TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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Uniqueness up to a positive affine transformation

Statement

Assume the strict-reference hypotheses and let $S$ be the normalized representation from Additive extensive representation of compound comparison. Let $T : X \to R$ be another additive extensive representation of the same comparisons: for every finite compound $C = \sum_{i} c_{i} e_{x_{i}}$ define $T (C) : = \sum_{i} c_{i} T (x_{i})$ , and require $C ⪯ D$ iff $T (C) \leq T (D)$ whenever $m (C) = m (D)$ . Then there are unique $a > 0$ and $b \in R$ such that $T (x) = a S (x) + b$ for every $x \in X$ . Conversely every such positive affine change represents the same comparisons.

Facts & Assumptions

[F1]

Additive extensive representation of compound comparison: $x \sim_{s} R_{S (x)}$ for every $x$ , with $S (x_{0}) = 0$ and $S (x_{1}) = 1$ .

[F2]

Signed comparison of finite coefficient maps and Spectator cancellation and signed-order algebra: signed equivalence is obtained by cancelling common finite coefficients; additive order comparisons extend consistently to finite signed maps.

[A1]

Comparison and stability for compounds: the chosen pair satisfies $x_{0} ⪯ x_{1}$ and not $x_{1} ⪯ x_{0}$ .

Proof

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

1.1F1F2algebra

Fix $x \in X$ and put $s = S (x)$ . The equivalence $x \sim_{s} R_{s}$ in [F1] says that the positive and negative parts of their coefficient difference compare in both directions. Since $T$ represents the same order, their finite $T$ -sums are equal. By its definition on finite compounds and linearity of finite sums, the signed value of $R_{s}$ is $(1 - s) T (x_{0}) + s T (x_{1})$ , including when $s \notin [0, 1]$ . Thus $T (x) = T (x_{0}) + [T (x_{1}) - T (x_{0})] S (x)$ .

2.1A1F1step 1.1algebra∎

Because $x_{0} ⪯ x_{1}$ but not conversely, the exact representation property of $T$ gives $T (x_{0}) < T (x_{1})$ . Hence $a : = T (x_{1}) - T (x_{0}) > 0$ and $b : = T (x_{0})$ give the asserted form. Conversely, multiplying both sides of any represented-value inequality by $a > 0$ and adding the same $b$ times the common compound mass preserves and reflects its order; therefore $a S + b$ represents the same comparisons. The parameters are unique by evaluating at $x_{0}$ and $x_{1}$ , where $S$ equals $0$ and $1$ .

Uniqueness up to a positive affine transformation

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources