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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.

A one-step scalar Kalman update

Example

Take a one-step scalar model with X0∼N(0,1), transition X1=X0+q1 for q1∼N(0,1), and observation Y1=X1+r1 for r1∼N(0,1), with the three Gaussian variables independent. At the observation y1=2, the predicted law is N(0,2), the innovation variance is 3, the gain is 2/3, and the selected filtering-kernel value is N(4/3,2/3).

Facts & Assumptions

Given: AC and the stated independent-noise linear-Gaussian model.

[A1]

AC is assumed through the Gaussian model and conditional-law theorem. The Axiom of Choice.

[F1]

Under AC, the Kalman recursion with positive innovation covariance defines an everywhere filtering-kernel version of the exact conditional state law. The finite-horizon linear-Gaussian Kalman filter is exact.

[F2]

The model permits singular positive-semidefinite initial, process, and measurement covariances with mutually independent centered Gaussian noises. Linear-Gaussian state-space models with independent noises.

[F3]

Every positive-semidefinite covariance defines a multivariate-normal law under AC. Multivariate normal law, including singular covariance.

Verification

1.1A1F2F3givenalgebra

Set m0∣0=0, P0∣0=1, F1=H1=1, and Q1=R1=1. The covariances are positive semidefinite and the noises satisfy [F2]. The prediction is m1∣0=0 and P1∣0=1⋅1⋅1+1=2. Thus S1=1⋅2⋅1+1=3>0, and for y1=2 the innovation is v1=2.

2.1A1F1step 1.1algebra∎

Solving the scalar innovation system gives K1=2/3. Therefore m1∣1=0+(2/3)2=4/3 and P1∣1=2−(2/3)3(2/3)=2/3. By [F1], the selected conditional-law version at y1=2 is exactly N(4/3,2/3).

Depends on

Used by

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Dependency tree · two levels

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