tagged scalable compounds. The represented state is

JΓ(x):=wΓe(Γ,x),m(JΓ(x))=wΓ.

Write J(x):=JΓ(x) when x already carries its type tag. All state notation in this framework's compound comparisons, including a bare x, denotes JΓ(x); a raw unit point mass is always written e(Γ,x). A separated pair is addition of represented states, not a unit product tag. Nonnegative multiples are pointwise and zero terms are omitted. For an encoded compound c, its aggregate composition is the finite amount function

ν(c):=∑(Γ,x)∈supp⁡(c)c(Γ,x)wΓνΓ;m(c)=∑p∈Pν(c)(p).

Fix the formal interpretation Π(c) as the finite separated preparation whose factor at each nonzero raw tag (Γ,x) is the amount c(Γ,x)/wΓ of that typed state, using the declared coordinate scaling transport. Let P:={Π(c):c is a finite nonnegative encoded compound} be the set of these preparations, with an invariant bookkeeping mass M:P→R≥0 satisfying M(Π(c))=m(c). As additional model data choose a binary relation Q⊆{(p,q)∈P2:M(p)=M(q)}. Define its pullback on encoded compounds by A≺B if and only if (Π(A),Π(B))∈Q. The transport, reversibility and A1–A7 assumptions below constrain this chosen relation; they do not construct it from coordinates. Merging or splitting repeated copies of the same preparation is reversible; permutation and reassociation preserve the preparation. Thus Π(JsΓ(sx)) and Π(sJΓ(x)) are the same declared scaled preparation and yield

JsΓ(sx)∼sJΓ(x),

although their raw point masses have different tags and are not equal. This reversible transport commutes with finite composition, nested type constructors, and scaled splitting. It is part of the precise formal scaling interpretation, not a deduction about physical accessibility from coordinate multiplication. A physical use must adopt the stated preparation maps and accessibility assumptions; fixed-composition models compare only preparations with the same ν, while a declared channel model may additionally compare different compositions of equal m.

This Definition restricts comparisons A≺B to m(A)=m(B); the coefficient-map supplier defines the mass, not the comparison relation. Write A∼B when both directions hold, and A≺≺B when only the forward direction holds. On any fixed simple type, dividing every represented state by the same positive constant wΓ gives its unit-point-mass comparison. Positive scaling and its inverse preserve and reflect order. An additive entropy density for this fixed type means a function S:Γ→R such that, for every two finite nonnegative unit-point-mass compounds C=∑jaje(Γ,xj) and D=∑kbke(Γ,yk) with ∑jaj=∑kbk, one has

C≺D⟺∑jajS(xj)≤∑kbkS(yk).

Empty sums are zero. Given such an S, define HS(C):=∑jajS(xj) on compounds supported on this type. Its represented-state values are HS(JΓ(x))=wΓS(x) and HS(∑jtjJΓ(xj))=∑jtjwΓS(xj) for finite nonnegative tj. These identities follow by substituting the raw coefficients in the finite sum. Representing only the comparisons e(Γ,x)≺e(Γ,y) is insufficient for the displayed compound equivalence. Existence of an additive entropy density is not assumed here; a later representation theorem must establish it under its full compound-comparison hypotheses. No independent rescaling across different types is assumed without a calibrator.

The common relation satisfies A1–A6 on all equal-mass finite compounds: (A1) reflexivity; (A2) transitivity; (A3) if A≺B, then A+C≺B+C for every compound C; (A4) if A≺B, then tA≺tB for every t>0; (A5) tJΓ(x) is equivalent to sJΓ(x)+(t−s)JΓ(x) for t>0 and 0<s<t, with the reversible scaled-copy transport above; (A6) for equal-mass A,B, fixed positive equal-mass spectators Z0,Z1, and every sequence ϵn>0 decreasing to zero, A+ϵnZ0≺B+ϵnZ1 for all n implies A≺B. These are hypotheses, not conclusions. No comparison hypothesis is included here.

For each i and x∈Γi, define the forward sector Ax:={y∈Γi:JΓi(x)≺JΓi(y)}. The simple-domain assumptions are: (A7) for x,y∈Γi and 0<t<1, the encoded compound tJΓi(x)+(1−t)JΓi(y) compares to JΓi(tx+(1−t)y); both have bookkeeping mass wΓi, and convexity puts the recombined coordinate in the chart; (S1) each x has a strict successor; (S2) Ax has exactly one supporting hyperplane at x, expressible as u−ux+Pi(x)⋅(v−vx)=0 with a finite-valued slope map Pi:Γi→Rni, and Pi is locally Lipschitz; (S3) the boundary of Ax relative to Γi is arcwise connected. A supporting hyperplane means that Ax lies in one of its closed half-spaces and meets the plane at x. The affine hull is the smallest affine subspace containing a set; relative interior is interior in that affine hull. Closure and boundary are taken in the indicated affine hull or in Γi, as named; boundary is closure intersected with closure of the complement in that relative topology.

This definition imposes no physical interpretation, units, or empirical claim. A contact theorem that uses joins must separately assume that each join is a simple domain satisfying these same forward-sector hypotheses.

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

Thermal joins and conditional contact hypotheses

Definition

For two simple domains Γi⊂Rni+1 and Γj⊂Rnj+1, write their coordinates (ui,vi) and (uj,vj). Their thermal-join domain is

Δij:={(u,vi,vj):∃ui,uj, (ui,vi)∈Γi, (uj,vj)∈Γj, u=ui+uj}.

The joined simple type is a distinct formal tag with amount function νΔij=νΓi+νΓj and weight wΔij=wΓi+wΓj, as defined in Forward-sector hypotheses for a simple state domain. Encode the separated pair by JΓi(x)+JΓj(y) and a joined coordinate Z by JΔij(Z). Their bookkeeping masses agree, and their aggregate composition functions agree. These encoded maps and their raw tags remain distinct. Positive coordinate-scaled copies use the reversible preparation transport of that definition; nested joins preserve their finite primitive amount function under reassociation and permutation. The empty type contributes the zero map.

Write Jij(x,y) for the join state with the summed energy coordinate and the unchanged work coordinates. A compound pair (x,y) and Jij(x,y) have the same total energy and work coordinates but are distinct tagged objects. Define x∼Ty exactly when JΓi(x)+JΓj(y) and JΔij(Jij(x,y)) are accessible in both directions. Nested joins use the declared reassociation and permutation identifications.

Assume every finite join is itself a simple domain satisfying Forward-sector hypotheses for a simple state domain. The contact hypotheses are mathematical assumptions on the common relation: (T1) (x,y)≺Jij(x,y); (T2) every join state has at least one reversible split into component states at its stated coordinates, and the equal-state split of any state into positive scaled copies of itself is reversible; (T3) ∼T is transitive; (T4) for every simple state x there are states x0,x1 in its domain with x0≺≺x≺≺x1 and x0∼Tx1; (T5) for every ordered pair of simple systems, every x in the first and every work coordinate v in the second's projection ρ(Γj):={v:∃u,(u,v)∈Γj}, there is a state y with work coordinate v and x∼Ty.

For the completed calibrator branch, add (C1) join coherence: whenever x∼Ty, the join state Jij(x,y) is thermally equivalent to each of x and y under the nested-join identifications. The coordinate sum in the definition alone does not assert C1. In later theorem statements, T1–T5 and C1 are explicitly listed as hypotheses; they are not treated as established physical laws.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

Relative interiors of finite-dimensional convex sets

Statement

For a nonempty set C, define aff⁡C={∑j=1Najcj:N≥1, cj∈C, aj∈R, ∑jaj=1}. With a fixed c0∈C, this is c0+span⁡{c−c0:c∈C}, where span consists of finite linear combinations. An affine Euclidean space is such a translate of a finite-dimensional linear subspace with its induced Euclidean distance. Relative balls use this distance; interior means containing a relative ball, and closure means every relative ball meets the set. Let C be a nonempty convex subset of a finite-dimensional affine Euclidean space H. Its relative interior ri⁡HC is the interior in H=aff⁡C. Then ri⁡HC is nonempty. If w∈ri⁡HC, y∈C‾ H, and 0≤t<1, then (1−t)w+ty∈ri⁡HC.

Facts & Assumptions

Given: A nonempty convex C⊆H in finite dimension. Affine dimension is the dimension of its translation space.

[F1]

A convex subset of Rm contains every line segment between two of its points: every finite convex combination of points of C belongs to C.

Proof

1.1givenF1construct

Let k=dim⁡(aff⁡C). If k=0, then C is a singleton and is open in its affine hull. If k>0, the definition of affine hull supplies k+1 affinely independent points of C spanning it. Their simplex lies in C by [F1]. Its barycenter has all k+1 barycentric coordinates positive; sufficiently small coordinate perturbations preserve positivity and their sum one. Thus a relative neighborhood of that barycenter lies in the simplex, proving ri⁡HC≠∅.

2.1givenF1choosealgebra∎

Fix w,y,t as in the Statement. The case t=0 is w∈ri⁡HC. Suppose 0<t<1 and put q=(1−t)w+ty. Choose r>0 so the relative ball BH(w,r) lies in C. Since y∈C‾ H, choose one y∗∈C with ∥y∗−y∥<(1−t)r/(4t). For every q′∈BH(q,(1−t)r/4), put w∗=(q′−ty∗)/(1−t). Then ∥w∗−w∥<r/2, so w∗∈C and q′=(1−t)w∗+ty∗∈C by [F1]. A relative neighborhood of q lies in C, so q∈ri⁡HC.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

Supporting hyperplanes at relative boundary points

Statement

For nonempty C, define aff⁡C={∑j=1Najcj:N≥1, cj∈C, aj∈R, ∑jaj=1}. An affine Euclidean space is a translate of a finite-dimensional real linear subspace with the induced Euclidean metric. A supporting hyperplane at z is {q:n⋅(q−z)=0} for a nonzero vector n, with z in the set and all its points in one closed half-space. Let C be a nonempty convex set in a finite-dimensional affine Euclidean space H of dimension k≥1, let aff⁡C=H, and put K=C‾ H. If z∈∂HK (that is, z∈K and every relative neighborhood of z meets H∖K), then there is a nonzero linear functional ℓ on the translation space of H such that ℓ(q−z)≤0 for all q∈C. If C⊂Rd has affine dimension at most d−2, then every z∈C has at least two distinct ambient supporting hyperplanes containing aff⁡C.

Facts & Assumptions

Given: The stated finite-dimensional affine Euclidean space and convex set.

[F3]

The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn: Euclidean coordinates identify linear functionals with vectors and give the usual unit sphere.

[F4]

A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value: a continuous real function on a nonempty compact metric space attains its minimum.

[F5]

Unit spheres in finite-dimensional normed spaces are sequentially compact: unit spheres in finite-dimensional normed spaces are sequentially compact without a choice principle.

[F6]

The rationals embed densely in the reals: rational coordinates are dense in real coordinates.

[F7]

Reciprocals of the positive integers decrease to zero supplies the positive real sequence ϵm=1/ιN(σ(m)), m∈N, decreasing to zero with the canonical-natural field convention.

Proof

1.1givenF3F6F7construct

Fix affine coordinates H≅Rk. Closure preserves convexity without choosing approximating sequences: for x,y∈K and any ϵ>0, choose two points of C within ϵ of them; their convex combination approximates (1−t)x+ty within ϵ, so this point lies in K. Enumerate rational k-tuples canonically by increasing sums of absolute numerators and positive denominators, breaking the finite ties lexicographically. Density [F6] and the open complement of K show that for each m∈N at least one enumerated point lies in B(z,ϵm)∖K. Let zm be the first such point. This is a uniquely specified sequence with zm→z, not a countable selection of arbitrary witnesses.

2.1F2F3F4step 1.1algebra

The nonempty compact set K∩B‾(zm,∥zm−z∥+1) contains z. By [F2] and [F4], squared distance to zm attains a minimum there, and every point outside that ball is farther than the candidate z. Thus a nearest point qm∈K exists. It is unique: if two distinct minimizers had squared distance dm2, their midpoint would have squared distance dm2−∥qm−qm′∥2/4, contradicting minimality and convexity. Consequently (qm) is also canonically determined. Since zm∉K, put um=(zm−qm)/∥zm−qm∥. Expanding the distance at qm+t(q−qm) and letting t↓0 gives um⋅(q−qm)≤0 for every q∈K. Also ∥qm−z∥≤2∥zm−z∥→0.

3.1F3F5step 2.1

Every um has norm one. The exact choice-free sequential-compactness supplier [F5] gives a subsequence converging to a unit vector u. This introduces no additional choice hypothesis.

4.1F3step 2.1step 3.1algebra

For every q∈K, pass to the limit in um⋅(q−qm)≤0 along that subsequence. Continuity of the Euclidean pairing and qm→z give u⋅(q−z)≤0. The unit vector u defines a nonzero supporting functional for K, hence for C, at z.

5.1givenF3algebra∎

If C⊂Rd has affine hull H of dimension at most d−2, write its direction space as the span of k≤d−2 independent vectors b1,…,bk. Its orthogonal complement is the solution space of the k homogeneous equations bj⋅n=0. Row reduction has at most k pivot variables, leaving at least d−k≥2 free variables. Assigning one free variable the value one and the others zero gives two independent solutions (their free-coordinate vectors are different unit vectors). Thus choose two nonparallel nonzero normals n1,n2 orthogonal to the direction space of H. For each j, the hyperplane {q:nj⋅(q−z)=0} contains H and therefore supports C at z. The two hyperplanes are distinct because their normals are not parallel.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

Interior rays from a unique supporting plane

Statement

Use the Euclidean distance ∥q−z∥; int⁡K consists of points containing a ball in K, closure consists of points whose every ball meets K, and ∂K=K‾∖int⁡K. Closed means equal to closure. A supporting hyperplane at x∈K is {q:n⋅(q−x)=0} with n≠0 and K contained in one closed half-space. Let K⊂Rd be closed and convex with nonempty interior, d≥1, and let x∈∂K. Suppose K has exactly one supporting hyperplane at x. Choose its inward unit normal n, so n⋅(q−x)≥0 for every q∈K. Then for every v with n⋅v>0, there is δ>0 such that x+tv∈int⁡K for every 0<t<δ.

Facts & Assumptions

Given: The closed convex set, boundary point, unique supporting hyperplane, and Euclidean inner product.

[F1]

Strictly separate a point from a nonempty closed convex Euclidean set: a point outside a nonempty closed convex Euclidean set admits a strict separating linear functional.

[F2]

The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn: Euclidean linear functionals are represented by vectors.

[F4]

A convex subset of Rm contains every line segment between two of its points: finite convex combinations of points in a convex set remain in it.

Proof

1.1givenF2

Let T={a(q−x):a≥0, q∈K}‾. This is a closed convex cone. If a nonzero functional ℓ is nonnegative on T, then it is nonnegative on K−x, so its zero plane through x supports K. Uniqueness in the hypothesis makes that plane the plane with normal n; the inward sign forces ℓ=cn⋅( ⋅ ) for some c>0.

2.1F1step 1.1algebra

The support at x gives T⊆{w:n⋅w≥0}. If v with n⋅v>0 were outside T, strict separation [F1] of v from the cone would give a nonzero functional ℓ nonnegative on T and negative at v: the separating bound is zero because 0∈T and T is a cone. Step 1.1 then makes ℓ a positive multiple of n, contradicting n⋅v>0. Thus the open half-space n⋅w>0 lies in T, and each such v is an interior point of T.

3.1step 2.1constructF2algebra

Fix such v. Since v∈int⁡T, choose d+1 affinely independent vectors in a small ball about v whose simplex contains a neighborhood of v. Approximate each vector by a nonzero cone generator gj=aj(qj−x) with aj>0, qj∈K. Sufficiently close approximations preserve affine independence and keep v in the simplex interior, so a neighborhood N of v lies in the simplex of the gj. For each w∈N, write w=∑jθj(w)aj(qj−x) with θj(w)≥0 and ∑jθj(w)=1. Put A=max⁡jaj>0. The nonnegative barycentric weights sum to one, so ∑jθj(w)aj≤A for every w∈N.

4.1step 3.1givenF4algebra∎

Choose δ>0 with δA<1. For 0<t<δ and each w∈N, x+tw=(1−t∑jθjaj)x+∑jtθjajqj is a convex combination of points of K, hence lies in K by [F4]. Therefore x+tN⊂K, so a neighborhood of x+tv lies in K and x+tv∈int⁡K.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

Finite convex functions are locally Lipschitz

Statement

Use the Euclidean norm ∥z∥=(∑j=1dzj2)1/2, and balls B(a,r)={z:∥z−a∥<r}. A set is open if it contains a ball about each of its points. Let U⊂Rd be open and convex, and let f:U→R be finite and convex, meaning f((1−t)x+ty)≤(1−t)f(x)+tf(y) for every x,y∈U and t∈[0,1]. For every x0∈U, there are a neighborhood V of x0 and L<∞ such that ∣f(x)−f(y)∣≤L∥x−y∥ for all x,y∈V. The same conclusion holds for finite concave functions by applying the result to −f.

Facts & Assumptions

Given: The open convex domain, finite-valued function, and Euclidean norm.

[F1]

A convex subset of Rm contains every line segment between two of its points: line segments and finite convex combinations of points in a convex set remain in that set.

Proof

1.1givenF1construct

Choose a closed cube centered at x0 whose concentric dilation by factor 2 lies in U. Every point of the cube is a convex combination of its finitely many vertices, so [F1] and convexity bound f above there by the largest vertex value M. If x lies in the cube, then 2x0−x lies in the dilated cube; convexity gives f(x0)≤(f(x)+f(2x0−x))/2, hence f(x)≥2f(x0)−M′, where M′ is the largest value of f at the finitely many vertices of the dilated cube, an upper bound there by the same convex-combination argument. Thus f is bounded above and below on a ball about x0.

2.1step 1.1F1algebra∎

Choose concentric balls B(x0,r)⊂B(x0,2r)⊂U on which m≤f≤M. For distinct x,y∈B(x0,r/2), extend the segment from x through y to z=y+(r/2)(y−x)/∥y−x∥, which lies in B(x0,2r). Put θ=∥y−x∥/(r/2+∥y−x∥), so y=(1−θ)x+θz. Convexity yields f(y)−f(x)≤θ(M−m)≤(2(M−m)/r)∥y−x∥. Repeating with x,y interchanged gives the reverse bound. Negation proves the concave case.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

Zero-start integral inequality

Statement

Use the Darboux integral on nondegenerate intervals and the convention ∫aaf=0. Let L≥0, T>0, and let f:[0,T]→[0,∞) be continuous. If f(t)≤L∫0tf(s) ds for every t∈[0,T], then f(t)=0 for every t∈[0,T].

Facts & Assumptions

Given: A nonnegative continuous function and the stated integral inequality.

[F1]

A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion: a continuous function on a compact interval is Riemann integrable.

Proof

1.1givenF1F2F3F4algebra

For a continuous function on [a,b], the Darboux definition gives m(b−a)≤∫abf≤M(b−a) whenever m≤f≤M, because every lower and upper sum has these bounds. For a<c<b, concatenating partitions of [a,c] and [c,b] adds both their lower sums and their upper sums. By integrability [F1], choose on each interval a lower sum within η of its integral and an upper sum within η of its integral, and refine these finitely many partitions together. The concatenated sums sandwich ∫abf within 2η of ∫acf+∫cbf; letting η↓0 proves additivity. This also holds when c is an endpoint under the zero-interval convention. Define reversed integrals by ∫baf=−∫abf. By [F1], the integral defining h(t)=∫0tf(s) ds exists. Continuity of f gives h′(t)=f(t) on (0,T): the difference quotient error is bounded by sup⁡∣s−t∣≤∣r∣∣f(s)−f(t)∣, which tends to zero as r→0. By [F4], choose a finite M with 0≤f≤M. Then ∣h(t)−h(s)∣=∣∫stf∣≤M∣t−s∣ for all s,t∈[0,T], which proves continuity at the endpoints as well as in the interior. The hypothesis gives h′≤Lh, so [F2]–[F3] yield g′(t)≤0 for g(t)=e−Lth(t) on (0,T). The mean value theorem makes g nonincreasing; since g(0)=0 and g≥0, g≡0 and h≡0.

2.1step 1.1given∎

For any t>0, the nonnegative continuous function f has integral h(t)=0. If f(t0)>0 at any t0>0, continuity gives a positive lower bound on a nonempty subinterval of [0,t0], contradicting h(t0)=0. At t=0 the assumed inequality gives f(0)≤0; nonnegativity gives f(0)=0. Hence f≡0.

LemmaStatement: Literature-sourcedProof: AI-adaptedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

Spectator cancellation without comparison completeness

Statement

Under A1–A6 of Forward-sector hypotheses for a simple state domain, for compounds A,B with m(A)=m(B) and every finite spectator compound C, A+C≺B+C if and only if A≺B. Consequently, for finite-support signed maps u,v of equal total coefficient, define u⪯sv by (v−u)−≺(v−u)+. This relation is well-defined, agrees with ≺ on nonnegative maps, is invariant under adding a common signed map, is transitive, and is closed under addition of two comparisons. No comparison hypothesis is used.

Facts & Assumptions

Given: Finite-support nonnegative compound maps, equal-mass comparisons, and A1–A6.

[A1]

Forward-sector hypotheses for a simple state domain: the compound relation is reflexive and transitive, composition-consistent, positively scale-invariant, admits splitting/recombination, and satisfies the stated fixed-spectator stability axiom. It does not assume CH.

[F1]

Tagged scalable compounds: compounds are finite-support nonnegative maps; their total coefficient masses add under composition and scale under nonnegative dilation. Compound mass is well-defined and additive also proves that zero mass is equivalent to the empty map; the spectator need not have the same mass as either base.

[F2]

Signed comparison for forward-sector relations: for equal-total finite-support maps u,v, u⪯sv means (v−u)−≺(v−u)+, with common signed translation leaving the difference unchanged.

[F3]

The canonical natural ι(n)=n⋅1F of a field and Canonical naturals are positive and strictly increasing give the canonical embedding ιN:N→R, whose successor values are positive and are finite sums of 1. Reciprocals of the positive integers decrease to zero gives the real sequence ϵk=1/ιN(σ(k))>0, k∈N, with ϵk→0.

Proof

1.1A1F1F3algebrainduction

Suppose A+C≺B+C and m(C)>0. For each k∈N, set n=σ(k) and r=ιN(n)>0, and write A/r=(1/r)A. The embedding's successor rule and finite induction show that the sum of n copies of A/r is r(A/r)=A, and likewise for B. Scaling the given comparison by 1/r gives A/r+C/r≺B/r+C/r. By composition consistency, replace the n copies of A/r one at a time by B/r, keeping the already replaced copies, the remaining copies, and C/r as spectators; after these finitely many replacements, A+ϵkC≺B+ϵkC. This holds for every natural index k. By [F3], ϵk>0 and ϵk→0, so stability with Z0=Z1=C gives A≺B. If m(C)=0, nonnegativity implies C=0 and the conclusion is the premise itself. Conversely, A≺B gives A+C≺B+C by consistency.

2.1A1F1F2step 1.1

For equal-total signed maps u,v, the difference d=v−u has finite support and total coefficient zero, so m(d+)=m(d−). If u,v are nonnegative, their pointwise common part k(x)=min⁡{u(x),v(x)} gives u=k+d− and v=k+d+. Cancellation in step 1.1 proves u≺v exactly when d−≺d+, which by [F2] is u⪯sv; adding a common signed map leaves d unchanged by [F2].

3.1A1F1F2step 1.1algebra∎

If u⪯sv and v⪯sw, [F2] identifies these with comparisons of (v−u)− to (v−u)+ and (w−v)− to (w−v)+. Consistency combines them. Their pointwise common minimum cancels by step 1.1, leaving (w−u)−≺(w−u)+, which [F2] identifies as u⪯sw. The same finite-sum argument proves closure under addition; reflexivity follows from reflexivity of the zero compound.

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Signed comparison for forward-sector relations

Definition

Let u,v be finite-support real coefficient maps on the tagged state labels of Tagged scalable compounds with equal total coefficient. For a typed state x, compound notation uses the represented map J(x) from Forward-sector hypotheses for a simple state domain; its raw unit point mass is e(Γ,x). Signed maps still use the ordinary total of their raw coefficients, exactly the bookkeeping mass of the represented nonnegative maps. Put d=v−u, d+(x)=max⁡(d(x),0) and d−(x)=max⁡(−d(x),0). Define u⪯sv to mean d−≺d+ under the compound relation of Forward-sector hypotheses for a simple state domain. Write u∼sv when both u⪯sv and v⪯su. These are formal signed comparison maps, not physical negative amounts. The positive and negative parts are finite-support compounds of equal mass. If u∼sv, then cu∼scv for every real c: positive c uses scaling, c=0 is reflexivity, and negative c swaps the positive and negative parts while equivalence supplies both comparison directions. This Definition introduces the signed notation; ordinary-order compatibility and the signed-order laws are proved separately.

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Scaling invariance of thermal equivalence

Statement

Under T1–T3 and the equal-state reversible split in Thermal joins and conditional contact hypotheses, if x∼Ty then sx∼Tty for every s,t>0.

Facts & Assumptions

Given: The contact and scaling assumptions for tagged states and their joins.

[A1]

Thermal joins and conditional contact hypotheses: each equal-state scaled pair is reversibly equivalent to its join, and thermal equivalence is transitive.

Proof

1.1A1given

Let x be any typed state and s>0. The state z=(1+s)x in the scaled type has the equal-state reversible split with fractions 1/(1+s) and s/(1+s); its two components are exactly x and sx. Thus (x,sx) is reversible with its join, which by definition is x∼Tsx. The same argument gives y∼Tty.

