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 finite adversarial gradient game can cycle

Example

Let $θ, φ \in R$ be a generator and discriminator parameter, and let the generator minimize and the discriminator maximize the shared payoff $L (θ, φ) = θ φ$ . Starting from any $(θ_{0}, φ_{0}) \neq (0, 0)$ , use one exact unit-step generator gradient descent update followed by one exact unit-step discriminator gradient ascent update, with the discriminator update evaluated at the new generator parameter:

θ_{n + 1} = θ_{n} - \partial_{θ} L (θ_{n}, φ_{n}), φ_{n + 1} = φ_{n} + \partial_{φ} L (θ_{n + 1}, φ_{n}) .

The resulting parameter sequence repeats after six steps and does not converge. This scalar bilinear game is a finite-parameter adversarial-game illustration, not the GAN cross-entropy objective or a neural GAN; it shows why a distribution-space optimum alone is not a finite-parameter training-convergence theorem.

Example references

Unrestricted distribution-space optimum and its scope

Goodfellow et al., §4.2, final paragraph on finite MLP parameterization, complete 9-page paper

Facts & Assumptions

Given: $θ_{n}, φ_{n} \in R$ , payoff $L (θ, φ) = θ φ$ , generator descent first, discriminator ascent second at the updated $θ$ , unit step size, and a nonzero initial pair.

[F1]

Optimal discriminator and Jensen–Shannon divergence in the unrestricted GAN game: the proved GAN optimum is for the unrestricted distribution-space objective, and it asserts no convergence for finite neural alternating optimization.

Verification

Given: The scalar payoff and exact alternating update rule in the Example, with $(θ_{0}, φ_{0}) \neq (0, 0)$ .

1.1givenalgebra

For $h \neq 0$ , $(L (θ + h, φ) - L (θ, φ)) / h = ((θ + h) φ - θ φ) / h = φ$ , so $\partial_{θ} L = φ$ . Likewise $(L (θ, φ + h) - L (θ, φ)) / h = θ$ , so $\partial_{φ} L = θ$ .

2.1givenstep 1.1algebra

Substitution into the update gives $θ_{n + 1} = θ_{n} - φ_{n}$ and $φ_{n + 1} = φ_{n} + θ_{n + 1} = θ_{n}$ . Thus $x_{n + 1} = M x_{n}$ for $x_{n} = (θ_{n}, φ_{n})^{T}$ and $M = (\begin{array}{cc} 1 & - 1 \\ 1 & 0 \end{array})$ .

3.1step 2.1algebra

Matrix multiplication gives $M^{2} = (\begin{array}{cc} 0 & - 1 \\ 1 & - 1 \end{array})$ and $M^{3} = - I$ . Hence $x_{n + 3} = - x_{n}$ and $x_{n + 6} = x_{n}$ for every $n$ .

4.1givenF1step 3.1algebra∎

Since $x_{0} \neq 0$ , the subsequence $(x_{6 k})_{k \geq 0}$ is constantly $x_{0}$ and $(x_{6 k + 3})_{k \geq 0}$ is constantly $- x_{0} \neq x_{0}$ . These subsequences have different limits, so $(x_{n})$ does not converge. The computed cycle is for this explicitly defined toy game; [F1] shows it supplies no conclusion about convergence of GAN neural training.

Depends on

Optimal discriminator and Jensen–Shannon divergence in the unrestricted GAN game

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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

Ian J. Goodfellow et al., Generative Adversarial Nets, arXiv:1406.2661v1 (standard reference, not scraped)