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 tail-swap involution on a concrete intersecting pair

Example

Take the identity system with

$P_{1} : (1, 0) \to (2, 1), P_{2} : (0, 1) \to (1, 2),$

where $P_{1}$ has step word NE and $P_{2}$ has step word EN. The two paths meet at the lattice point $(1, 1)$ .

Facts & Assumptions

Given: the intersecting pair above.

[L1]

The intersecting-system involution swaps the tails at the first canonical intersection point and changes the permutation by a transposition (Tail-swapping is a sign-reversing involution on the intersecting systems).

Verification

technique · direct

1.1

The first common point of $P_{1}$ and $P_{2}$ is $(1, 1)$ , reached after the first step in each path.

given

2.1

Splitting at $(1, 1)$ , the prefixes are N and E, and the tails are E and N; swapping the tails therefore gives the new pair NN from $(1, 0)$ to $(1, 2)$ and EE from $(0, 1)$ to $(2, 1)$ .

L1step 1.1

3.1

The new pair carries the transposed endpoint assignment, and applying the same tail swap at $(1, 1)$ again returns the original pair. That is exactly the involution property of [L1] in this concrete case.

L1step 2.1∎

Remarks

The example shows why the meeting point has to be selected canonically. A different intersection choice would not necessarily be undone by a second application.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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, §10.13 (standard reference, not scraped)