LemmaStatement: 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)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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 $x_{0} ≺ ≺ x_{1}$ with $x_{0} \sim_{T} x_{1}$ , and let $x$ satisfy $x_{0} ≺ ≺ x ≺ ≺ x_{1}$ . For $0 \leq λ \leq 1$ , write $R_{λ}$ for the compound state $((1 - λ) x_{0}, λ x_{1})$ , omitting a zero coefficient at the endpoints. Then there is a unique $λ \in (0, 1)$ with $R_{λ} \sim x$ 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 $x_{0} ≺ ≺ x ≺ ≺ x_{1}$ .

1.1A1F2algebra

For $0 \leq λ < μ \leq 1$ , the signed difference $R_{μ} - R_{λ} = (μ - λ) (J (x_{1}) - J (x_{0}))$ has positive part $(μ - λ) J (x_{1})$ and negative part $(μ - λ) J (x_{0})$ . Since $x_{0} ≺ x_{1}$ , positive scaling and [F2] give $R_{λ} ≺ R_{μ}$ . A reverse comparison would, by [F2] and inverse positive scaling, give $x_{1} ≺ x_{0}$ , contrary to strictness. Thus the reference states are strictly ordered by parameter.

1.2F1F3given

For $0 < λ < 1$ , [F1] makes $(1 - λ) x_{0}$ and $λ x_{1}$ thermally equivalent. By the definition of thermal equivalence, their compound is reversible with its join state. The equal-state split makes $((1 - λ) x, λ x)$ 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 $R_{0} = x_{0}$ and $R_{1} = x_{1}$ directly. Hence $R_{λ}$ and $x$ are comparable for every $λ \in [0, 1]$ .

2.1A1F2step 1.1choose

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

3.1A1F2step 1.2step 2.1

If $s < 1$ , then $s + 1 / n \notin L$ for every sufficiently large $n$ . Step 1.2 gives comparability and the definition of $L$ excludes $R_{s + 1 / n} ≺ x$ , so $x ≺ R_{s + 1 / n}$ . The identity $R_{s + 1 / n} + (1 / n) x_{0} = R_{s} + (1 / n) x_{1}$ and stability yield $x ≺ R_{s}$ . If $s = 1$ , the strict strip gives $x ≺ R_{1}$ ; together with step 2.1 this would imply $x_{1} ≺ x$ , contrary to $x ≺ ≺ x_{1}$ . Thus $s < 1$ , and step 2.1 and the preceding stability comparison give $R_{s} \sim x$ .

4.1F2step 1.1step 3.1algebra∎

The equivalence $R_{s} \sim x$ also excludes $s = 0$ , because $R_{0} = x_{0}$ and $x_{0} \sim̸ x$ , and excludes $s = 1$ by strictness of $x ≺ ≺ x_{1}$ . If $R_{t} \sim x$ for another $t$ , transitivity gives $R_{s} \sim R_{t}$ ; step 1.1 then forces $s = t$ . Hence the parameter is unique and lies in $(0, 1)$ .

A unique reference coordinate inside a thermal strip

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources