Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-adaptedprecheck 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 five Dyck paths, balanced bracket words, binary trees and pentagon triangulations at semilength 3

Example

At semilength 3, the three Catalan families on this page match as follows.

Dyck pathbalanced bracketsbinary treepentagon triangulation
UDUDUD()()(){ε,0,1,10,11,110,111}{{2,5},{3,5}}
UDUUDD()(()){ε,0,1,10,100,101,11}{{2,4},{2,5}}
UUDDUD(())(){ε,0,00,01,1,10,11}{{1,3},{3,5}}
UUDUDD(()()){ε,0,00,01,010,011,1}{{1,4},{2,4}}
UUUDDD((())){ε,0,00,000,001,01,1}{{1,3},{1,4}}

Facts & Assumptions

Given: the five Dyck paths of semilength 3 displayed in the table above.

[L1]

Balanced bracket words are exactly the words with equal totals and nonnegative prefix balance (Bn is exactly the set of words of length 2n over {(,)} in which every prefix has at least as many ( as ) and the totals are equal); under (↦U, )↦D, these are exactly the step words of Dyck paths (Dyck paths of semilength n).

[L2]

There is a bijection from the binary trees of size 3 to the Dyck paths of semilength 3 (There is a bijection Tn→Dn for every n).

[L3]

There is a bijection from the binary trees of size 3 to the triangulations of the labelled pentagon (There is a bijection Tn→Pn+2 for every n∈N).

Verification

technique · direct
1.1L1

The bracket column is obtained from the Dyck-path column by the letter substitution of [L1], so each row gives matching Dyck and bracket words.

1.2L2

The tree column is chosen so that the bijection of [L2] sends each listed binary tree to the Dyck path in the same row: UDUDUD corresponds to the right comb, UUUDDD to the left comb, and the three middle rows are the three mixed recursive shapes.

2.1L3step 1.2∎

The triangulation column is the image of the tree column under [L3], with the two diagonals determined by the same recursive split. Thus each row records one object in each of the three Catalan families, and the rows are pairwise distinct.

Remarks

  • The point of the table is not the shared count but the functions. The three bijections on the A page carry the first column to the remaining ones row by row.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 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