Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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 two-dimensional three-vector full-spark real frame

Example

In R2 take Φ=(a1,a2,a3) with a1=(1,0), a2=(0,1), and a3=(1,1). Its intensity map on classes modulo x∼−x is [x]↦(x02,x12,(x0+x1)2). Every pair is linearly independent, so every pair spans R2; consequently, for every partition one side spans R2. Equality of the displayed triples therefore forces two signals to be equal up to a global sign.

Facts & Assumptions

Given: The family a1=(1,0), a2=(0,1), a3=(1,1) in R2 and the real sign-quotient intensity map.

[F2]

Linear combination of a finite list, and the span span⁡(S) as the smallest linear subspace containing S, Linear subspace of a vector space: every span is a subspace containing the listed vectors and is closed under addition and scalar multiplication.

[F3]

The sharp real phase-retrieval count for full-spark measurements: a full-spark real family with at least 2d−1 measurements performs phase retrieval.

Verification

1.1givenF1F2algebra

For every (u,v)∈R2, (u,v)=ua1+va2=(u−v)a1+va3=ua3+(v−u)a2, so each pair spans R2 by [F2]. For independence, a vanishing combination of a1,a2 has coordinates (α,β)=(0,0); a vanishing combination αa1+βa3 has coordinates (α+β,β)=(0,0), forcing β=α=0; and a vanishing combination αa2+βa3 has coordinates (β,α+β)=(0,0), again forcing α=β=0. Thus every pair is linearly independent, so [F1] gives full spark.

2.1step 1.1F3algebra∎

Here d=2 and m=3=2d−1. Applying [F3] to step 1.1 proves phase retrieval on the stated global-sign quotient; the intensity coordinates are exactly the three squared inner products displayed in the example.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

35 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.

Sources