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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Nesting and comparison of simple-system states

Statement

Under Forward-sector hypotheses for a simple state domain, the sectors are nested: if $y \in A_{x}$ , then $A_{y} \subseteq A_{x}$ ; equality holds when $y \in \partial A_{x}$ , and $A_{y} \subset {int}_{Γ_{i}} A_{x}$ when $y \in {int}_{Γ_{i}} A_{x}$ . For every $x, y \in Γ_{i}$ , either $x ≺ y$ or $y ≺ x$ (possibly both). Strict comparison is exactly interior membership: $x ≺ ≺ y$ if and only if $y \in {int}_{Γ_{i}} A_{x}$ .

Facts & Assumptions

Given: A1–A7 and S1–S3 on a simple domain $Γ_{i}$ .

[A1]

Forward-sector hypotheses for a simple state domain: $A_{x}$ 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 $w \in \partial A_{x}$ , then $A_{w} = A_{x}$ .

[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 $(1 - t) w + t q$ lies in the relative interior for every $0 \leq t < 1$ .

Proof

1.1A1F3algebra

If $y \in A_{x}$ , transitivity gives $A_{y} \subseteq A_{x}$ . If also $y \in \partial A_{x}$ , [F3] gives equality. Suppose instead $y \in int A_{x}$ . If $A_{y}$ met $\partial A_{x}$ at $z$ , then $z \in A_{y}$ and [F3] would give $A_{z} = A_{x}$ . Since $z \in A_{y}$ , transitivity gives $A_{z} \subseteq A_{y} \subseteq A_{x}$ , so $A_{y} = A_{x}$ . But S2 puts $y$ on $\partial A_{y}$ , contradicting $y \in int A_{x}$ . Thus $A_{y}$ avoids $\partial A_{x}$ and, being a subset of the closed sector $A_{x}$ , lies in $int A_{x}$ .

1.2A1F1F2F3

Suppose neither $y \in A_{x}$ nor $x \in A_{y}$ . Choose $z \in int A_{x}$ using [F2]. The segment from $y$ to $z$ lies in $Γ_{i}$ and meets $\partial A_{x}$ at a point $w$ strictly between its endpoints. By [F3], $A_{w} = A_{x}$ , so $w ≺ z$ . The collinear implication in [F1], applied to $w = t y + (1 - t) z$ , gives $y ≺ w ≺ z$ . Hence every $z \in int A_{x}$ lies in $A_{y}$ . Reversing $x, y$ gives $int A_{y} \subseteq A_{x}$ .

2.1F1F2F4step 1.2

Because $A_{x}$ is full-dimensional by [F2], its affine hull is the ambient space. Fix $w \in int A_{x}$ . For any $q \in A_{x}$ , relative closedness gives $q \in {\overline{A_{x}}}^{Γ_{i}}$ , and [F4] puts $(1 - t) w + t q$ in $int A_{x}$ for every $0 \leq t < 1$ . Letting $t ↑ 1$ shows $q \in {\overline{int A_{x}}}^{Γ_{i}}$ . The reverse inclusion follows because $A_{x}$ is relatively closed. Hence $A_{x} = {\overline{int A_{x}}}^{Γ_{i}}$ ; the same holds for $A_{y}$ . Relative closedness from [F1] therefore extends the two interior inclusions in step 1.2 to $A_{x} \subseteq A_{y}$ and $A_{y} \subseteq A_{x}$ . Equality would put $x \in A_{y}$ , contradicting the supposition. Thus at least one of $x ≺ y$ or $y ≺ x$ always holds.

3.1A1F1step 1.1algebra∎

If $x ≺ y$ , then $y \in A_{x}$ . If $y$ is a boundary point, step 1.1 gives $A_{x} = A_{y}$ , hence $x ≺ y$ and $y ≺ x$ . Otherwise step 1.1 gives $A_{y} \subset int A_{x}$ , and $x \notin A_{y}$ because S2 places $x$ on $\partial A_{x}$ . Thus $x ≺ ≺ y$ . Conversely, if $y \in int A_{x}$ , step 1.1 and S2 show $x \notin A_{y}$ , so $x ≺ ≺ y$ .

Nesting and comparison of simple-system states

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources