ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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.

Simple pendulum phase portrait

Example

Assume ${A C}_{ω}$ . On $T^{*} S^{1}$ consider the normalized pendulum

$H (q, p) = \frac{p^{2}}{2} + 1 - \cos q .$

For $0 < E < 2$ its energy curve describes oscillation; for $E > 2$ its two components describe rotations in opposite directions; and $E = 2$ is the singular separatrix through the unstable equilibrium.

Facts & Assumptions

Given: $q$ is taken modulo $2 π$ and $p \in R$ .

[F1]

Hamilton's equations give $\dot{q} = p$ and $\dot{p} = - \sin q$ . Hamilton equations in canonical cotangent coordinates.

Verification

technique · cases

1.1

The level equation is $p^{2} = 2 (E - 1 + \cos q)$ . Its critical points solve $p = 0$ and $\sin q = 0$ : $(0, 0)$ is a minimum of energy $0$ , while $(π, 0)$ is a saddle of energy $2$ .

givenalgebra

2.1

Assume $0 < E < 2$ . The allowed angles form a proper interval about $q = 0$ ; the positive and negative square-root branches meet at two turning points $p = 0$ , producing a closed oscillatory curve. By [F1], the sign of $p$ is the direction of angular travel and reverses at those endpoints.

assume-case oscillationF1step 1.1

2.2

Assume $E > 2$ . The right-hand side is strictly positive for every $q$ . The two graphs $p = \pm \sqrt{2 (E - 1 + \cos q)}$ are disjoint circles over $S^{1}$ , and [F1] gives rotation with fixed sign.

assume-case rotationF1step 1.1

2.3

Assume $E = 2$ . The two branches meet at the saddle and form the homoclinic separatrix; the level is singular and is not a regular Liouville torus.

assume-case separatrixF1step 1.1

3.1

Finally $E = 0$ gives only the stable equilibrium, while $E < 0$ gives the empty level because $H \geq 0$ . These alternatives and steps 2.1--2.3 exhaust all energies.

step 1.1step 2.1step 2.2step 2.3cases-exhaustive∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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

Ana Cannas da Silva, Lectures on Symplectic Geometry (standard reference, not scraped)