2.1A1step 1.1given∎

Since x∼Ty, symmetry gives y∼Tx. Transitivity with x∼Tsx yields y∼Tsx; symmetry then gives sx∼Ty. Transitivity with y∼Tty now yields sx∼Tty.

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A unique reference coordinate inside a thermal strip

Statement

Let a simple domain and its finite thermal joins satisfy Forward-sector hypotheses for a simple state domain and Thermal joins and conditional contact hypotheses. Let x0≺≺x1 with x0∼Tx1, and let x satisfy x0≺≺x≺≺x1. For 0≤λ≤1, write Rλ for the compound state ((1−λ)x0,λx1), omitting a zero coefficient at the endpoints. Then there is a unique λ∈(0,1) with Rλ∼x in the common accessibility relation.

Facts & Assumptions

Given: The compound axioms A1–A6, thermal-join axioms T1–T3 with the equal-state split, strict strip endpoints, and join closure.

[A1]

Forward-sector hypotheses for a simple state domain: compounds have the preorder, consistency, positive scaling, splitting, and stability assumptions; CH is not part of this definition.

[F1]

Thermal joins and conditional contact hypotheses and Scaling invariance of thermal equivalence: an equal-state split is reversible, thermal equivalence is transitive, and scaling either member of a thermal-equilibrium pair preserves thermal equivalence.

[F2]

Signed comparison for forward-sector relations and Spectator cancellation without comparison completeness: equal-total signed maps compare by their positive and negative parts; cancellation is valid under A1–A6 without CH.

[F3]

Nesting and comparison of simple-system states: every pair of states in each finite join domain is comparable.

Proof

Given: [A1], [F1]–[F3], and x0≺≺x≺≺x1.

1.1A1F2algebra

For 0≤λ<μ≤1, the signed difference Rμ−Rλ=(μ−λ)(J(x1)−J(x0)) has positive part (μ−λ)J(x1) and negative part (μ−λ)J(x0). Since x0≺x1, positive scaling and [F2] give Rλ≺Rμ. A reverse comparison would, by [F2] and inverse positive scaling, give x1≺x0, contrary to strictness. Thus the reference states are strictly ordered by parameter.

1.2F1F3given

For 0<λ<1, [F1] makes (1−λ)x0 and λx1 thermally equivalent. By the definition of thermal equivalence, their compound is reversible with its join state. The equal-state split makes ((1−λ)x,λx) reversible with its join state and with x. Both join states belong to the join of two scaled copies of the same simple domain, so [F3] compares them. The endpoint cases use R0=x0 and R1=x1 directly. Hence Rλ and x are comparable for every λ∈[0,1].

2.1A1F2step 1.1choose

Let L={λ∈[0,1]:Rλ≺x}. The strict endpoint comparison x0≺x puts 0 in L. Step 1.1 and transitivity make L downward closed. Let s=sup⁡L. If s>0, then s−1/n∈L for every sufficiently large n. Since Rs−1/n+(1/n)x1=Rs+(1/n)x0, consistency yields Rs+(1/n)x0≺x+(1/n)x1. Stability with bases Rs,x and fixed spectators x0,x1 gives Rs≺x. If s=0, this comparison is the given x0≺x.

3.1A1F2step 1.2step 2.1

If s<1, then s+1/n∉L for every sufficiently large n. Step 1.2 gives comparability and the definition of L excludes Rs+1/n≺x, so x≺Rs+1/n. The identity Rs+1/n+(1/n)x0=Rs+(1/n)x1 and stability yield x≺Rs. If s=1, the strict strip gives x≺R1; together with step 2.1 this would imply x1≺x, contrary to x≺≺x1. Thus s<1, and step 2.1 and the preceding stability comparison give Rs∼x.

4.1F2step 1.1step 3.1algebra∎

The equivalence Rs∼x also excludes s=0, because R0=x0 and x0≁x, and excludes s=1 by strictness of x≺≺x1. If Rt∼x for another t, transitivity gives Rs∼Rt; step 1.1 then forces s=t. Hence the parameter is unique and lies in (0,1).

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Convexity and relative closedness of forward sectors

Statement

Under Forward-sector hypotheses for a simple state domain, for every simple state x, the forward sector Ax={y∈Γi:x≺y} is convex and closed relative to Γi. For collinear y=tx+(1−t)z, 0<t<1, y≺z implies x≺y, and x≺z implies x≺y.

Facts & Assumptions

Given: A1–A7 and S1–S3 on a nonempty open convex domain Γi.

[A1]

Forward-sector hypotheses for a simple state domain: A1–A7 include consistency, scaling, splitting, convex-combination access, and stability for positive spectators.

[F1]

Relative interiors of finite-dimensional convex sets: a nonempty convex set has nonempty relative interior, and segments from a relative-interior point to a closure point, excluding the closure endpoint, remain in the relative interior.

[F2]

Spectator cancellation without comparison completeness: any finite common spectator cancels when the two bases have equal mass, under A1–A6 without CH.

Proof

1.1A1algebra

If x≺y and x≺z, splitting gives x∼(tx,(1−t)x). Consistency, positive scaling, and transitivity then give (tx,(1−t)x)≺(ty,(1−t)z). The convex-combination access hypothesis in [A1] gives (ty,(1−t)z)≺ty+(1−t)z, so the convex combination belongs to Ax.

1.2A1F2algebra

Let y=tx+(1−t)z with 0<t<1. The convex-combination access hypothesis in [A1] gives (tx,(1−t)z)≺y. Splitting gives y∼(ty,(1−t)y). If y≺z, scaling by 1−t and adding ty gives (ty,(1−t)y)≺(ty,(1−t)z). Compose and cancel the common (1−t)z by [F2], then scale by 1/t, to get x≺y. If instead x≺z, scaling by 1−t and adding tx gives (tx,(1−t)x)≺(tx,(1−t)z)≺y; recombination gives x≺y.

2.1A1F1step 1.1algebra

Take y∈Ax‾ Γi. If y∈Ax, the claim is immediate. Otherwise choose w∈ri⁡Ax by [F1]. Because Γi is open, choose z=w+(1+η)(y−w)∈Γi for some η>0. For every integer n>1+η, yn=(1+1/n)y−z/n=w+(1−η/n)(y−w) lies in ri⁡Ax by [F1], so x≺yn. Also y=nn+1yn+1n+1z. The convex-combination access hypothesis in [A1] gives (nn+1yn,1n+1z)≺y; scaling by (n+1)/n and using the reversible split of y gives (yn,z/n)≺(y,y/n).

3.1A1step 2.1algebra∎

Consistency adds the same spectator z/n to x≺yn, and transitivity with step 2.1 gives (x,z/n)≺(y,y/n) for every n>1+η. Apply the stability hypothesis in [A1] to bases x,y, fixed positive equal-mass represented spectators z,y, and the tail sequence ϵm=1/(m+n0) with n0>1+η. It follows that x≺y. Thus every relative closure point belongs to Ax.

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Every forward sector has nonempty interior

Statement

Under Forward-sector hypotheses for a simple state domain, every forward sector Ax has nonempty interior in its ambient simple domain Γi.

Facts & Assumptions

Given: A1–A7 and S1–S3 on an open convex Γi⊂Rd, d=ni+1.

[A1]

Forward-sector hypotheses for a simple state domain: each sector is supported at every state by exactly one hyperplane, and every state has a strict successor.

[F1]

Convexity and relative closedness of forward sectors: Ax is convex and closed relative to Γi; if y∈Ax, then Ay⊆Ax by transitivity.

[F2]

Supporting hyperplanes at relative boundary points: a convex set has a support at its relative boundary in its affine hull, and a set of affine dimension at most d−2 has at least two distinct ambient supports through each of its points.

[F3]

Relative interiors of finite-dimensional convex sets: a convex set has nonempty relative interior, and every segment from a relative-interior point to a closure point, except for the closure endpoint, stays in the relative interior.

Proof

1.1A1F2

Suppose Ax has empty ambient interior. If d=1, S1 supplies a strict successor y≠x; convexity puts the nondegenerate interval between x and y in Ax, contradicting empty interior. Thus take d≥2. The affine hull of Ax has dimension at most d−1. If its dimension were at most d−2, [F2] would give two distinct supporting hyperplanes at x, contradicting S2. Hence its affine hull is a hyperplane H; this H supports Ax at every point, including x.

1.2A1F1F2

By S1 choose y with x≺≺y. Then y∈Ax and x∉Ay. Transitivity gives Ay⊆Ax⊆H. If Ay had affine dimension at most d−2, [F2] would again give multiple ambient supports at y, contrary to S2. Thus aff⁡Ay=H.

1.3F1F2F3choose

Choose w∈ri⁡HAy by [F3]. The segment from w to x is contained in H∩Γi by convexity of both sets, starts in Ay, and ends outside Ay. Its intersection with Ay is closed by relative closedness [F1], so it has a last point z before x. Points immediately after z on this segment lie outside Ay and, because Ay is closed in the open set Γi, outside Ay‾ H. Thus z∈∂HAy‾ H. The support statement [F2] supplies a nonzero functional on H whose zero hyperplane supports Ay at z.

2.1A1F1step 1.3algebra∎

Extend this functional to the ambient Euclidean space by taking its normal vector in the direction space of H. Its zero hyperplane contains the within-H support and is not H itself. Since z∈Ay, transitivity gives Az⊆Ay, so the extended plane supports Az at z; the plane H also supports Az. These are distinct, contradicting S2 at z. Therefore Ax has nonempty ambient interior.

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Nesting and comparison of simple-system states

Statement

Under Forward-sector hypotheses for a simple state domain, the sectors are nested: if y∈Ax, then Ay⊆Ax; equality holds when y∈∂Ax, and Ay⊂int⁡ΓiAx when y∈int⁡ΓiAx. For every x,y∈Γi, either x≺y or y≺x (possibly both). Strict comparison is exactly interior membership: x≺≺y if and only if y∈int⁡ΓiAx.

Facts & Assumptions

Given: A1–A7 and S1–S3 on a simple domain Γi.

[A1]

Forward-sector hypotheses for a simple state domain: Ax is defined by x≺y, every state is in its own sector, and each support plane is unique.

[F1]

Convexity and relative closedness of forward sectors: sectors are convex and relatively closed; the collinear implications hold.

[F2]

Every forward sector has nonempty interior: every sector has nonempty ambient interior.

[F3]

Forward orientation and the Lipschitz boundary graph: if w∈∂Ax, then Aw=Ax.

[F4]

Relative interiors of finite-dimensional convex sets: if w is in the relative interior of a convex set and q lies in its closure, then (1−t)w+tq lies in the relative interior for every 0≤t<1.

Proof

1.1A1F3algebra

If y∈Ax, transitivity gives Ay⊆Ax. If also y∈∂Ax, [F3] gives equality. Suppose instead y∈int⁡Ax. If Ay met ∂Ax at z, then z∈Ay and [F3] would give Az=Ax. Since z∈Ay, transitivity gives Az⊆Ay⊆Ax, so Ay=Ax. But S2 puts y on ∂Ay, contradicting y∈int⁡Ax. Thus Ay avoids ∂Ax and, being a subset of the closed sector Ax, lies in int⁡Ax.

1.2A1F1F2F3

Suppose neither y∈Ax nor x∈Ay. Choose z∈int⁡Ax using [F2]. The segment from y to z lies in Γi and meets ∂Ax at a point w strictly between its endpoints. By [F3], Aw=Ax, so w≺z. The collinear implication in [F1], applied to w=ty+(1−t)z, gives y≺w≺z. Hence every z∈int⁡Ax lies in Ay. Reversing x,y gives int⁡Ay⊆Ax.

2.1F1F2F4step 1.2

Because Ax is full-dimensional by [F2], its affine hull is the ambient space. Fix w∈int⁡Ax. For any q∈Ax, relative closedness gives q∈Ax‾ Γi, and [F4] puts (1−t)w+tq in int⁡Ax for every 0≤t<1. Letting t↑1 shows q∈int⁡Ax‾ Γi. The reverse inclusion follows because Ax is relatively closed. Hence Ax=int⁡Ax‾ Γi; the same holds for Ay. Relative closedness from [F1] therefore extends the two interior inclusions in step 1.2 to Ax⊆Ay and Ay⊆Ax. Equality would put x∈Ay, contradicting the supposition. Thus at least one of x≺y or y≺x always holds.

3.1A1F1step 1.1algebra∎

If x≺y, then y∈Ax. If y is a boundary point, step 1.1 gives Ax=Ay, hence x≺y and y≺x. Otherwise step 1.1 gives Ay⊂int⁡Ax, and x∉Ay because S2 places x on ∂Ax. Thus x≺≺y. Conversely, if y∈int⁡Ax, step 1.1 and S2 show x∉Ay, so x≺≺y.

