ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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.

A positive-definite quadratic ellipsoid is a regular level set

Example

Let $A$ be a symmetric positive-definite real $m \times m$ matrix and put $F (x) = ⟨ A x, x ⟩$ . The ellipsoid $F^{- 1} (1)$ is a regular level set, and $T_{x} F^{- 1} (1) = (A x)^{⊥} .$

Facts & Assumptions

Given: A symmetric positive-definite matrix $A$ and the quadratic function $F$ .

[L1]

Positive definiteness means $⟨ A x, x ⟩ > 0$ for every $x \neq 0$ (Positive definite, negative definite, semidefinite, and indefinite quadratic forms, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L2]

The power and product rules give continuous partial derivatives for $F (x) = ⟨ A x, x ⟩$ ; the Jacobian and continuous-partials theorem therefore give $D F (x) h = ⟨ A h, x ⟩ + ⟨ A x, h ⟩ = 2 ⟨ A x, h ⟩$ for symmetric $A$ (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

[L3]

At a regular level point, the level is locally a graph and its tangent space is the derivative kernel (A regular level set is locally a $C^{k}$ graph of dimension $m - n$ , The tangent space to a regular level set).

Verification

technique · direct

1.1

If $F (x) = 1$ , then $x \neq 0$ by [L1], and $A x \neq 0$ because $A x = 0$ would give $F (x) = 0$ .

givenL1

2.1

By [L2], $D F (x) (A x) = 2 ∥ A x ∥_{2}^{2} > 0$ , so $D F (x)$ is a nonzero functional and hence surjective onto $R$ .

step 1.1L2

3.1

Therefore $1$ is a regular value, and [L3] gives $T_{x} F^{- 1} (1) = \ker D F (x) = {h : ⟨ A x, h ⟩ = 0} = (A x)^{⊥}$ .

step 2.1L3

4.1

The calculation also shows that no singular point can occur on the asserted level; positive definiteness and the level value $1$ exclude the only possible degeneracy $x = 0$ .

step 1.1step 3.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

48 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

J. M. Lee, Introduction to Smooth Manifolds, regular-level examples (standard reference, not scraped)