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

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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.

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A map from hexagon triangulations to size-four binary trees that is not injective

Statement refuted

Equal Catalan counts do not make a natural-looking rule injective. Define

$f : P_{6} ⟶ T_{4}$

by sending a triangulation $T$ of the labelled hexagon to the canonical comb tree determined only by its closing-side split index $k$ :

if $k = 2$ , take the tree whose left subtree has size $0$ and right subtree has size $3$ ;
if $k = 3$ , take the tree whose subtrees have sizes $1$ and $2$ ;
if $k = 4$ , take the tree whose subtrees have sizes $2$ and $1$ ;
if $k = 5$ , take the tree whose left subtree has size $3$ and right subtree has size $0$ ;

and in every case fill each nonzero subtree by the right comb of the required size.

Facts & Assumptions

Given: the two triangulations $T_{1} = {{2, 4}, {2, 5}, {2, 6}}, T_{2} = {{2, 6}, {3, 5}, {3, 6}} .$

[L1]

Every triangulation of the hexagon has a unique closing-side split index $k$ (For $m \geq 3$ and a triangulation $T$ of the $m$ -gon there is a unique $k$ with $1 < k < m$ such that ${1, k}$ and ${k, m}$ are both chords of $T$ or sides, and $T$ splits along $k$ ).

[L2]

A function is injective when equal outputs force equal inputs (Injection, surjection, bijection).

Counterexample

technique · direct

1.1

Both $T_{1}$ and $T_{2}$ are triangulations of the labelled hexagon, and both have the same closing-side split index $k = 2$ : the side ${2, 6}$ is present in each, and no index smaller than $2$ is available.

L1given

2.1

By the definition of $f$ , both triangulations therefore map to the same canonical size-four comb tree, namely the tree with empty left subtree and right comb of size $3$ . So $f (T_{1}) = f (T_{2})$ .

step 1.1

3.1

The input triangulations are distinct because ${2, 4} \in T_{1}$ but ${2, 4} \notin T_{2}$ . Hence equal outputs do not force equal inputs, so [L2] shows that $f$ is not injective.

L2step 2.1∎

Remarks

The failure is deliberate: the rule remembers only the top split and then replaces the two sides by canonical combs, so it discards most of the triangulation.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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

D. Guichard, An Introduction to Combinatorics and Graph Theory, §3.5 (standard reference, not scraped)