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Forward orientation and the Lipschitz boundary graph

Statement

Under Forward-sector hypotheses for a simple state domain, each simple domain has one orientation sign εi∈{−1,1}, constant across the domain, such that at fixed work coordinate v, x≺y exactly when εi(uy−ux)≥0. In coordinates oriented so εi=1, for each x∈Γi the projection ρx of ∂ΓiAx onto work coordinates is open and connected; the boundary is the graph (ux(v),v) of a C1 function with locally Lipschitz derivative

Dux(v)=−Pi(ux(v),v).

If y∈∂Ax, then Ay=Ax and their boundary graphs agree.

Facts & Assumptions

Given: A1–A7 and S1–S3 on an open convex simple domain Γi.

[A1]

Forward-sector hypotheses for a simple state domain: each Ax is supported at x by a unique finite-slope plane; Pi is locally Lipschitz.

[F1]

Convexity and relative closedness of forward sectors: Ax is convex and closed relative to Γi, and its collinear order implications hold.

[F2]

Every forward sector has nonempty interior: every Ax has nonempty ambient interior.

[F3]

Supporting hyperplanes at relative boundary points: a convex set has a support at each relative boundary point of its closure in its affine hull.

[F4]

Interior rays from a unique supporting plane: at a full-dimensional convex set's boundary point with a unique support, every direction strictly toward the supporting half-space's interior enters the set's interior for all sufficiently small positive steps.

[F5]

Finite convex functions are locally Lipschitz: every finite convex function on an open convex set is locally Lipschitz.

[F6]

Zero-start integral inequality: a nonnegative continuous function bounded by a constant times its zero-start integral is identically zero.

[F9]

Newton–Leibniz needs only continuity on [a,b], differentiability on (a,b), and a Riemann-integrable extension of the interior derivative: a continuous function on a closed interval, differentiable in its interior with an integrable derivative extension, has endpoint increment equal to the integral of that derivative.

[F10]

A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion: a continuous real function on a compact interval is Riemann integrable.

Proof

Given: [A1] and [F1]–[F10]. Write a support plane's inward unit normal with energy component nonzero, as ensured by its finite-slope form.

1.1A1F1F2F4

To apply [F4] to a sector at a boundary point z, choose r>0 with B‾(z,r)⊂Γi and put K=Ax∩B‾(z,r). This is globally closed and convex, and agrees with Ax near z. It has nonempty interior: join z to an interior point of Ax and take points sufficiently close to z. Supporting planes at z coincide for K and Ax: a local supporting inequality extends along every segment from z to a point of Ax. Any such plane also supports Az⊆Ax at z, so S2 makes it unique. Thus [F4] applies to K and its interior conclusions hold for Ax locally. For fixed x=(ux,vx), the vertical section Ix:={u:(u,vx)∈Ax} is a relatively closed convex interval by [F1]. The support plane at x confines it to one side of ux. Its inward normal has nonzero energy component, so [F4] gives a nontrivial interior vertical segment on that side.

1.2A1F6F8F9F10givenconstruct

Let u,w be two C1 graph solutions of Dq=−Pi(q,v) on a common open domain, and suppose u(v0)=w(v0). Since Pi is locally Lipschitz, choose a convex neighborhood of (u(v0),v0) on which one Lipschitz constant L works. Continuity of u,w lets us choose a convex ball about v0 whose closure lies in the common domain and whose two graph images stay in that neighborhood; take any v in a smaller concentric ball. Put d=v−v0 and γ(t)=v0+td for 0≤t≤1. By [F8], r(t)=u(γ(t))−w(γ(t)) is C1 and r′(t)=[−Pi(u(γ(t)),γ(t))+Pi(w(γ(t)),γ(t))]⋅d. The local Lipschitz bound for Pi gives ∣r′(t)∣≤L∥d∥∣r(t)∣. The derivative is continuous, hence integrable by [F10]; [F9] gives r(t)=∫0tr′(s) ds; hence h(t):=∣r(t)∣≤L∥d∥∫0th(s) ds. Apply [F6] on [0,1] to obtain h=0. Thus u=w near v0. Along any polygonal path in the common open domain, the set of path parameters at which equality holds is closed by continuity and open by this local result; since the path interval is connected and the set is nonempty, equality holds along the whole path. Every open connected subset of Euclidean space is polygonally connected: points reachable from a fixed point by finitely many segments in small balls form a nonempty relatively open and relatively closed set. Therefore equality extends throughout each common connected domain.

2.1A1F1F4F7step 1.1construct

The convex, relatively closed interval Ix contains a nontrivial segment on the inward side by step 1.1 and [F4]. Suppose it does not contain the whole open Γi-fiber on that side. Its endpoint b toward that side then lies inside Γi; relative closedness puts b in Ix, so the interval from x to b is a compact nondegenerate interval by [F7]. Let x1 be its midpoint. Since x1∈Ax, transitivity gives Ax1⊆Ax. The supporting plane of Ax1 has nonzero energy coefficient, so its vertical section lies wholly on one side of x1; being contained in Ix, it lies in one half of [x,b] and has at most half its length. It is nonempty and closed relative to [x,b], hence compact. Repeat: take the midpoint xk+1 of the current compact interval Ixk; then xk≺xk+1, so the next section is nested inside the current one and again has at most half its length. These intervals have nonempty total intersection: otherwise their open complements would cover the first compact interval, a finite subcover would do so by [F7], and nesting would make one interval empty. Their lengths tend to zero, so the intersection has a unique point x∞, still in the first interval contained in Γi. The midpoint recursion and this unique intersection require no selection from an arbitrary countable family. Relative closedness puts x∞ in every Axk, hence Ax∞⊆Axk by transitivity. Its vertical section is therefore only {x∞}, contradicting the nontrivial interior vertical segment at x∞ supplied by [F4]. Thus Ix contains the whole open fiber on its inward side.

3.1F1step 2.1algebra

All states on a fixed work-coordinate fiber have the same orientation. If x is oriented upward and a state z above it were oriented downward, the full upward section from step 2.1 gives x≺z, and the full downward section from z gives z≺w for a state w below x, contradicting the support at x. If a state y below an upward x were oriented downward, choose z above x with x≺z; the first collinear implication in [F1], applied to x between y and z, gives y≺x, contradicting the downward support at y. The reversed argument handles a downward x.

4.1A1F2F3F4step 3.1

By step 3.1, orientation is constant on each fixed-work fiber. The fiber orientation is locally constant in v. At any x, choose one vertical point just inside the supported half-space and one just outside; by [F4] the inside point can be chosen in int⁡Ax. Openness of Γi and of the interior and exterior regions preserves these memberships on nearby fibers. Each such fiber therefore meets ∂Ax between these two energy values. A support of Ax at that boundary point also supports the point's own sector and so has its S2 orientation; the inside/outside placement fixes that orientation. Thus nearby fibers have the same sign as the fiber through x. Since Γi is convex, it is connected; a locally constant two-valued sign is constant.

5.1step 3.1step 4.1algebra

Adopt the coordinate convention εi=1. Step 2.1 and the common orientation give x≺y exactly for uy≥ux at fixed v, with strict comparison for uy>ux.

6.1A1F1F2F3F4step 5.1

Put ρx=πv(∂Ax). At a boundary point z∈∂Ax, [F3] supplies a support to Ax. Because Az⊆Ax, that plane also supports Az and equals its unique S2 plane. Its finite slope gives a nonvertical inward direction; [F4] supplies nearby vertical points inside Ax, while the opposite side is outside by the support. Neighborhoods of one inside and one outside point show every nearby work fiber meets ∂Ax. Hence ρx is open. It is connected because S3 makes ∂Ax arcwise connected and projection is continuous. By step 5.1, every vertical fiber over ρx has exactly one boundary energy; call it ux(v).

7.1A1F1F5F8step 6.1

For v,v′∈ρx, convexity of Ax puts every convex combination of their boundary states in Ax. On any sufficiently small convex ball in ρx, the unique-boundary property and upward orientation make ux the least energy in each fiber. Therefore ux(tv+(1−t)v′)≤tux(v)+(1−t)ux(v′), so ux is finite and convex there. It is locally Lipschitz by [F5]. The inward support at (ux(v),v) gives ux(v′)−ux(v)≥−Pi(ux(v),v)⋅(v′−v). The support at (ux(v′),v′) gives ux(v′)−ux(v)≤−Pi(ux(v′),v′)⋅(v′−v). Local Lipschitz continuity of ux and Pi makes the difference between these linear bounds O(∥v′−v∥2). Thus ux is differentiable with derivative −Pi(ux(v),v), and that derivative is locally Lipschitz. Hence ux∈C1 with locally Lipschitz derivative.

8.1A1F1step 7.1step 1.2∎

If y∈∂Ax, then Ay⊆Ax. The boundary graphs agree at vy and solve the same differential equation. The set of v∈ρy where v∈ρx and uy(v)=ux(v) is nonempty, open by step 1.2, and closed by continuity and relative closedness of ∂Ax. Connectedness of ρy makes it all of ρy. This shows ρy⊆ρx. The same argument, using relative closedness of ∂Ay, shows ρy is relatively closed in connected ρx; it is relatively open because both projections are open. Hence ρx=ρy and the boundary graphs agree. In particular x∈∂Ay, so y≺x; together with y∈Ax and transitivity this gives Ax=Ay.

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Reference-strip gluing proves comparison for scaled copies

Statement

Assume Forward-sector hypotheses for a simple state domain and T1–T4 of Thermal joins and conditional contact hypotheses for every finite join. For any one simple type Γi, every two finite compounds of its scaled copies with the same total coefficient mass are comparable. Thus CH holds on all multiple scaled products of that type. No entropy across different material types is asserted here.

Facts & Assumptions

Given: The stated forward-sector and thermal-contact hypotheses; every thermal join remains a simple domain with these hypotheses.

[A1]

Forward-sector hypotheses for a simple state domain: A1–A7, S1–S3, and the tagged finite-compound conventions hold, without an assumed CH.

[A2]

Thermal joins and conditional contact hypotheses: T1–T4 provide simple thermal joins, reversible equilibrium splits, thermal scaling, and a thermal endpoint pair surrounding each state.

[F1]

A unique reference coordinate inside a thermal strip: every state in a strip with strict thermally equivalent endpoints is equivalent to a unique reference mixture with parameter in (0,1).

[F2]

Nesting and comparison of simple-system states: one-copy comparisons hold on each finite join domain; strict comparison is interior membership, and forward sectors are nested.

[F3]

Signed comparison for forward-sector relations and Spectator cancellation without comparison completeness: signed comparison is defined from positive/negative parts; it is translation invariant, additive, and cancellable under A1–A6.

[F4]

Forward orientation and the Lipschitz boundary graph: each simple domain has one constant fixed-work energy orientation; arbitrarily close states on a vertical segment are strictly ordered in that orientation.

Proof

1.1A1A2F2

By T4, the open strips Σ(a,b):={x:a≺≺x≺≺b} with a≺≺b and a∼Tb cover Γi. They are open: by [F2], a≺≺x is x∈int⁡Aa, and x≺≺b is x∉Ab; the latter set is open because Ab is relatively closed.

1.2A1F1F2F3F4construct

Let Σ(a,b) and Σ(c,d) overlap. Their intersection is open. Choose two states e≺≺f in a small vertical segment contained in the overlap, using [F4]. Let se,sf be their coordinates in the first strip and te,tf in the second, supplied by [F1]. If sf<se, the signed difference Rse−Rsf=(se−sf)(J(b)−J(a)) and positive scaling and signed compatibility [A1, F3] give Rsf≺Rse, so f≺e by the two reference equivalences; if sf=se, those references coincide and give f∼e. Both contradict e≺≺f. Hence sf>se, and the same argument gives tf>te.

2.1F1F3step 1.2algebra

The four reference equivalences give, in signed comparison, f−e∼s(sf−se)(b−a),f−e∼s(tf−te)(d−c). Put r=(sf−se)/(tf−te)>0. Additivity and positive scaling in [F3] give d−c∼sr(b−a). Applying the same algebra to the equivalences for e gives c∼sa+(se−rte)(b−a). Consequently a state in the second strip with coordinate t has first-chart coordinate se+r(t−te). The change of chart is affine with strictly positive slope.

3.1F3step 2.1algebra

The same calculation covers a finite chain of overlapping strips: after each overlap, express the new strip's endpoint states as signed-equivalent affine maps in the existing reference pair. The local strip lemma gives every state z of the added strip a coordinate sz∈R with J(z)∼s(1−sz)J(a)+szJ(b) in the existing chart. Signed addition, positive scaling, translation, and equivalence under negative scaling are valid by [F3]; hence induction along a finite overlap chain gives one such common chart for every state in the union.

4.1A1F5step 1.1step 3.1construct

Take any finite list of target states in Γi and join them in order by finitely many straight segments, which lie in Γi by convexity. The resulting connected path is a finite union of closed bounded segments, hence compact by [F5]. The open strips from step 1.1 cover it, so compactness gives a finite subcover. The intersection graph of those strips meeting the path is connected: otherwise their unions would partition the connected path into two disjoint nonempty relatively open sets. Order a spanning tree from one root strip; each later strip overlaps an earlier one in an open set and can be glued by step 3.1. Thus every target state has a common signed chart J(z)∼s(1−sz)J(a)+szJ(b) with a≺≺b.

