ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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.

The rank-two polynomial-module action

Example

For $n = 2$ , the polynomial action of the operator from Demazure–Lusztig operators satisfy the affine type-A relations satisfies $T_{1} \cdot 1 = Q, T_{1} \cdot X_{1} = X_{2}, T_{1} \cdot X_{2} = Q X_{1} + (Q - 1) X_{2} .$

Facts & Assumptions

Given: $A = Z [Q, Q^{- 1}] [X_{1}^{\pm 1}, X_{2}^{\pm 1}]$ , $s_{1}$ interchanging $X_{1}, X_{2}$ , and the polynomial action defined by the operator lemma.

[F1]

The type-A weight lattice, Laurent ring, and divided differences: $D_{1} (f) = (f - s_{1} (f)) / (1 - X_{1} / X_{2})$ , so $D_{1} (1) = 0$ , $D_{1} (X_{1}) = - X_{2}$ , and $D_{1} (X_{2}) = X_{2}$ .

[F2]

The rational skew group algebra for type A: $K ⋊ S_{n}$ acts on $K$ by $(a [w]) \cdot f = a w (f)$ .

[F3]

Demazure–Lusztig operators satisfy the affine type-A relations: $T_{i} = Q [s_{i}] + (Q - 1) ([1] - [s_{i}]) / (1 - X_{i} / X_{i + 1})$ and it preserves $A$ .

Verification

Given: The rank-two Laurent algebra, the skew-group action in [F2], and the operator in [F3].

1.1F1F2F3algebra

Under the action of [F2], the operator in [F3] acts by $T_{1} \cdot f = Q s_{1} (f) + (Q - 1) (f - s_{1} (f)) / (1 - X_{1} / X_{2}) = Q s_{1} (f) + (Q - 1) D_{1} (f)$ .

2.1F1step 1.1algebra∎

Since $s_{1} (1) = 1$ and $D_{1} (1) = 0$ by [F1], step 1.1 gives $T_{1} \cdot 1 = Q$ . For $X_{1}$ , [F1] gives $T_{1} \cdot X_{1} = Q X_{2} - (Q - 1) X_{2} = X_{2}$ ; for $X_{2}$ it gives $T_{1} \cdot X_{2} = Q X_{1} + (Q - 1) X_{2}$ . These outputs lie in $A$ , as required.

Depends on

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