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ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)
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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.

The alias classes in a 23−1 fraction

Example

Take k=3 with factors A,B,C and the positive generator equation xAxBxC=+1, conventionally written I=ABC. The four runs, in (A,B,C) order, are (−1,−1,+1), (−1,+1,−1), (+1,−1,−1), and (+1,+1,+1). The complete equal-column alias classes are

{I,ABC},{A,BC},{B,AC},{C,AB}.

In the unrestricted model with all eight factorial terms, each pair contributes through its coefficient sum. Separating a main effect from its aliased interaction therefore requires additional runs or an explicit sparsity or hierarchy assumption.

Facts & Assumptions

Given: k=3, the generator equation xAxBxC=+1, the four runs displayed in the Example, and the unrestricted fixed-design model with columns for all subsets of {A,B,C}.

[F1]

Regular two-level fractions, signed defining relations, and resolution: a regular fraction is specified by independent generator equations; H is generated by their signed words, and equal-column aliases are columns that agree on every run.

[F2]

The defining subgroup gives the exact alias relations: A and B have equal columns exactly when (+1,A△B)∈H.

[F3]

Aliased factorial columns prevent separate coefficient identification: if two distinct columns satisfy cA=σcB, their contribution depends only on σβA+βB, so the coefficients are not separately identifiable in the unrestricted model.

Verification

technique · enumerate the fraction and multiply its one defining word by every effect word
1.1givenF1constructchoose

Choose xA,xB∈{−1,+1} freely and set xC=xAxB. This gives exactly the four runs listed in the Example, one for each of the four choices of (xA,xB), and every run satisfies xAxBxC=+1.

2.1F1F2step 1.1algebra

Here H={I,ABC}. By [F2], effects have equal columns exactly when their symmetric difference is either ∅ or {A,B,C}. Pairing each subset with its complement gives precisely {I,ABC}, {A,BC}, {B,AC}, and {C,AB}; these four pairs exhaust the eight subsets.

3.1givenF3step 2.1algebra∎

In each listed pair the columns are equal, so [F3] shows that the corresponding terms enter the fitted vector through βI+βABC, βA+βBC, βB+βAC, or βC+βAB, respectively. The design by itself does not identify the two coefficients within any pair separately.

Sources

Oehlert §18.2 and NIST/SEMATECH §§5.3.3.4.1–5.3.3.4.3 give the standard I=ABC half-fraction and its alias notation. The four-row enumeration and coefficient ambiguity are verified locally.

Depends on

Used by

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Sources