5.1A1F3step 4.1algebra∎

Write the arbitrary scaled-copy compounds as A=∑jαjJrjΓi(rjxj),B=∑kβkJhkΓi(hkyk), with rj,hk>0 and nonnegative coefficients, omitting zero terms. The declared reversible transport in [A1] gives JrjΓi(rjxj)∼rjJΓi(xj) and similarly on the other side; ordinary positive scaling and composition therefore reduce them to the unscaled base-type states with effective amounts aj=αjrj and bk=βkhk. Their masses are respectively rjwΓi and hkwΓi before this transport, so equal total mass M implies L=M/wΓi=∑jaj=∑kbk. If M=0, both maps are empty and reflexivity applies. Otherwise use step 4.1 for the finite list of base states xj,yk, with common-chart coordinates sj,tk. Signed addition and splitting [F3] make the transported compounds equivalent to RqA=(L−qA)J(a)+qAJ(b) and RqB=(L−qB)J(a)+qBJ(b), where qA=∑jajsj and qB=∑kbktk. If qA<qB, their signed difference has positive part (qB−qA)J(b) and negative part (qB−qA)J(a), so a≺b and positive scaling compare them; equality gives reflexivity, and qB<qA gives the reverse comparison. Transitivity transfers this order through the reversible transports to the original distinct scaled tags. Hence CH holds for every allowed scaled-copy pair, without identifying those tags or dropping their scale factors.

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Thermal joins force a common mathematical energy orientation

Statement

Let a family of simple domains satisfy Forward-sector hypotheses for a simple state domain, and suppose every pair has an admitted simple thermal join satisfying T1 and T2 of Thermal joins and conditional contact hypotheses. Then the fixed-work orientation signs from Forward orientation and the Lipschitz boundary graph agree throughout the family. One common affine proof-coordinate convention u′=εu, ε∈{−1,1}, therefore makes each fixed-work accessibility order increase with u′ and preserves the energy-sum formula for joins. This is a mathematical coordinate convention; it does not change a physical energy quantity or assert a sign of a measured temperature.

Facts & Assumptions

[A1]

Forward-sector hypotheses for a simple state domain: the component charts are open, represented states have their positive type weights, and the relation is composition-consistent and transitive.

[A2]

Thermal joins and conditional contact hypotheses: T1 sends every admissible split to its join; T2 supplies, for each join state, at least one admissible split accessible in both directions to that join state, at the same summed energy and work coordinates.

[F1]

Forward orientation and the Lipschitz boundary graph: each component domain has a constant sign εΓ with fixed-work comparison equivalent to εΓ(uy−ux)≥0.

[F2]

Spectator cancellation without comparison completeness: any finite common spectator cancels when the two bases have equal mass; its mass need not equal the base mass.

Proof

Given: [A1], [A2], [F1], and [F2].

1.1A1A2F1construct

Suppose two component signs are opposed. Choose a join state Z and a reversible split (x,y) by T2 in [A2]. Write x=(ux,vx) and y=(uy,vy). Openness in [A1] permits a sufficiently small nonzero δ with x′=(ux+δ,vx) and y′=(uy−δ,vy) in their respective charts. Choose the sign of δ so εΓxδ>0; opposed signs imply εΓy(−δ)>0. Thus [F1] gives x≺≺x′ and y≺≺y′. These pairs have the same summed energy and work coordinates and the same represented aggregate composition and mass.

2.1A1A2F1F2step 1.1∎

If (x′,y′)≺(x,y), consistency with y≺y′ gives (x′,y)≺(x′,y′)≺(x,y). Cancel the common represented spectator y by [F2] to get x′≺x, contrary to step 1.1. But T1 gives (x′,y′)≺Z, and the chosen T2 split gives Z≺(x,y), so transitivity produces exactly this forbidden reverse comparison. Opposed signs are impossible. Every pair has an admitted join, hence one common sign ε works for the whole family. The affine change u′=εu preserves openness and convexity and sends u=ux+uy to u′=εux+εuy; its positive orientation describes the same accessibility relation. No physical energy, unit or calibration premise is redefined.

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Thermal classes either fill the domain or split an adiabat

Statement

Let Γ⊂Rn+1 be a simple domain satisfying the forward-sector hypotheses and T1–T4. Let S be the additive extensive entropy supplied by Additive extensive representation of compound comparison after CH for its finite scaled products is proved by Reference-strip gluing proves comparison for scaled copies. Then either all pairs of states in Γ are thermally equivalent, or there are x,y on one entropy level (equivalently, one adiabat ∂Ax by Nesting and comparison of simple-system states) with x̸∼Ty.

Facts & Assumptions

Given: The same-type accessibility relation, the thermal relation, the hypotheses in the Statement, and the current entropy representation.

[A1]

Forward-sector hypotheses for a simple state domain: A1–A7 and S1–S3 hold; the sector theorem supplies a constant fixed-work energy orientation for each type.

[A2]

Thermal joins and conditional contact hypotheses: T1–T4 hold and thermal equivalence is transitive.

[F1]

Reference-strip gluing proves comparison for scaled copies: every equal-mass finite compound comparison of this simple type is complete.

[F2]

Additive extensive representation of compound comparison: under that comparison property, an additive extensive entropy represents every same-type finite-compound comparison.

[F3]

Forward orientation and the Lipschitz boundary graph: each simple domain has a constant orientation sign for fixed-work energy comparisons; in coordinates oriented in that sign, sector boundaries are the stated graphs.

[F4]

Finite convex functions are locally Lipschitz: a finite concave entropy is continuous on the open domain.

[F6]

Nesting and comparison of simple-system states: a boundary point y∈∂Ax has Ay=Ax; strict accessibility is interior membership. Together with [F2], this identifies entropy levels with sector boundaries.

[F7]

Thermal joins force a common mathematical energy orientation: T1 and the reversible-split existence in T2 force a common orientation sign; the common affine proof-coordinate change preserves all energy-sum joins and does not redefine physical energy.

Proof

1.1A1F1F2F3F7

Use [F7]'s common mathematical energy orientation for the proof coordinates. Apply [F2] using [F1] after dividing represented coefficient maps by the fixed positive type weight wΓ, as specified in [A1]. Multiply the resulting unit-state entropy by wΓ to evaluate represented states; this positive factor preserves all same-type order comparisons. Write S(x) for that represented-state value. The convex-combination access hypothesis in [A1] and additivity imply concavity: for 0<t<1, accessibility of (tx,(1−t)y) to tx+(1−t)y gives tS(x)+(1−t)S(y)≤S(tx+(1−t)y). By [F3], S is strictly increasing in energy at fixed work coordinate.

1.2A1F3F4algebra

Fix a work coordinate v and let Jv:={u:(u,v)∈Γ}, an open interval. The restriction sv(u)=S(u,v) is finite, concave, continuous by [F4], and strictly increasing by [F3]. For an interior u, concavity makes the right secant slopes nonincreasing as the secant endpoint moves right, and the left secant slopes monotone as the endpoint approaches u; hence the one-sided derivative limits sv,+′(u) and sv,−′(u) exist. Choose a<u<b in Jv: the secants to a,b bound those limits, so they are finite, and concavity gives s−′≥s+′. Monotonicity gives nonnegative slopes. If s+′(u)=0, then for every z>u concavity bounds (sv(z)−sv(u))/(z−u)≤s+′(u)=0, contradicting strict increase. Thus s−′≥s+′>0. Define Θ(u,v)=[1/sv,−′(u),1/sv,+′(u)], a nonempty compact interval of positive finite reciprocal slopes.

1.3A1A2F1F2F3F4F6

Suppose the second alternative fails, so each adiabat contains one thermal-equivalence class. By [F1]–[F2], Ax={y:S(y)≥S(x)}. Continuity of S and strict increase in energy at fixed work coordinate imply that every point with S=σ has arbitrarily nearby points with larger and smaller entropy; hence each entropy level Eσ={x:S(x)=σ} is a forward-sector boundary. The supposition makes each Eσ lie in one thermal class. Let R=S(Γ) and, for each thermal class Cα, put Rα={S(x):x∈Cα}. These sets partition R: two distinct classes cannot share a level under the supposition. The range R is an interval, since any two points of the convex domain are joined by a segment and the continuous entropy takes every intermediate value along it.

2.1A1A2F1F2step 1.2

Fix x=(u1,v1) and y=(u2,v2) of this same simple type, and let U=u1+u2. The admissible energy splits form the open interval D={w:(U−w,v1)∈Γ, (w,v2)∈Γ}, which contains u2. Define g(w)=sv1(U−w)+sv2(w). For the join state J(x,y), T2 gives an admissible split q whose pair is reversible with the join. For every other split p, T1 gives p≺J(x,y)≺q, so p≺q. The pair states p,q are finite compounds of two copies of this same simple type; [F1]–[F2] and their entropy representation therefore give g(p)≤g(q). Thus q maximizes g. If the initial split x,y is reversible with the join, it is equivalent to q; entropy representation gives equal entropy, so it too maximizes g. Conversely, if u2 maximizes g, then g(u2)=g(q), and [F1]–[F2] make the initial pair reversible with q; composing with q∼J(x,y) shows x∼Ty. For a concave function at an interior point, maximality is equivalent to g+′(u2)≤0≤g−′(u2): right secants are bounded above by the right derivative and left secants below by the left derivative. Here g+′(u2)=−sv1,−′(u1)+sv2,+′(u2),g−′(u2)=−sv1,+′(u1)+sv2,−′(u2). The two inequalities are exactly sv1,+′(u1)≤sv2,−′(u2) and sv2,+′(u2)≤sv1,−′(u1), which say the positive slope intervals intersect; reciprocal reverses their endpoints, so this is equivalent to Θ(x)∩Θ(y)≠∅.

3.1A2F5step 2.1

Let Tα be one thermal class. The compact intervals Θ(x), x∈Tα, pairwise intersect by step 2.1. For any finite subfamily, pairwise intersection implies common intersection: the largest left endpoint is no greater than the smallest right endpoint, since the intervals attaining those two endpoints intersect. Fix x0∈Tα; the sets Θ(x)∩Θ(x0) are closed subsets of the compact interval Θ(x0) and have the finite-intersection property. Compactness [F5] gives a common point, so Iα:=⋂x∈TαΘ(x) is a nonempty compact interval. If distinct classes had intersecting common intervals, representatives x,y would have intersecting Θ intervals and would be thermally equivalent by step 2.1, a contradiction. Thus the intervals Iα for distinct classes are disjoint and strictly ordered on the real line; use this as the order of the classes.

4.1A1F3F4F5step 1.2step 1.3step 3.1

For each work coordinate v, strict continuity of sv maps the open interval Jv onto an open interval Rv⊂R. On Rv, the level-to-class map is nondecreasing: as u increases, concavity makes both one-sided slopes nonincreasing, so both endpoints of Θ(u,v) are nondecreasing. If two such intervals Θ are disjoint, monotonicity of both endpoints forces the earlier interval to lie strictly below the later one; by step 3.1, this is the class order. The local class maps agree on overlaps of Rv because each entropy level lies in one class. The open intervals Rv cover R. Any compact subinterval of R has a finite subcover by these intervals by [F5]; its intersection graph is connected, since otherwise the connected subinterval would be split into two disjoint nonempty relatively open sets. Chaining the local orders through overlaps proves that the class map is nondecreasing on all of R.

5.1A2F2step 4.1∎

Fix σ∈Rα and x∈Eσ. T4 gives x0≺≺x≺≺x1 with x0∼Tx1. Entropy representation gives S(x0)<σ<S(x1), and their levels belong to the same class Rα. The global nondecreasing order from step 4.1 forces every level between them, including an open neighborhood of σ, into Rα. Thus each class set is open in the interval R. Since these disjoint open sets partition connected R, only one nonempty class exists. Under the supposition, all states are thermally equivalent, proving the dichotomy.

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Contact calibrators under explicit join coherence

Statement

Let Γ1,Γ2 be simple domains in a family of finite thermal joins satisfying A1–A7, S1–S3, T1–T5, and C1 of Forward-sector hypotheses for a simple state domain and Thermal joins and conditional contact hypotheses. Then there are x0≺≺x1 in Γ1 and y0≺≺y1 in Γ2 such that

(x0,y1)∼(x1,y0).

Moreover, for every fixed finite composition of scaled copies of Γ1 and Γ2, any two equal-mass states of that same composition are comparable (CH). The conclusion is conditional on the explicitly stated join-coherence and universal-range assumptions C1 and T5.

Facts & Assumptions

Given: The forward-sector and contact hypotheses for the whole join-closed family.

[A1]

Forward-sector hypotheses for a simple state domain: A1–A7 and S1–S3 hold on each member of the family, including finite joins.

[A2]

Thermal joins and conditional contact hypotheses: T1–T5 hold for every ordered pair in the family, and C1 makes an equilibrated joined state thermally equivalent to each equilibrium constituent.

