ExampleConstruction: AI-generatedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 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 reflection bijection applied to $U D D U D U$

Example

Take the diagonal path with step word UDDUDU. Its height sequence is

$0, 1, 0, - 1, 0, - 1, 0.$

The first visit to the level $- 1$ is at index $3$ . Reflecting the initial segment through the line $y = - 1$ changes the first four heights to

$- 2, - 3, - 2, - 1,$

so the reflected path has step word DUUUDU and runs from $(0, - 2)$ to $(6, 0)$ .

Facts & Assumptions

Given: the path UDDUDU.

[L1]

If $a > c$ and $b > c$ , reflection sends a path from $(0, a)$ to $(n, b)$ that first visits level $c$ at $τ$ to the path with heights $h^{'} (i) = 2 c - h (i)$ for $i \leq τ$ and $h^{'} (i) = h (i)$ for $i \geq τ$ ; this is a bijection onto the paths from $(0, 2 c - a)$ to $(n, b)$ (Reflecting the initial segment at the first visit to level $c$ ).

[L2]

The Catalan count at semilength $3$ is $C_{3} = (\binom{6}{3}) - (\binom{6}{4}) = 5$ ( $C_{n} + (\binom{2 n}{n + 1}) = (\binom{2 n}{n})$ , The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

Verification

technique · direct

1.1

The path UDDUDU starts at height $0$ , ends at height $0$ , and first reaches the level $- 1$ at the index $3$ .

given

2.1

Reflecting the heights $0, 1, 0, - 1$ through the line $y = - 1$ gives $- 2, - 3, - 2, - 1$ , so the reflected step word is DUUUDU; applying the same reflection to DUUUDU returns UDDUDU.

L1step 1.1

3.1

The count behind the example agrees with [L2]: there are $(\binom{6}{3}) = 20$ diagonal paths from $(0, 0)$ to $(6, 0)$ , $(\binom{6}{4}) = 15$ of them touch the level $- 1$ , and the remaining $5$ are the Dyck paths of semilength $3$ .

L2∎

Remarks

The reflected path is not a Dyck path; that is the whole point. The bijection removes exactly the paths that touch the forbidden level.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

25 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

C. Krattenthaler, "Lattice Path Enumeration", ch. 10 of the Handbook of Enumerative Combinatorics, §10.3 (standard reference, not scraped)

The reflection bijection applied to UDDUDUUDDUDU