Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge 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 small expected cost can coexist with frequent finite-horizon violations

Example

Consider a one-state CMDP with actions safe and risky, deterministic self-transition, zero reward, discount γ=1/10, cost 0 for safe, cost 1 for risky, and cost threshold b=1/8. The stationary policy chooses risky with probability 1/10 at every decision, with independent action draws in the first seven decisions. Its expected discounted cost is 1/9<1/8, so it is feasible, while the probability of at least one cost during those seven decisions is 5,217,031/10,000,000>1/2.

Example references

Facts & Assumptions

Given: One state s, two actions, deterministic self-transition, γ=1/10, cost c(s,safe,0,s)=0, cost c(s,risky,0,s)=1, threshold b=1/8, and policy probability 1/10 for risky at each step.

[F1]

Finite discounted constrained Markov decision process: a finite CMDP's transition/reward masses and bounded transition-cost signals determine finite-time state and transition masses.

[F2]

Expected discounted-cost feasibility constraint: Jcπ(μ) is the discounted sum of the finite-time expected costs, and feasibility means Jcπ(μ)≤b.

Verification

1.1F1F2F3givenalgebra

At every time t, the state is s and the expected one-step cost is 1/10. Thus by [F2] and [F3], Jcπ(μ)=∑t=0∞(1/10)t(1/10)=(1/10)/(1−1/10)=1/9.

2.1F2step 1.1algebra

Since 1/9<1/8=b, the policy is feasible under the expected discounted-cost constraint.

3.1givenalgebra∎

The probability that all seven independent action draws are safe is (9/10)7=4,782,969/10,000,000. Hence the probability of at least one cost is 1−(9/10)7=5,217,031/10,000,000>1/2, as claimed.

Depends on

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