[F1]

Reference-strip gluing proves comparison for scaled copies: CH holds on every finite scaled product of copies of any simple type in the family.

[F2]

Additive extensive representation of compound comparison: under that CH, each simple type and its scaled products have an additive extensive entropy representing accessibility.

[F3]

Thermal classes either fill the domain or split an adiabat: for each simple join, either every pair of its states is thermally equivalent or one of its adiabats contains two thermally inequivalent states.

[F4]

Nesting and comparison of simple-system states and Forward orientation and the Lipschitz boundary graph: a common adiabat has identical forward sectors; fixed-work accessibility is ordered by energy, and each component adiabat has one state at each work coordinate in its projection.

[F5]

Spectator cancellation without comparison completeness and Signed comparison for forward-sector relations: signed equivalence is preserved by finite addition and cancellation of common spectators.

[F8]

Thermal joins force a common mathematical energy orientation: T1–T2 and arbitrary-spectator cancellation force a common mathematical energy orientation; the common affine change preserves the join energy-sum rule.

Proof

Given: [A1]–[A2], [F1]–[F5], and [F8].

1.1A1F1F2F4F8

Work in [F8]'s common affine mathematical energy orientation, without redefining any physical energy or calibration. Apply [F1] and [F2] to each simple type, including every finite join, using the positive normalization by its fixed weight in [A1]. Throughout, entropy at a state means its represented-state value: wΓ times its unit-point-mass representation. This normalization preserves order and finite scalar amounts. The convex-combination access hypothesis in [A1] makes each resulting entropy concave, since the comparison of a scaled pair with its convex-combination state translates under the representation to the concavity inequality. By [F4], each entropy is strictly increasing with energy at fixed work coordinates.

1.2A2F4choose

Use T5 to choose xˉ∈Γ1 and yˉ∈Γ2 with xˉ∼Tyˉ. Let ϕ=J12(xˉ,yˉ) and consider its adiabat ∂Aϕ. T2 gives every Z on that adiabat at least one split Z=J12(xZ,yZ) with xZ∼TyZ.

2.1A2F4step 1.2choose

Suppose a join state Z on ∂Aϕ has two distinct equilibrium splits (x(1),y(1)) and (x(2),y(2)). By the definition of ∼T, both pairs are reversible with Z, hence mutually accessible compounds. Their work coordinates are fixed by Z and their total energies equal its energy. If the first-component energies agreed, both component states in the splits would agree, so distinctness forces unequal first-component energies. If ux(1)<ux(2), fixed-work comparison [F4] gives x(1)≺≺x(2); conservation of total energy gives uy(2)<uy(1), hence y(2)≺≺y(1). Taking (x0,x1,y0,y1)=(x(1),x(2),y(2),y(1)) gives (x0,y1)∼(x1,y0) because both split compounds are reversible with Z. If the first energies occur in the opposite order, swap the two splits and obtain the same calibrator form. Therefore, if no calibrator exists, every Z∈∂Aϕ has a unique equilibrium split.

3.1A1A2F1F4F5step 1.2step 2.1algebra

Assume no calibrator exists. For any equilibrium split (xZ,yZ) of Z∈∂Aϕ, reversibility with Z and the common adiabat give (xZ,yZ)∼(xˉ,yˉ) as compounds. By [F1], xZ and xˉ are comparable; [F4] says their sectors coincide exactly when they are equivalent. If xZ≺≺xˉ, pair reversibility gives (xˉ,yˉ)≺(xZ,yZ). Consistency adds yˉ to xZ≺xˉ, and cancellation [F5] gives yˉ≺yZ. If yZ≺yˉ, consistency and the pair comparison instead cancel to give xˉ≺xZ, a contradiction. Thus yˉ≺≺yZ, making a calibrator. The case xˉ≺≺xZ is symmetric: it gives yZ≺≺yˉ, again a calibrator. Since neither is allowed under the assumption, xZ∼xˉ. Adding spectators to the two directions and cancelling by [F5] then gives both yˉ≺yZ and yZ≺yˉ, so yZ∼yˉ. Thus every split on this adiabat has components in the corresponding component adiabats.

4.1A2F4step 3.1

Under the no-calibrator assumption, fix a first-component work coordinate v1 in the projection of ∂Axˉ. The boundary-graph result [F4] gives exactly one state x∗ on that component adiabat at v1. Every split of a join state on ∂Aϕ with first work coordinate v1 therefore has first component x∗. Each split is thermal, so x∗∼TyZ for its second component. T3 makes all such yZ thermally equivalent, and C1 makes each joined state J(x∗,yZ)=Z thermally equivalent to x∗. Hence all join states on this adiabat over fixed v1 lie in one thermal class. Fixing a second-component work coordinate instead gives the corresponding conclusion for first components.

5.1A2F4step 4.1construct

Let Ω be the open connected work-coordinate projection of ∂Aϕ from [F4]. From any point of Ω, the set reachable by finitely many coordinate-axis segments contained in open boxes in Ω is relatively open, and its complement is also relatively open; connectedness makes the reachable set all of Ω. Along each such segment, step 4.1 gives equivalence of the corresponding join states; T3 composes these finitely many equivalences. Thus the whole adiabat ∂Aϕ is one thermal class. Since the argument works for every starting join state, every join adiabat is one thermal class.

6.1A2F3step 1.2step 2.1step 5.1choose

Still assuming no calibrator, apply [F3] to each join system. Its alternative with two thermally inequivalent states on one adiabat is excluded by step 5.1, so all states of each join system are thermally equivalent. For any x,x′∈Γ1, use T5 with one allowed second work coordinate to choose y,y′ with x∼Ty and x′∼Ty′. C1 relates each component to its joined state, and all join states are in one thermal class; T3 therefore gives x∼Tx′. The same argument gives one thermal class in Γ2. Now take Z=ϕ∈∂Aϕ and a split (x,y) supplied by T2. Choose nonzero sufficiently small δ so (ux+δ,v1)∈Γ1 and (uy−δ,v2)∈Γ2, possible because the domains are open. These states form a distinct split of the same Z. Since the original split has x∼Ty and each component type is one thermal class, T3 gives thermal equivalence of the changed pair as well. This contradicts uniqueness of the split from step 2.1. Therefore a calibrator exists.

7.1F2F5step 6.1algebra

Fix a calibrator x0≺≺x1, y0≺≺y1, whose existence is step 6.1. Choose same-type entropy representatives S1,S2 from [F2], with positive gaps h1=S1(x1)−S1(x0) and h2=S2(y1)−S2(y0). Normalize S~i=Si/hi, so both calibrator increments are one, and put s(x)=S~1(x)−S~1(x0) and t(y)=S~2(y)−S~2(y0). Write u=J(x0)+J(y1) and v=J(x1)+J(y0). The calibrator says u∼v, so v−u=(J(x1)−J(x0))−(J(y1)−J(y0)) is equivalent to zero; translation and cancellation [F5] give (J(x1)−J(x0))∼s(J(y1)−J(y0)). For any state x, the signed map dx=J(x)−(1−s(x))J(x0)−s(x)J(x1) has total mass and S~1-value zero. Its positive and negative parts have equal mass and entropy, so [F2] makes them equivalent. Hence J(x)∼s(1−s(x))J(x0)+s(x)J(x1); the same argument applies to states of the second type. These are formal signed maps when a coefficient lies outside [0,1].

8.1A1F2F5step 7.1algebra∎

For a state pair (x,y), put λ=(s(x)+t(y))/2 and δ=(s(x)−t(y))/2. Adding the signed reductions from step 7.1 makes the pair equivalent to (1−λ)(J(x0)+J(y0))+λ(J(x1)+J(y1)) plus δ[(J(x1)−J(x0))−(J(y1)−J(y0))]. The calibrator increment equivalence from step 7.1 makes the latter term equivalent to zero for either sign of δ. Thus each pair reduces to its reference map at parameter λ. For any finite compound A with amounts M1,M2 of the two types, add and scale these single-state reductions. Writing q1(A),q2(A) for the amount-weighted sums of s(x),t(y), respectively, gives A∼s(M1−q1(A))J(x0)+q1(A)J(x1)+(M2−q2(A))J(y0)+q2(A)J(y1). Using (J(x1)−J(x0))∼s(J(y1)−J(y0)), this is equivalent to M1J(x0)+M2J(y0)+Q(A)(J(x1)−J(x0)), where Q(A)=q1(A)+q2(A). If the total mass is zero, both compounds are empty and reflexivity compares them. Otherwise, for same-composition A,B, their base terms agree. If Q(A)<Q(B), the positive and negative parts of the reference-map difference are (Q(B)−Q(A))J(x1) and (Q(B)−Q(A))J(x0), so x0≺x1 and positive scaling compares the maps forward. If the Q values agree the maps coincide; if Q(A)>Q(B) the reverse comparison holds. Translation, addition, and ordinary-order compatibility [F5] transfer the result to A,B. This proves CH for every fixed finite composition of the two types and their scaled copies.

Remarks

The proof uses C1 in steps 4.1 and 6.1 and T5 in steps 1.2 and 6.1. It does not establish the calibrator from the bare set-valued join formula or from T1–T4 alone. All conclusions remain conditional mathematical consequences of the listed hypotheses.

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Calibration of an arbitrary family of simple-system entropies

Statement

Let I be a nonempty set of primitive simple-system labels containing a distinguished meter label m. Assume every finite separated product is explicitly represented by its finitely many tagged elementary factors; positive scaling and finite product are the only identifications. Suppose the join-closed family satisfies the hypotheses of Contact calibrators under explicit join coherence, and fix a strict meter pair z0≺≺z1 in Γm. Assume the Axiom of Choice. Then one can choose, for every i∈I, an additive constant and positive scale for its same-type entropy so that

S((t1i1,x1),…,(tnin,xn))=∑ℓ=1ntℓSiℓ(xℓ)

is additive, extensive, and characterizes every same-composition comparison on every finite separated product. No conclusion is asserted for transitions that change composition.

Facts & Assumptions

Given: The explicit finite-factor family, common contact branch, strict meter pair, and Axiom of Choice.

[A1]

The Axiom of Choice: every family of nonempty sets admits a choice function; it selects the reference states, entropy representatives, and calibrator data used below.

[F1]

Contact calibrators under explicit join coherence: for each pair of simple types in the join-closed family, strict calibrator endpoints exist and their reversible pair relation identifies the two reference increments in signed comparison.

[F2]

Additive extensive representation of compound comparison: once a type and its scaled products have CH, a finite real additive extensive entropy represents their equal-mass comparisons.

[F3]

Signed comparison for forward-sector relations and Spectator cancellation without comparison completeness: signed comparison is translation-invariant and additive, agrees with ordinary comparison on nonnegative equal-mass maps, and permits common-spectator cancellation.

Proof

Given: [A1], [F1]–[F3].

1.1A1F1F2choose

Use the positive type normalization in Forward-sector hypotheses for a simple state domain whenever invoking [F2]; state entropies evaluate represented maps J(x), and their unit-label entropy density is Si(x)/wΓi. Apply [F2] to the meter type and choose its entropy Sm so Sm(z0)=0 and Sm(z1)=1. Set the meter reference to rm=z0, and use [A1] to choose a reference state ri and one same-type entropy representative for every i≠m. For each i≠m, use [F1] to choose calibrator states ai,bi∈Γi and ci,di∈Γm such that ai≺≺bi, ci≺≺di, and (ai,di)∼(bi,ci); also choose ri∈Γi. For the meter label set am=cm=z0 and bm=dm=z1; the two pairs (z0,z1) and (z1,z0) are the same two-copy meter compound up to permutation. Write Si0 for the selected entropy representative, with Sm0=Sm, and put Δi=Si0(bi)−Si0(ai)>0 and μi=Sm(di)−Sm(ci)>0. Define Si(x):=μiΔi(Si0(x)−Si0(ri)). Then Si(ri)=0 and Si(bi)−Si(ai)=μi, matching the meter entropy increment across the chosen calibrator endpoints. For i=m, Δm=μm=1, so this agrees with the fixed meter normalization.

2.1A1step 1.1construct

For a finite separated compound A with factors (tℓiℓ,xℓ), define S(A):=∑ℓtℓSiℓ(xℓ). The sum is finite and invariant under reordering; concatenating compounds adds the sums, and scaling every amount by t>0 multiplies it by t. The explicit-factor convention in the Statement therefore makes S well-defined, additive, and extensive.

