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.

Exact evaluation of a one-use BSC dependence-testing bound

Example

For a one-use BSC with crossover p=0.1, uniform input, and M=2, the dependence-testing upper bound on achievable average error is exactly 0.35.

Facts & Assumptions

Given: The BSC transition probabilities W(y∣x)=0.9 for y=x and 0.1 for y≠x, uniform input, and M=2.

[F1]

Finite-DMC dependence-testing achievability bound: a deterministic code exists whose average error is at most Eexp⁡(−[ı(X;Y)−ln⁡((M−1)/2)]+).

[F2]

Finite discrete memoryless channels: a BSC row is a probability vector, and its output law under an input vector is the matrix product PW.

Verification

1.1F2givenalgebra

Under the uniform input, the output is uniform. Therefore ı(X;Y)=ln⁡(0.9/0.5)=ln⁡1.8 with probability 0.9 and ı(X;Y)=ln⁡(0.1/0.5)=ln⁡0.2 with probability 0.1.

2.1F1step 1.1casesalgebra∎

The threshold is ln⁡((M−1)/2)=ln⁡(1/2). On the correct-output event, the integrand is exp⁡[−(ln⁡1.8−ln⁡(1/2))]=1/3.6=5/18, contributing 0.9(5/18)=1/4. On the crossover event, ln⁡0.2<ln⁡(1/2), so the positive part is zero and the integrand is one, contributing 0.1. Thus the expectation is 1/4+1/10=7/20=0.35, and [F1] supplies a code with average error at most this value.

Depends on

Used by

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Dependency tree · two levels

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Sources