Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-12
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 concrete calculation of (L1L2)L3=L1(L2L3)

Example

Let L1={a}, L2={ε,b}, and L3={a,bb} over the alphabet {a,b}. Then (L1L2)L3=L1(L2L3)={aa,abb,aba,abbb}.

Facts & Assumptions

Given: L1={a}, L2={ε,b}, and L3={a,bb}.

[L1]

Language concatenation is LK={uv:uL and vK} by Language concatenation, powers, and Kleene star.

[L2]

Language concatenation is associative by Language concatenation is associative.

Verification

technique · direct
1.1

Using [L1], we have L1L2={aε,ab}={a,ab}. Therefore (L1L2)L3={aa,abb,aba,abbb}.

givenL1
1.2

Again by [L1], L2L3={εa,εbb,ba,bbb}={a,bb,ba,bbb}. Therefore L1(L2L3)={aa,abb,aba,abbb}.

givenL1
2.1

The two explicit calculations in steps 1.1 and 1.2 agree, so this example realizes the identity of [L2] as the four-word set {aa,abb,aba,abbb}.

L2step 1.1step 1.2

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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