3.1A1F1F2F3step 1.1step 2.1algebra

Fix a composition with amounts Mi for finitely many labels and let A,B be two states of that same composition. For a state x of type i, put si(x):=Si(x)−Si(ai)μi. The same-type entropy representation and the CH for type i show J(x)∼s(1−si(x))J(ai)+si(x)J(bi). Indeed, the signed difference has total coefficient zero and Si-value zero, so its positive and negative parts have equal mass and entropy and therefore compare both ways by [F2]. The calibrator relation gives (J(bi)−J(ai))∼s(J(di)−J(ci)) by taking the signed difference of (ai,di)∼(bi,ci) and canceling common terms with [F3]. Add and scale these identities over the finitely many factors of A. With qi(A):=∑ℓ: iℓ=itℓsi(xℓ), the resulting signed map is A∼s∑iMiJ(ai)+∑iqi(A)(J(di)−J(ci)), where the sums range over the finite support of the composition. The same holds for B. Since they have the same composition, the reference base cancels, and B−A is signed-equivalent to the finite meter map D:=∑i(qi(B)−qi(A))(J(di)−J(ci)). Extend the meter entropy linearly to raw finite signed coefficient maps using density Sm(z)/wΓm at a meter raw tag (Γm,z); in particular its value on J(z) is Sm(z). Then Sm(D)=∑i(qi(B)−qi(A))μi=S(B)−S(A). The positive and negative parts of D have equal mass. If S(A)≤S(B), the same-type entropy representation [F2] for finite meter products gives D−≺D+, hence D compares in the forward signed direction. Translation, addition, and ordinary-order compatibility [F3] transfer this to A≺B. Conversely, if A≺B, the same signed equivalences and translation in [F3] give D−≺D+. The meter representation in [F2] then forces Sm(D)=S(B)−S(A)≥0. Reversing the entropy inequality gives the reverse comparison. Thus S characterizes every same-composition comparison. Each particular comparison uses only finitely many labels; no infinite sum or uncountable product occurs.

4.1A1step 1.1step 2.1algebra∎

The meter normalization fixes the multiplicative unit. Changing the chosen reference state for type i changes only its additive constant; on a finite compound these constants contribute amount-weighted conserved-type offsets. This proves the stated freedom and completes the construction.

Remarks

AC is used only for simultaneous reference-state, entropy-representative, and calibrator choices over the arbitrary label set I. Cross-composition monotonicity requires the additional cost and offset hypotheses developed later.

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Decorated finite-chain costs with catalysts

Definition

Let G be a set of tagged system types closed under finite composition T×T′ and positive scaling tT. Scaling is associative and acts componentwise on products: s(tT)=(st)T and t(T×T′)=(tT)×(tT′). Each type T has a nonempty state set XT of finite tagged compounds, every state in XT is represented by the coefficient-map interpretation of Forward-sector hypotheses for a simple state domain and has the fixed aggregate composition ν(a)=νT of its declared type, using the amount-function interpretation in Forward-sector hypotheses for a simple state domain, and has total coefficient mass m(T)=wT≥0, and νT×T′=νT+νT′, νtT=tνT, with m(T×T′)=m(T)+m(T′) and m(tT)=t m(T). Matching scalar masses alone do not identify aggregate composition. The entropy σT:XT→R is the finite additive extensive entropy from Calibration of an arbitrary family of simple-system entropies; in particular σtT(ta)=tσT(a).

For each pair of types prescribe the product map πT,T′:XT×XT′→XT×T′ by π(a,b)=a+b on represented coefficient maps, and require it to be surjective: XT×T′={a+b:a∈XT,b∈XT′}. Thus each product state admits a factorization, without requiring it to be unique. For t>0, prescribe the scaling bijection st,T:XT→XtT by st,T(a)=ta, with inverse coefficient scaling by 1/t. The type identifications and the precise commuting conditions are specified below. Require σT×T′(a+b)=σT(a)+σT′(b) and σtT(ta)=tσT(a); entropy is therefore independent of any nonunique product factorization. Include the formal empty unit type 1, with X1={0}, where 0 is the empty coefficient map, m(1)=0, σ1(0)=0, T×1=T, and t1=1. A transition edge from type T to type T′ is a pair a∈XT, b∈XT′ such that the underlying equal-mass compounds satisfy a≺b under Forward-sector hypotheses for a simple state domain. Thus an edge can exist only when m(T)=m(T′). Its cost is the finite real number σT′(b)−σT(a).

A decorated finite chain from Γ to Γ′ consists of a catalyst type C∈G, an integer N≥1, types T0=Γ×C,T1,…,TN=Γ′×C, and states ak∈XTk, bk+1∈XTk+1 with an edge ak≺bk+1 for every 0≤k<N. Its cost is the finite real sum

c=∑k=0N−1(σTk+1(bk+1)−σTk(ak)).

Define the ordered extended real line R‾=R∪{−∞,+∞} by placing −∞ below every real and +∞ above every real. Finite addition retains its usual meaning; a real added to either infinity leaves that infinity, and equal-sign infinities add to themselves. Opposite-sign infinite sums are undefined. Positive multiplication preserves each infinity, and negation exchanges the infinities. For a nonempty real set bounded below, its infimum is its greatest real lower bound; completeness ensures existence. Define F:G×G→R‾ by taking F(Γ,Γ′) as the infimum of all such witness costs. Set F=+∞ if there is no witness and F=−∞ if the witness costs are unbounded below. The infimum is over real witness costs, so no subtraction of extended-real infima is used. The structural identifications are product permutation and reassociation, T×1=T, 1T=T, s(tT)=(st)T, t(T×R)=(tT)×(tR) and t1=1. Their state maps are the identity on the corresponding represented coefficient maps: finite addition is commutative and associative, adds zero for the unit, and commutes with coefficient scaling. Thus the displayed product and scaling maps commute with these identifications, which preserve ν, m, σ and accessibility.

Additional reversible identifications, if used, are separate specified data: a set E⊆G2 of pairs (T,R) with νT=νR and, for each pair, a bijection rT,R:XT→XR such that σR(rT,R(a))=σT(a) and a∼rT,R(a) for every a∈XT. Include identity pairs with identity maps and close these data under inverse and composition, with rR,T=rT,R−1 and rR,H∘rT,R=rT,H. Require closure under positive scaling and product with another specified pair (D,H), with rtT,tR(ta)=trT,R(a) and rT×D,R×H(a+d)=rT,R(a)+rD,H(d). The latter must hold independently of every nonunique product factorization. These requirements give exactly the commuting product/scaling diagrams. They preserve aggregate composition and mass; replacing a state by its image preserves and reflects every comparison with any compound, by mutual accessibility, consistency and transitivity. Their maps and all structural maps must agree on overlapping identifications. One may take only the structural data, with no additional pairs. The calibration theorem does not supply additional identifications: these are hypotheses to check whenever chosen. Reversible equal-state splitting is an accessibility assertion; it does not identify every state of a product with a single unsplit state. The cost identities and inequalities require a separate proof.

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Algebraic properties of decorated-chain costs

Statement

For the cost F of Decorated finite-chain costs with catalysts, F(Γ,Γ)=0, F is positively homogeneous, invariant under adding the same catalyst to both endpoints, and satisfies the triangle inequality whenever the extended-real sum on its right is defined. It is subadditive under composition whenever F(Γ1,Γ1′)+F(Γ2,Γ2′) is defined. Every pair of actual forward and reverse decorated-chain witnesses has nonnegative total cost. A finite reverse witness therefore bounds all forward costs below; F=−∞ can occur only when the reverse direction has no witness.

Facts & Assumptions

Given: A1–A6 on accessible finite compounds, the additive extensive entropy σ, and the decorated-chain definition.

[A1]

Forward-sector hypotheses for a simple state domain: accessibility is reflexive, transitive, consistent under common spectators, and positively scalable; stability is available.

[F1]

Decorated finite-chain costs with catalysts: each chain has a finite real cost, and F is its infimum, with empty infimum +∞.

[F2]

Calibration of an arbitrary family of simple-system entropies: σ is additive and extensive on every separated finite compound and represents same-type comparisons.

Proof

1.1A1F1F2construct

A reflexive identity edge from any nonempty state of Γ to itself has cost zero, so F(Γ,Γ)≤0. For a decorated closed chain, start with the compound containing every edge input a0,…,aN−1. Apply the edge a0≺b1 with all other edge inputs as spectators, then apply a1≺b2 with the already produced b1 and all still-unused inputs as spectators, and continue through aN−1≺bN. Consistency and transitivity give ∑k=0N−1ak≺∑k=1Nbk. Since TN=T0 and the intermediate types occur once on each side, these two compounds have the same total type up to the declared reassociation and permutation. Their entropy difference is exactly the chain cost. The same-type entropy representation [F2] makes this difference nonnegative. Hence every closed-chain witness costs at least zero and F(Γ,Γ)=0.

1.2A1F1F2construct

Take finite witnesses for (Γ1,Γ1′) and (Γ2,Γ2′). Run the first chain with a fixed state of Γ2 and its second catalyst as spectators; then run the second chain with the first chain's output state and first catalyst as spectators. Consistency allows both sequences, and each fixed spectator contributes zero entropy difference. The resulting witness from Γ1×Γ2 to Γ1′×Γ2′ has the sum of the two costs. If both infima are finite real numbers, choose witnesses within ϵ/2 of each infimum and let ϵ↓0. If one infimum is −∞ and the other is finite, use a fixed finite witness for the latter and arbitrarily negative witnesses for the former; the product costs are unbounded below. If both are −∞, choose both costs arbitrarily negative, so the product infimum is −∞. If either is +∞ and the other is not −∞, the right side is +∞ and the inequality is automatic. The mixed extended sum +∞+(−∞) is undefined and is excluded by the Statement.

1.3A1F1F2

A witness for F(Γ×C0,Γ′×C0) is a witness for F(Γ,Γ′) with enlarged catalyst C0×C, so F(Γ,Γ′)≤F(Γ×C0,Γ′×C0). Conversely, add one fixed state of the nonempty catalyst C0 to every edge of an arbitrary witness for F(Γ,Γ′). Its cost is unchanged, proving the reverse inequality. Thus F(Γ×C0,Γ′×C0)=F(Γ,Γ′).

2.1A1F1F2step 1.1

Scaling every state, type, and catalyst in a witness by t>0 multiplies its cost by t because σtT(ta)=tσT(a). This gives F(tΓ,tΓ′)≤tF(Γ,Γ′); applying the same argument with scale 1/t gives equality, including the extended values.

3.1A1F1F2step 1.1construct∎

Concatenate a witness from Γ to Γ′ with one from Γ′ to Γ′′, adding the other chain's fixed catalyst state as spectator to each edge. The result is a witness from Γ to Γ′′ whose cost is the sum. Taking finite approximations proves the triangle inequality when both costs are finite; an unbounded-below term with a finite witness in the other direction gives the corresponding −∞ case; if a term is +∞ and the sum is defined, the inequality is automatic. For an actual forward witness and an actual reverse witness with possibly different catalysts, add the reverse catalyst's fixed state to every forward edge and the forward catalyst's fixed state to every reverse edge. Consistency supplies these padded edges, and additivity makes each edge cost unchanged. The padded chains now have the same catalyst, so concatenate them into a closed chain; step 1.1 gives c+d≥0. Thus any fixed finite reverse witness of cost d bounds every forward witness below by −d.

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Catalyst-chain cost characterizes cross-type transitions

Statement

Assume the calibrated entropy family of Calibration of an arbitrary family of simple-system entropies, the cross-type accessibility axioms A1–A6 of Forward-sector hypotheses for a simple state domain, and the decorated cost F of Decorated finite-chain costs with catalysts. For every pair of types Γ,Γ′ with m(Γ)=m(Γ′), and states x∈XΓ, y∈XΓ′,

x≺y⟺σΓ(x)+F(Γ,Γ′)≤σΓ′(y).

The infimum defining F need not be attained. If F=+∞, the right side is false; if F=−∞, the right side is true.

Facts & Assumptions

Given: The complete additive extensive entropy on every finite type, the finite-chain cost, and A1–A6 including stability.

[A1]

Forward-sector hypotheses for a simple state domain: consistency permits running accessible edges with common spectators; stability removes fixed positive spectators whose common scale tends to zero.

[F1]

Calibration of an arbitrary family of simple-system entropies: every type and finite separated product has a finite real additive extensive entropy representing same-composition comparisons; its Statement fixes a strict meter pair z0≺≺z1.

[F2]

Decorated finite-chain costs with catalysts: F is the infimum of the real costs of finite chains with an arbitrary common catalyst.

[F3]

Algebraic properties of decorated-chain costs: every actual decorated-chain witness has a finite real additive cost, and costs have no negative closed cycles.

[F4]

Spectator cancellation without comparison completeness: common finite compound spectators can be cancelled under A1–A6.

[F5]

Reciprocals of the positive integers decrease to zero: with natural index k, the real sequence ϵk=1/ιN(σ(k)) is positive, strictly decreasing and converges to zero; its Facts F2–F3 explicitly interpret the Statement's denominator k+1 as ιN(σ(k)).

Proof

1.1F1F2

If x≺y, the one-edge witness with an identity catalyst has cost σΓ′(y)−σΓ(x). Hence F(Γ,Γ′)≤σΓ′(y)−σΓ(x), which is the forward inequality, including the case F=+∞ where no such transition can exist.

