ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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.

Reset bound for a biased two-state memory

Physical scope

A symbolic finite-memory calculation under the reset theorem's physical assumptions; it is not a measured reset process.

Example

Let the retained-register set be the singleton $R = {r_{0}}$ and let the memory be a bit with probability $p \in (0, 1)$ of state 1. The register carries no information.

Facts & Assumptions

Given: A finite memory law and the endpoint assumptions of Conditional work bound for finite memory reset.

[F1]

Finite memory and retained-register reset model: $a_{r} = \sum_{m} p (m, r)$ , $q (m ∣ r) = p (m, r) / a_{r}$ for $a_{r} > 0$ , and $H (M ∣ R) = \sum_{r : a_{r} > 0} a_{r} [- \sum_{m : q (m ∣ r) > 0} q (m ∣ r) \log q (m ∣ r)]$ , with natural logarithms and entropy in nats.

[F2]

Conditional work bound for finite memory reset: $W \geq k_{B} T_{b} H (M ∣ R)$ under its stated endpoint entropy, fixed-temperature bath, unchanged endpoint energy, and total entropy nondecrease assumptions.

Verification

Given: $p \in (0, 1)$ and $T_{b} > 0$ .

1.1givenF1algebra

Since $R = {r_{0}}$ , its marginal is $a_{r_{0}} = 1$ , and the conditional law is $q (1 ∣ r_{0}) = p$ , $q (0 ∣ r_{0}) = 1 - p$ . The entropy definition therefore gives $H (M ∣ R) = H (M) = - p \log p - (1 - p) \log (1 - p)$ .

2.1F2step 1.1algebra∎

Therefore the theorem gives $W \geq k_{B} T_{b} [- p \log p - (1 - p) \log (1 - p)]$ for this preparation. It reports no work measurement and makes no claim about an apparatus.

Reset bound for a biased two-state memory

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources