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 ballot problem with three votes for $A$ and two for $B$

Example

The ten orderings of three $A$ votes and two $B$ votes are:

$A A A B B, A A B A B, A A B B A, A B A A B, A B A B A, A B B A A, B A A A B, B A A B A, B A B A A, B B A A A .$

Exactly two of them, AAABB and AABAB, keep candidate $A$ strictly ahead after every vote.

Facts & Assumptions

Given: $p = 3$ and $q = 2$ .

[L1]

The ballot theorem gives $(p + q) N = (p - q) (\binom{p + q}{p})$ (Bertrand's ballot problem: for $p > q \geq 0$ the orderings in which the first candidate is strictly ahead throughout satisfy $(p + q) N = (p - q) (\binom{p + q}{p})$ ).

[L2]

For $p \geq q \geq 0$ , if $N$ counts the orderings in which the first candidate is never behind, then $(p + 1) N = (p - q + 1) (\binom{p + q}{q})$ (The weak ballot count: for $p \geq q \geq 0$ the orderings in which the first candidate is never behind satisfy $(p + 1) N = (p - q + 1) (\binom{p + q}{q})$ ).

Verification

technique · direct

1.1

The ten words displayed above are exactly the words of length $5$ with three $A$ letters and two $B$ letters, so there are $(\binom{5}{3}) = 10$ of them.

given

2.1

Reading the lead $A - B$ after each vote shows that only AAABB and AABAB stay strictly positive at every stage, so $N = 2$ .

step 1.1

3.1

This agrees with [L1], since $(3 + 2) N = (3 - 2) (\binom{5}{3})$ reads $5 N = 10$ and therefore $N = 2$ .

L1step 2.1

4.1

For the weak form with $p = q = 2$ , the orderings AABB and ABAB are exactly the ones in which $A$ is never behind, so the weak count is $2$ ; that is $C_{2}$ and agrees with [L2].

L2∎

Remarks

The strict and weak counts differ because ties are allowed only in the second statement. At this size the difference is already visible.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

22 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 ballot problem with three votes for AA and two for BB