CounterexampleConstruction: 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.

Exchangeability failure can destroy split coverage

Statement refuted

The split-conformal lower coverage bound holds for arbitrary calibration and test distributions, even when their joint law is not exchangeable.

Facts & Assumptions

Given: The feature is fixed at $x_{0}$ . The single calibration response is $Y_{1} = 0$ almost surely and the test response is $Y_{2} = 1$ almost surely. The training sigma-field is trivial, the response space is ${0, 1}$ , the score is $s (x, y) = y$ , $m = 1$ , and $α = 1 / 2$ .

[F1]

The split algorithm takes the $k$ th calibration score for $k = ⌈ (m + 1) (1 - α) ⌉ \leq m$ . Split conformal prediction set

[F2]

The split coverage theorem requires conditional exchangeability of calibration and test examples given training. Split conformal marginal coverage

Counterexample

1.1givenF2

The calibration and test examples are not exchangeable: the event that the first response is $0$ and the second is $1$ has probability $1$ , while the permuted event has probability $0$ .

2.1givenF1step 1.1∎

Here $k = ⌈ 2 (1 - 1 / 2) ⌉ = 1$ , so $q = Y_{1} = 0$ by [F1]. Thus $C (x_{0}) = {y \in {0, 1} : y \leq 0} = {0}$ , whereas $Y_{2} = 1$ almost surely. Coverage is $0$ , refuting the claim without exchangeability.

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