ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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.

Superextremal RN absence of radial horizon

Example

Let $M > 0$ and $Q^{2} = q_{e}^{2} + q_{m}^{2} > M^{2}$ . In the selected ingoing RN model, $f (r) > 0$ for every $r > 0$ , so no constant- $r$ null hypersurface arises from a radial root. Every point has a future radial null path reaching arbitrarily large $r$ , and the escape set is the whole manifold with empty relative boundary. The escape terminology uses the specified interface Causal solutions, Killing horizons, escape and metric extensions. This is not a collapse solution or a cosmic-censorship counterexample.

Facts & Assumptions

Given: $M > 0$ , $Q^{2} > M^{2}$ , and the selected ingoing RN metric and orientation.

[F1]

Selected ingoing RN extension and time orientation: $g = - f d v^{2} + 2 d v d r + r^{2} γ$ , $g^{- 1} (d r, d r) = f$ , and $n = \partial_{v} + (f / 2) \partial_{r}$ is future null.

[F2]

The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive: an integral of a continuous integrand has that integrand as its derivative.

[L1]

Causal solutions, Killing horizons, escape and metric extensions, Definition (d): the escape set uses continuous future piecewise- $C^{1}$ causal paths, with finite $C^{1}$ subdivisions on compact parameter segments and nonzero future causal piece tangents. A single such path must reach arbitrarily large values of the specified radius; the escape boundary is relative to the selected model, without a conformal completion.

[A1]

The radial domain is $r > 0$ ; lengths are measured in the stated geometric units. The claim concerns only the radial-root and escape notions just specified.

Verification

1.1F1A1algebra

For each $r > 0$ , $r^{2} f (r) = r^{2} - 2 M r + Q^{2} = (r - M)^{2} + (Q^{2} - M^{2}) > 0.$ Therefore $f (r) > 0$ and the quadratic has no real root. Since $g^{- 1} (d r, d r) = f$ by [F1], every positive constant- $r$ level has spacelike conormal, not a null conormal. This proves there is no positive radial root or constant- $r$ null level in this example.

1.2F1F2L1A1algebra

Fix any $r_{0} > 0$ , $v_{0}$ and sphere point $ω_{0}$ . For $r \geq r_{0}$ put $v (r) = v_{0} + \int_{r_{0}}^{r} \frac{2}{f (s)} d s .$ The integrand is smooth and positive, and [F2] gives $v^{'} (r) = 2 / f (r)$ . The curve $γ (r) = (v (r), r, ω_{0})$ has tangent $2 f (r)^{- 1} \partial_{v} + \partial_{r} = 2 f (r)^{- 1} n$ , a positive multiple of the future null field in [F1]. Reparameterizing by $s = r - r_{0} \in [0, \infty)$ gives the path domain in [L1]. It is defined for every finite $r \geq r_{0}$ and reaches arbitrarily large radii under [L1]. Thus every point belongs to the escape set.

2.1L1step 1.1step 1.2∎

Step 1.2 includes every point in the escape set; its complement is empty, so its relative boundary is empty. Step 1.1 separately excludes a constant- $r$ null level. These are algebraic facts about this model and do not imply censorship or collapse conclusions.

Remarks

The current vocabulary fixes the exact path class and relative escape boundary used by this verification. No global event horizon is defined or claimed.

Superextremal RN absence of radial horizon

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources