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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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$∣ P_{n + 2} ∣ = C_{n}$

Statement

For every natural number $n$ , the set $P_{n + 2}$ of triangulations of the labelled $(n + 2)$ -gon is finite and has cardinality

$∣ P_{n + 2} ∣ = C_{n} .$

Facts & Assumptions

Given: a natural number $n$ .

[L1]

There is a bijection $T_{n} \to P_{n + 2}$ (There is a bijection $T_{n} \to P_{n + 2}$ for every $n \in N$ ).

[L2]

$∣ T_{n} ∣ = C_{n}$ ( $∣ T_{n} ∣ = C_{n}$ ).

[F1]

If $A$ is finite and $f : A \to B$ is a bijection, then $B$ is finite and $∣ A ∣ = ∣ B ∣$ (The cardinality $∣ A ∣$ of a finite set).

Proof

technique · direct

1.1

The bijection of [L1] identifies $T_{n}$ with $P_{n + 2}$ .

2.1

The set $T_{n}$ has cardinality $C_{n}$ by [L2], so [F1] transports that cardinality along the bijection of step 1.1 and yields $∣ P_{n + 2} ∣ = C_{n}$ .

L2F1step 1.1∎

Remarks

At $n = 4$ this says that the labelled hexagon has $C_{4} = 14$ triangulations. The companion page writes them out in full.

Depends on

Used by

All fourteen triangulations of the labelled hexagon Example

Dependency tree · two levels

19 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, Exercise 3.5.5 (standard reference, not scraped)

∣Pn+2∣=Cn∣Pn+2​∣=Cn​