Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-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.

Parareal reproduces the first k fine-grid nodes after k sweeps

Statement

For the exact-arithmetic recurrence in Finite-budget Parareal correction sweeps, let x0 be the initial state and xn+1=Fn(xn) its sequential fine trajectory. For every completed sweep k and every 0≤n≤min⁡{k,N}, Unk=xn. Consequently, if the requested sweep budget K≥N is fully completed, the returned table contains the complete sequential fine trajectory in row K.

Facts & Assumptions

Given: The exact-arithmetic recurrence is defined on the rows considered; each row called completed is formed from successful evaluations.

[F1]

Fine and coarse propagators on a common macro grid: the sequential fine trajectory is xn+1=Fn(xn) from x0.

[F2]

Finite-budget Parareal correction sweeps: U0k=x0 for every sweep, the initial row is coarse-propagated, and each correction is Un+1k+1=Gn(Unk+1)+Fn(Unk)−Gn(Unk). The algorithm returns every fully completed row and may stop early after a propagation failure or time-budget expiry.

Proof

1.1givenF1F2algebra

At sweep k=0, the only claimed node is n=0, and U00=x0=x0 by initialization and the definition of the fine trajectory [F1].

1.2givenF1F2inductionalgebra

Suppose the claim holds at sweep k. At sweep k+1, U0k+1=x0 by [F2]. Induct on n from 0 to min⁡{k,N−1}; if Unk+1=xn, then n≤k also gives Unk=xn, so [F2] yields Un+1k+1=Gn(xn)+Fn(xn)−Gn(xn)=Fn(xn)=xn+1 by [F1]. Thus the claim holds through node min⁡{k+1,N}.

2.1F2step 1.2inductionalgebra∎

Induction on k proves the first assertion. If the requested rows 0,…,K all complete and K≥N, applying it at k=K gives UnK=xn for every 0≤n≤N; by [F2], row K is in the returned table.

Depends on

Used by

Dependency tree · two levels

4 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