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Additive extensive representation of compound comparison

Statement

Let $X$ be nonempty and suppose its finite-support compounds carry the comparison structure of Comparison and stability for compounds over the complete ordered field $R$ . If there are $x_{0}, x_{1} \in X$ with $x_{0} ⪯ x_{1}$ and not $x_{1} ⪯ x_{0}$ , define $S (x) : = \sup L_{x}$ using Reference entropy supremum is finite and attained. Then $S : X \to R$ is normalized by $S (x_{0}) = 0$ , $S (x_{1}) = 1$ , and for all equal-mass compounds $A = \sum_{i} a_{i} e_{x_{i}}$ and $B = \sum_{j} b_{j} e_{y_{j}}$ ,

$A ⪯ B ⟺ \sum_{i} a_{i} S (x_{i}) \leq \sum_{j} b_{j} S (y_{j}) .$

Thus $S$ is an additive extensive representation on finite compounds. If no strict pair exists, CH instead makes every equal-mass pair of compounds equivalent, and any constant state function gives a (degenerate) additive representation. No concavity, continuity, or physical law follows without additional hypotheses.

Facts & Assumptions

[A1]

Comparison and stability for compounds: equal-mass compounds satisfy the preorder, consistency, scaling, stability and comparison hypotheses; the state set is nonempty.

[F1]

Signed comparison of finite coefficient maps and Spectator cancellation and signed-order algebra: signed comparisons are translation-invariant and additive, and agree with ordinary comparison for nonnegative compounds.

[F2]

Reference entropy supremum is finite and attained: in the strict-reference case, each $S (x) = \sup L_{x}$ is finite and $x \sim_{s} R_{S (x)}$ .

[F3]

Ordering the two-state reference mixtures: $R_{λ} ⪯_{s} R_{μ}$ iff $λ \leq μ$ for the chosen strict reference pair.

[F4]

Tagged scalable compounds and Compound mass is well-defined and additive: each state denotes its unit-mass point compound $e_{x}$ ; every compound is the finite sum of its strictly positive support coefficients, or the empty sum. Nonnegative scaling is pointwise, $0 C = 0$ , and a nonnegative compound has zero mass exactly when it is empty.

Proof

Given: [A1] and [F1]–[F4].

1.1F2F3

In the strict-reference case, [F3] gives $R_{λ} ⪯_{s} e_{x_{0}}$ exactly for $λ \leq 0$ , and $R_{λ} ⪯_{s} e_{x_{1}}$ exactly for $λ \leq 1$ . Their suprema are therefore $S (x_{0}) = 0$ and $S (x_{1}) = 1$ .

1.2A1F1F2F4algebra

Let $C = \sum_{i} a_{i} e_{x_{i}}$ have mass $M$ and set $q_{C} : = \sum_{i} a_{i} S (x_{i})$ . Discard any zero terms in this presentation using [F4]. If the support is empty, $C = 0$ , $M = q_{C} = 0$ and $C_{R} = 0$ , so signed reflexivity gives the claimed equivalence. Otherwise each remaining coefficient is strictly positive. For each remaining constituent, [F2] gives $e_{x_{i}} \sim_{s} R_{S (x_{i})}$ ; positive scaling and finite repeated use of composition combine these into $C \sim_{s} C_{R}$ , where $C_{R} : = \sum_{i} a_{i} R_{S (x_{i})} = (M - q_{C}) e_{x_{0}} + q_{C} e_{x_{1}}$ . This calculation is valid for signed coefficients because the signed relation is additive and translation-invariant by [F1].

2.1A1F1F3step 1.2

Apply step 1.2 to both $A$ and $B$ , and write $q_{A}, q_{B}$ for the resulting scalar sums. Since $A \sim_{s} A_{R}$ and $B \sim_{s} B_{R}$ , transitivity gives $A ⪯ B$ exactly when $A_{R} ⪯_{s} B_{R}$ . Their masses agree, and $B_{R} - A_{R} = (q_{B} - q_{A}) (e_{x_{1}} - e_{x_{0}})$ ; [F3] therefore gives $A_{R} ⪯_{s} B_{R}$ exactly when $q_{A} \leq q_{B}$ . This proves the representation formula.

3.1A1F1F4algebra∎

In the no-strict-pair case, take any $x, y \in X$ . By [F4] their unit compounds have equal mass one, so CH gives one of $x ⪯ y$ or $y ⪯ x$ ; if exactly one held it would be a strict pair, so both hold. Fix $r \in X$ . Let $C$ have mass $M$ . If $M = 0$ , [F4] gives $C = 0$ , and reflexivity handles this empty compound without applying the positive-scaling axiom at zero. If $M > 0$ , use the canonical finite decomposition over the support from [F4]; every coefficient there is positive. Scale each constituent equivalence $e_{x} \sim e_{r}$ by its positive coefficient and compose the finitely many comparisons to obtain $C \sim M e_{r}$ . Thus equal-mass $A, B$ are both empty when their common mass is zero, and otherwise each is equivalent to the same $M e_{r}$ ; hence they are equivalent in both cases. For a constant $S (x) = c$ , both compound values equal $c M$ , including zero for empty compounds, so the representation equivalence holds.

Additive extensive representation of compound comparison

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources