ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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 Fibonacci generating function and Binet formula over $Q (\sqrt{5})$

Example

Let $(F_{n})$ be the Fibonacci sequence and put

$ϕ = \frac{1 + \sqrt{5}}{2}, \hat{ϕ} = \frac{1 - \sqrt{5}}{2} .$

Then, in $Q ⟦ x ⟧$ ,

$\sum_{n \geq 0} F_{n} x^{n} = \frac{x}{1 - x - x^{2}},$

and, in the splitting field $Q (\sqrt{5})$ ,

$F_{n} = \frac{ϕ^{n} - {\hat{ϕ}}^{n}}{\sqrt{5}} (n \geq 0) .$

Facts & Assumptions

Given: The Fibonacci initial values and recurrence.

[L1]

The Fibonacci sequence satisfies $F_{0} = 0$ , $F_{1} = 1$ , and $F_{n + 2} = F_{n + 1} + F_{n}$ (The Fibonacci sequence $F_{0} = 0, F_{1} = 1$ and Lucas sequence $L_{0} = 2, L_{1} = 1$ ).

[L2]

Multiplication by the reciprocal recurrence denominator converts a recurrence into its finite numerator (A coefficient sequence is eventually linearly recurrent if and only if its formal generating function is rational).

[L3]

Over a characteristic-zero splitting field, distinct characteristic roots give a unique linear combination of their powers (Over a named splitting field in characteristic zero, repeated characteristic roots give polynomial-times-exponential closed forms).

[L4]

The factors $t - λ$ of the characteristic polynomial correspond to the factors $1 - λ x$ of the reciprocal denominator (Reciprocal-root convention: $χ (t) = \prod_{i} (t - λ_{i})^{m_{i}}$ corresponds to $Q (x) = \prod_{i} (1 - λ_{i} x)^{m_{i}}$ ).

Verification

technique · direct

1.1givenL1L2algebra

If $F (x) = \sum_{n \geq 0} F_{n} x^{n}$ , coefficient extraction using [L1] gives $(1 - x - x^{2}) F (x) = x$ ; [L2] therefore gives the displayed rational generating function.

1.2L4algebra

The polynomial $t^{2} - t - 1$ factors as $(t - ϕ) (t - \hat{ϕ})$ in $Q (\sqrt{5}) [t]$ , in agreement with [L4].

2.1step 1.2L1L3algebra

By [L3], $F_{n} = A ϕ^{n} + B {\hat{ϕ}}^{n}$ . The equations $A + B = F_{0} = 0$ and $A ϕ + B \hat{ϕ} = F_{1} = 1$ give $A = 1 / \sqrt{5}$ and $B = - 1 / \sqrt{5}$ .

3.1step 2.1∎

Substitution in step 2.1 proves Binet's formula, including $n = 0$ and $n = 1$ .

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

R. P. Stanley, Enumerative Combinatorics, vol. 1, 2nd ed., Example 4.1.2 (standard reference, not scraped)

The Fibonacci generating function and Binet formula over Q(5)Q(5​)