Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated
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 prompt-only reward offset leaves Bradley–Terry comparisons and policy unchanged

Example

For a fixed prompt x, let Yx be finite and nonempty, let qx be a probability vector with qx(y)>0 for every y∈Yx, let β>0, and let rx:Yx→R. For any constant cx∈R, set rx′(y)=rx(y)+cx for every y∈Yx. Then every Bradley–Terry comparison probability at x and the unique Gibbs-optimal policy on Yx are unchanged.

Example references

Finite Bradley–Terry preference model: comparison probabilities depend on reward differences for a fixed prompt.

Finite KL-regularized policy improvement: defines the finite objective being optimized.

Finite-action Gibbs policy theorem: supplies the normalized policy formula.

Rafailov et al., §5.1 and Theorem 1

Purdue PrefOpt course notes, §2 and §3e

Facts & Assumptions

Given: A finite response set Yx, full-support reference qx, positive β, reward rx, and constant cx.

[F1]

Finite Bradley–Terry preference model: Pr⁡(a≻b∣x)=σ(rx(a)−rx(b)) for distinct responses.

[F2]

Finite KL-regularized policy improvement: the finite objective is Fqx,rx,β(p)=∑ypyrx(y)−βD(p∥qx).

[F3]

The finite-action Gibbs policy uniquely maximizes reward minus KL: the Gibbs policy has unnormalized weights qx(y)erx(y)/β and partition sum Zx.

Verification

1.1F1givenalgebra

For any distinct a,b∈Yx, rx′(a)−rx′(b)=(rx(a)+cx)−(rx(b)+cx)=rx(a)−rx(b).

2.1F2F3givenalgebra∎

The objective being optimized is the one in [F2], and [F3] gives its normalized weights. The shifted unnormalized weight is qx(y)erx′(y)/β=ecx/βqx(y)erx(y)/β, so Zx′=ecx/βZx. Dividing each shifted weight by Zx′ cancels the common factor and gives px∗′(y)=px∗(y) for every y.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources