ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge 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.

A stable coupled model has an unstable explicit fast microstep

Example

Consider $x^{'} = - x$ , $z^{'} = x - 8 z$ from $(x_{0}, z_{0}) = (1, 1)$ . Its exact solution for arbitrary initial data is $x (t) = x_{0} e^{- t}, z (t) = e^{- 8 t} z_{0} + \frac{x_{0}}{7} (e^{- t} - e^{- 8 t}) .$ It tends to zero as $t \to \infty$ , so the continuous linear system is asymptotically stable, meaning every solution tends to the origin. Apply Frozen-slow explicit Euler with fast microsteps with $H = 3 / 5$ , $m = 2$ , and $h = 3 / 10$ . The method gives $x^{+} = 2 / 5$ , $z_{1} = - 11 / 10$ , and $z_{2} = 46 / 25$ . At fixed frozen $x$ , a fast-coordinate perturbation is multiplied per microstep by $1 - 8 h = - 7 / 5$ , whose magnitude exceeds one; the exact homogeneous fast perturbation instead decays.

Facts & Assumptions

Given: The linear vector field, initial data, macro step, and microstep count specified above.

[F1]

The assigned algorithm updates the slow coordinate once by explicit Euler and then performs $m$ fast Euler updates using the original frozen slow state (Frozen-slow explicit Euler with fast microsteps).

[F2]

The exponential satisfies $e^{0} = 1$ and $e^{a + b} = e^{a} e^{b}$ , has derivative $e^{t}$ , is increasing, and obeys $1 + t \leq e^{t}$ for $t \geq 0$ (The real exponential function and the number $e$ by a power series, The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ , The exponential function is strictly increasing, The exponential function is smooth and $(\exp)^{'} = \exp$ , $1 + x \leq \exp (x)$ for every real $x$ , hence $(1 - p)^{m} \leq \exp (- m p)$ ).

[A1]

In this example $f (x, z) = - x$ and $g (x, z) = x - 8 z$ .

Proof

1.1F2A1algebra

Differentiating the displayed formula with [F2] gives $x^{'} = - x$ and $z^{'} = x - 8 z$ , and at $t = 0$ it gives the arbitrary initial pair $(x_{0}, z_{0})$ . Thus it is the exact solution. For $t \geq 0$ , [F2] gives $e^{t} \geq 1 + t > 0$ and $e^{8 t} \geq 1 + 8 t > 0$ , while the addition formula gives $e^{- t} = 1 / e^{t}$ and $e^{- 8 t} = 1 / e^{8 t}$ . Therefore $0 < e^{- t} \leq 1 / (1 + t)$ and $0 < e^{- 8 t} \leq 1 / (1 + 8 t)$ . Given $ε > 0$ , both bounds are below $ε$ whenever $t > \max {1 / ε, 1 / (8 ε)}$ , so both exponentials and then both solution components tend to zero.

1.2F1F2A1algebra

By [F1] and [A1], $x^{+} = 1 + (3 / 5) (- 1) = 2 / 5$ . The fast updates use frozen $x = 1$ , so $z_{1} = 1 + (3 / 10) (1 - 8) = - 11 / 10$ and $z_{2} = - 11 / 10 + (3 / 10) (1 - 8 (- 11 / 10)) = 46 / 25$ . Subtracting two fast recurrences with the same frozen $x$ gives $δ z_{j + 1} = (1 - 8 h) δ z_{j} = - 7 / 5 δ z_{j}$ . The exact homogeneous fast equation $w^{'} = - 8 w$ has solution $w (t) = e^{- 8 t} w (0)$ by [F2]. Since $e^{8 h} \geq 1 + 8 h > 1$ , its reciprocal $e^{- 8 h}$ lies in $(0, 1)$ , so the exact homogeneous multiplier decays.

2.1F1step 1.1step 1.2algebra∎

Step 1.1 directly shows every exact solution tends to zero, whereas step 1.2 gives an expanding fast perturbation in each explicit microstep. The successful step costs one $f$ and two $g$ evaluations, which by itself makes no wall-clock speedup claim.

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