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.

Parcel derivatives and rest-energy shift

Example

Let $e_{0} \in C^{2} (X, R)$ be any specified single-marker state function on a nonempty open state domain $X \subset (0, \infty) \times R$ . Fix a mass per marker $m \geq 0$ with units $k g m a r k e r^{- 1}$ and $c > 0$ with units $m s^{- 1}$ , and define $e_{m} (v, σ_{p}) : = e_{0} (v, σ_{p}) + m c^{2} .$ Then $e_{m} \in C^{2} (X, R)$ . The derived temperature and pressure are unchanged, while $H_{m} = H_{0} + m c^{2}$ , $μ_{m} = μ_{0} + m c^{2}$ , $ε_{m} = ε_{0} + n m c^{2}$ , and $h_{m} = h_{0} + n m c^{2}$ . For parcel totals, $E_{m} (S, V, N) = E_{0} (S, V, N) + N m c^{2}$ . This is a mathematical change of the state function by a constant rest-energy offset, not an empirical statement about a material.

Facts & Assumptions

Given: The state domain, state function, marker volume and entropy coordinates from Rest fluid state, frame and geometric current definitions, and the differential identities of Local Gibbs, Euler, parcel first law and response identities.

[F1]

Rest fluid state, frame and geometric current definitions: $T = e_{σ_{p}}$ , $p = - e_{v}$ , $ε = n e$ , $H = e + p v$ , $h = ε + p$ , and $μ = H - T σ_{p}$ .

[F2]

Local Gibbs, Euler, parcel first law and response identities: the parcel energy $E (S, V, N) = N e (V / N, S / N)$ has differential $d E = T d S - p d V + μ d N$ .

[A1]

$m \geq 0$ and $c > 0$ are fixed parameters, so the added energy per marker is the constant $m c^{2}$ on $X$ .

Verification

1.1F1A1algebra

Since $m c^{2}$ is constant in $(v, σ_{p})$ , $(e_{m})_{σ_{p}} = (e_{0})_{σ_{p}}$ and $- (e_{m})_{v} = - (e_{0})_{v}$ . Thus the derived temperature and pressure are unchanged.

2.1F1step 1.1A1algebra

The definitions give $H_{m} = e_{m} + p_{m} v = H_{0} + m c^{2}$ and $μ_{m} = H_{m} - T_{m} σ_{p} = μ_{0} + m c^{2}$ .

2.2F1step 1.1A1algebra

With $n = 1 / v$ unchanged, $ε_{m} = n e_{m} = ε_{0} + n m c^{2}$ and $h_{m} = ε_{m} + p_{m} = h_{0} + n m c^{2}$ .

3.1F2step 2.1A1algebra∎

At fixed $(S, V, N)$ , the parcel construction gives $E_{m} = E_{0} + N m c^{2}$ . Differentiating adds only $m c^{2} d N$ to $d E_{0}$ ; hence at fixed marker count it changes neither the entropy derivative $T$ nor the pressure-work derivative $- p$ . This agrees with the chemical-potential shift in step 2.1 and the parcel first law in [F2].

Parcel derivatives and rest-energy shift

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources