Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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.

The cycle lemma on the word (1,+1,+1,1,+1,1,+1)

Example

Let

a=(1,+1,+1,1,+1,1,+1),

a word of length 7 and weight 1. Its seven cyclic shifts and their partial sums are:

shiftpartial sums
(1,+1,+1,1,+1,1,+1)1,0,1,0,1,0,1
(+1,+1,1,+1,1,+1,1)1,2,1,2,1,2,1
(+1,1,+1,1,+1,1,+1)1,0,1,0,1,0,1
(1,+1,1,+1,1,+1,+1)1,0,1,0,1,0,1
(+1,1,+1,1,+1,+1,1)1,0,1,0,1,2,1
(1,+1,1,+1,+1,1,+1)1,0,1,0,1,0,1
(+1,1,+1,+1,1,+1,1)1,0,1,2,1,2,1

Facts & Assumptions

Given: the two words above and the weight-two word (+1,+1,+1,1).

[L1]

If every letter of a length-m integer word is at most 1 and its weight is 1, then exactly one starting index gives a cyclic shift whose nonempty partial sums are all positive (The cycle lemma (Dvoretzky–Motzkin): if every ai1 and a=k1, then exactly k of the m cyclic shifts of a have all partial sums positive).

[L2]

The cycle-lemma Catalan count gives (23+1)C3=(73) ((2n+1)Cn=(2n+1n), a second derivation of the Catalan count).

Verification

technique · direct
1.1

In the table, only the second row has all partial sums strictly positive, so the word a has exactly one good shift.

given
2.1

Deleting the leading +1 from that good shift gives (+1,1,+1,1,+1,1), which is the Dyck word UDUDUD of semilength 3.

step 1.1
2.2

This agrees with [L1] and [L2]: [L1] predicts one good shift, and [L2] reads 7C3=(73)=35, so 75=35.

L1L2step 1.1
3.1

The weight-two word (+1,+1,+1,1) has exactly the two good shifts (+1,+1,+1,1) and (+1,+1,1,+1), so the general statement is visible too: the number of good shifts is the weight.

given

Remarks

  • The good shift is the second row because the page's convention counts strict positivity of every partial sum, not nonnegativity.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

38 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