1.2F1F2F3F5choose

Conversely assume the displayed entropy inequality. It is impossible when F=+∞. Fix an arbitrary natural index n∈N, use ϵn from [F5] and choose a decorated witness of real cost cn≤Δσ+hϵn, where Δσ=σΓ′(y)−σΓ(x) and h=σ(z1)−σ(z0)>0 is the entropy gap of a fixed strict meter pair from [F1]. Such witnesses exist from the definition of infimum, also if F=−∞. Write their types as T0=Γ×C,T1,…,TN=Γ′×C, their edges as ak≺bk+1, and their endpoints as a0=(x~,c0), bN=(y~,d0).

2.1A1F1F3F5step 1.2algebra

Put δn=ϵn. In one finite compound type form Ln=(x,bN,b1,…,bN−1,δnz0),Rn=(a0,y,a1,…,aN−1,δnz1). These have the same composition: both contain Γ, Γ′, the catalyst C, every intermediate type T1,…,TN−1, and one scaled meter state, up to reassociation and permutation. Additivity of σ and the definition of chain cost give σ(Rn)−σ(Ln)=Δσ−cn+δnh≥0. Thus the same-composition entropy representation [F1] gives Ln≺Rn.

3.1A1F4step 2.1

Starting from Rn, use consistency on a0≺b1 with y,δnz1,a1,…,aN−1 as spectators. The result contains b1 in place of a0. Next use a1≺b2 with y,δnz1,b1,a2,…,aN−1 as spectators, and continue in order. At stage k, the current compound contains ak, the earlier outputs b1,…,bk, all later inputs ak+1,…,aN−1, and the unchanged y,δnz1. After the last edge, transitivity gives Rn≺Rn′, where Rn′=(b1,…,bN,y,δnz1). Compose this with Ln≺Rn from step 2.1. The compounds Ln and Rn′ share the finite spectator b1+⋯+bN; cancel it by [F4]. The result is x+δnz0≺y+δnz1 for this n. Since n was arbitrary, this comparison holds for every natural index; the chain witnesses need not be chosen simultaneously.

4.1A1F5step 3.1∎

By [F5], δn↓0, stability in [A1] with bases x,y and fixed positive spectators z0,z1 yields x≺y. This proves the reverse implication without assuming that any infimum is attained.

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Cost-zero type offsets and their difference space

Definition

For the type set G and cost F of Decorated finite-chain costs with catalysts, define the relation Γ≈0Γ′ by F(Γ,Γ′)=F(Γ′,Γ)=0. Let P:=G×G be the set of raw type pairs. Define a relation on raw pairs by

(A,B)≡(A′,B′)⟺∃C such that A×B′×C≈0A′×B×C.

The separate well-definedness lemma proves that ≈0 is an equivalence compatible with composition and positive scaling, that ≡ is an equivalence on P, and that the following operations are well-defined on its classes. After that proof, write VF:=P/≡. Define addition by [(A,B)]+[(D,E)]:=[(A×D,B×E)], negation by −[(A,B)]:=[(B,A)], and positive scaling by t[(A,B)]:=[(tA,tB)]. Set zero to [(1,1)], where 1 is the empty unit type. Complete scalar multiplication for all t∈R by 0[(A,B)]:=[(1,1)] and, for t<0, t[(A,B)]:=[((−t)B,(−t)A)]; here −t>0, so only the already specified positive scaling of types is used. The same lemma establishes that these operations make VF a real vector space; no vector-space law is assumed by this Definition.

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The cost-zero offset construction is a real vector space

Statement

For the construction in Cost-zero type offsets and their difference space, ≈0 is an equivalence compatible with product and positive scaling; VF is a real vector space and its displayed operations are representative-independent.

Facts & Assumptions

Given: The decorated cost algebra, type composition and scaling, and the formal pair relation in the Definition.

[F1]

Cost-zero type offsets and their difference space: types are identified at zero cost in both directions, and a pair difference is identified after adding a common catalyst type.

[F2]

Algebraic properties of decorated-chain costs: costs satisfy nonnegative closed-cycle bounds, triangle inequalities, translation invariance, and positive homogeneity.

[A1]

Forward-sector hypotheses for a simple state domain: reversible splitting and recombination are allowed for every positive scaled amount.

Proof

1.1F1F2algebra

The relation ≈0 is reflexive because F(Γ,Γ)=0, and symmetric by definition. Suppose Γ≈0Γ′ and Γ′≈0Γ′′. The triangle inequality gives both directed costs between Γ and Γ′′ at most zero. For any ϵ>0, choose a finite reverse witness of cost d<ϵ; such a witness exists because the reverse infimum is at most zero; a value below ϵ must exist by the greatest-lower-bound convention, even if that infimum were initially −∞. Every forward witness of cost c closes with it, so [F2] gives c+d≥0, hence c>−ϵ. Letting ϵ↓0 shows every forward witness has nonnegative cost and its infimum is at least zero; the same argument reverses the direction. Both infima are therefore zero, proving transitivity. The same argument proves product compatibility: subadditivity makes both product costs at most zero, and any finite reverse witness bounds every forward product witness below by a quantity tending to zero. Positive homogeneity gives scaling compatibility directly. Thus ≈0 is an equivalence compatible with product and positive scaling, and the quotient is a commutative monoid.

1.2A1F1F2

Reversible splitting [A1] gives accessible transitions in both directions between (s+t)Γ and sΓ×tΓ. Extensivity makes each such edge have zero entropy cost. Any reverse witness bounds all forward costs below by [F2]'s closed-cycle inequality, so both infima are zero. Hence (s+t)Γ≈0sΓ×tΓ. The empty type 1 is an identity under composition; its reassociation and permutation identifications have zero cost by the type conventions.

2.1F1step 1.1algebra

The pair relation ≡ is reflexive using unit catalyst 1 and symmetric by swapping the two sides. Suppose (A,B)≡(A′,B′) via catalyst C, and (A′,B′)≡(A′′,B′′) via catalyst D. Their witnesses are A×B′×C≈0A′×B×C,A′×B′′×D≈0A′′×B′×D. Multiply the first equivalence by B′′×D, the second by B×C, and compose. Reassociating and permuting gives A×B′′×(B′×C×D)≈0A′′×B×(B′×C×D), which witnesses (A,B)≡(A′′,B′′). Hence ≡ is an equivalence relation.

3.1F1F2step 1.1step 2.1

Suppose (A,B)≡(A′,B′) via C, so A×B′×C≈0A′×B×C. For an arbitrary pair (D,E), multiply this witness by D×E. Reassociation gives (A×D)×(B′×E)×C≈0(A′×D)×(B×E)×C, exactly the witness for (A×D,B×E)≡(A′×D,B′×E). Apply the same argument to a change of representative of (D,E) and use transitivity; addition is representative-independent. Scaling a catalyst witness by t>0 and using product compatibility from step 1.1 proves that positive scalar multiplication is representative-independent. Product commutativity and associativity give the additive laws, and the empty type gives (A×1,B×1)=(A,B). Swapping a pair gives an inverse: its sum with (A,B) is (A×B,B×A), equivalent to (1,1) with unit catalyst because the two product types agree up to permutation and have zero cost in both directions. Thus the classes form an abelian group with zero [(1,1)].

4.1F1step 1.2step 3.1

For t>0, the displayed operation t(A,B)=(tA,tB) satisfies t(v+w)=tv+tw because scaling distributes over type composition, and t(sv)=(ts)v for s>0 by associativity of type scaling. Also 1v=v. The scalar-addition law for positive s,t follows from step 1.2: (s+t)[A,B]=[(s+t)A,(s+t)B]=[sA×tA,sB×tB]=s[A,B]+t[A,B]. The zero scalar is defined as the additive identity, and (−t)v is defined as t(−v), where −v swaps the pair.

5.1F1step 4.1step 3.1algebra∎

The positive scalar laws extend to negative scalars by swapping pairs. For the scalar-addition law with opposite signs, write s−t with s,t≥0. If s≥t, positive scalar addition gives sv=(s−t)v+tv, so cancellation in the additive group gives (s−t)v=sv−tv; if t>s, apply the same identity to t=(t−s)+s and negate. Cases with zero are immediate. For scalar composition, write each nonzero scalar as its sign times a positive magnitude; the defined action swaps the pair once for each negative sign and scales by the product of magnitudes, so (rs)v=r(sv), with zero cases immediate. For distributivity over vector addition, positive scalars distribute by step 4.1 and a negative scalar distributes by taking the additive inverse of that identity. Together with the abelian-group laws in step 3.1, these are all real-vector-space axioms.

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No-sinks cost offsets give a weak global entropy

Statement

Assume the Axiom of Choice. Let F be the decorated cost of Decorated finite-chain costs with catalysts, and assume the no-sinks condition

F(Γ,Γ′)<+∞⟺F(Γ′,Γ)<+∞

for every type pair. Let VF be the real difference space of Cost-zero type offsets and their difference space. Then there is an additive extensive offset B:G→R such that

−F(Γ′,Γ)≤B(Γ)−B(Γ′)≤F(Γ,Γ′).

If every admitted cross-type transition x≺y from Γ to Γ′ satisfies the calibrated cost inequality of Catalyst-chain cost characterizes cross-type transitions, then σΓ(x)+B(Γ) is nondecreasing across every such transition. Only this weak inequality is asserted; no-gap equalities and strict extensions are not concluded.

Facts & Assumptions

Given: The no-sinks condition, A1–A6 for cross-type transitions, the finite-chain costs, and the Axiom of Choice.

[A1]

The Axiom of Choice: adopt AC for the Hahn–Banach dominated-extension and basis-extension steps in the vector-space supplier.

[F1]

Cost-zero type offsets and their difference space and The cost-zero offset construction is a real vector space: VF is a real vector space of formal type differences, with composition additive and positive scaling homogeneous.

[F2]

Algebraic properties of decorated-chain costs: F(Γ,Γ)=0, F is translation-invariant, positively homogeneous and subadditive, finite reverse witnesses bound forward costs, and costs have no negative cycles.

[F3]

Infinite-valued no-sinks Hahn–Banach sandwich and conditional offsets: under AC, a sublinear p:V→R∪{+∞} with p(0)=0 and symmetric finite domain admits a linear L with −p(−v)≤L(v)≤p(v).

[F4]

Catalyst-chain cost characterizes cross-type transitions: every admitted transition x≺y satisfies σΓ′(y)−σΓ(x)≥F(Γ,Γ′).

Proof

Given: [A1], [F1]–[F4], and the no-sinks condition.

1.1A1F2given

If F(Γ,Γ′)<+∞, no-sinks gives a finite reverse witness of some real cost c. Concatenating it with any forward witness makes a closed chain, so [F2] gives forward cost at least −c. Thus F(Γ,Γ′)>−∞ whenever it is finite; all cost values are in R∪{+∞}.

2.1F1F2step 1.1algebra

For a pair class v=[Γ,Γ′]∈VF, define p(v):=F(Γ,Γ′). Suppose (Γ,Γ′)≡(Λ,Λ′) via C, so P=Γ×Λ′×C≈0Q=Λ×Γ′×C. Put D=Γ′×Λ′×C. Since P≈0Q, the triangle inequality gives F(P,D)≤F(Q,D) and F(Q,D)≤F(P,D); the costs have no −∞ by step 1.1, so these inequalities imply equality. Catalyst invariance from [F2] gives F(P,D)=F(Γ,Γ′) and F(Q,D)=F(Λ,Λ′). Therefore p is independent of the pair representative.

2.2F1F2givenstep 1.1

The identity F(Γ,Γ)=0 gives p(0)=0. Homogeneity of F gives p(tv)=tp(v) for t>0, and its product subadditivity gives p(v+w)≤p(v)+p(w). The no-sinks condition says p(v)<+∞ exactly when p(−v)<+∞. By step 1.1, p has no −∞ value.

3.1A1F1F3step 2.2algebra

Apply [F3] to VF and p. It gives a real-linear L:VF→R with −p(−v)≤L(v)≤p(v) for all v. Put B(Γ)=L([Γ,1]), where 1 is the empty unit type. Since [Γ×Γ′,1]=[Γ,1]+[Γ′,1], B is additive. Since [tΓ,1]=t[Γ,1] (the unit type satisfies t1=1), B(tΓ)=tB(Γ) for t>0. Also [Γ,Γ′]=[Γ,1]−[Γ′,1], so applying the sandwich to this pair gives the displayed two-sided cost bound.

4.1F4step 3.1algebra∎

For an admitted transition, [F4] gives σΓ′(y)−σΓ(x)≥F(Γ,Γ′), while step 3.1 gives B(Γ)−B(Γ′)≤F(Γ,Γ′). Subtracting the latter from the former yields [σΓ′(y)+B(Γ′)]−[σΓ(x)+B(Γ)]≥0. Thus the offset entropy is nondecreasing across every admitted transition.

Remarks

The no-sinks condition is an explicit hypothesis about cross-type reachability. AC is used by Infinite-valued no-sinks Hahn–Banach sandwich and conditional offsets and is carried here. The result does not assert a no-gap condition, a unique offset, or an exact entropy principle for all transitions.

5 · Examples, experiments and counterexamples

None yet.

Sources