Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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.

Minimum, maximum, and bounded shifts of stopping times

Statement

If σ,τ are stopping times, then σ∧τ and σ∨τ are stopping times. If c∈N0, then τ+c is a stopping time for (Fn), with ∞+c=∞. More generally, if ρ is stopping for Gn=Fn+c, then ρ+c is stopping for (Fn). The earlier shift (τ−c)+ is stopping for (Fn+c) but need not be stopping for (Fn). Finally, for every deterministic N∈N0, τ∧N is a bounded stopping time.

Facts & Assumptions

Given: The hypotheses, objects, and conventions in the Statement.

[F1]

Discrete stopping time supplies the finite-horizon test.

[F2]

Equivalent event tests for a discrete stopping time supplies the complementary tests.

Proof

1.1F1

The identities {σ∧τ≤n}={σ≤n}∪{τ≤n},{σ∨τ≤n}={σ≤n}∩{τ≤n} prove the first two claims by F1.

1.2F1

If n<c, {τ+c≤n}=∅; if n≥c, it is {τ≤n−c}∈Fn−c⊆Fn. If ρ is stopping for G, the same event is in Gn−c=Fn.

1.3F1

For the earlier shift, {(τ−c)+≤n}={τ≤n+c}∈Fn+c, so it is stopping for the shifted filtration. The right side need not lie in Fn, which is why no unshifted assertion is made.

2.1F1F2∎

A deterministic N is a stopping time, so step 1.1 makes τ∧N a stopping time, and it is bounded by N. All empty and infinite-value cases follow from the displayed identities.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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