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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.

Two indistinguishable laws leave error on an unseen atom

Example

Let X={a,b} with the discrete sigma-algebra and feature marginal PX(a)=PX(b)=1/2, and observe only the labeled pair (a,0). Under P0, the labels at a,b are 0,0; under P1, they are 0,1. The common learner input consists of this labeled pair and the full marginal PX. Let H={g:X→{0,1}}, and let Q:H→[0,1] satisfy ∑g∈HQ(g)=1, selected from that common input, so it is the same under both laws. For g∈H, define RP0(g)=121{g(a)≠0}+121{g(b)≠0},RP1(g)=121{g(a)≠0}+121{g(b)≠1}. The expected population risks satisfy max⁡j∈{0,1}∑g∈HQ(g)RPj(g)≥1/4.

Two-law diagram

Each panel gives feature, label, and mass. The input at a is observed in both panels; the feature masses agree, while only the unobserved label at b changes.

diagram failed to render:
\begin{tikzpicture}
\draw (0,0) rectangle (3.5,1.8);
\node at (1.75,1.45) {$P_0$};
\node at (0.9,0.9) {$a:0$};
\node at (2.6,0.9) {$b:0$};
\node at (0.9,0.35) {$1/2$};
\node at (2.6,0.35) {$1/2$};
\draw (4,0) rectangle (7.5,1.8);
\node at (5.75,1.45) {$P_1$};
\node at (4.9,0.9) {$a:0$};
\node at (6.6,0.9) {$b:1$};
\node at (4.9,0.35) {$1/2$};
\node at (6.6,0.35) {$1/2$};
\end{tikzpicture}

Facts & Assumptions

Given: The finite feature set {a,b}, common masses 1/2 at each point, observed label 0 at a, and the two deterministic label laws in the Example.

[F1]

The unlabeled marginal alone does not identify labels: for an unseen atom of mass q>0, the maximum expected risk under two label laws differing only at that atom is at least q/2 for any common predictor distribution.

[F2]

Population and empirical risk: population risk is the expected loss under the joint law; for the finite zero-one laws in this Example it is the sum of each atom mass times its error probability.

Verification

1.1givenF1algebra

Here the unseen atom is b and q=PX(b)=1/2. The two laws agree at the observed point a and have the same PX, so the theorem in [F1] applies and gives a maximum expected risk of at least q/2=1/4.

2.1givenF2step 1.1algebra∎

Directly, let p=Q{g∈H:g(b)=1}. The two expected risks have the same nonnegative contribution from a. Their contributions at b are respectively p/2 under P0 and (1−p)/2 under P1, so their sum is at least 1/2 and their maximum is at least 1/4.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.