Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-09
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.

Epsilon closure and word transitions on three states

Example

Let Q={p,q,r}, Σ={a,b}, start p, and F={r}, where the displayed states and letters are distinct. Specify the entire transition table by

Stateεab
p{q}
q{q,r}
r{r}

Then ECl({p})={p,q} and ECl()=. The state sets from p on ε,a,aa,ab,b,aba are, respectively, {p,q},{q,r},{q,r},{r},,. Thus a,aa,ab are accepted and ε,b,aba are rejected.

Facts & Assumptions

Given: The nine table entries, start p, and final set {r} above; write E=ECl.

[F1]

Epsilon closure is finite saturation; the empty-word value is the initial closure and each next letter takes a successor union followed by closure. Acceptance tests nonempty intersection with the final set (Epsilon-NFAs, finite epsilon closure, and extended word transition).

Verification

1.1

The table gives a subset of Q at each of its nine state/label pairs, so the transition is total on the required domain. The finite sets, start and finals meet the epsilon-NFA requirements. Starting from {p}, saturation adds q in one stage and adds nothing thereafter, since neither q nor r has an epsilon successor. Thus E({p})={p,q}; more generally E(T)=T{q} if pT, and E(T)=T otherwise. In particular E()=, E({q,r})={q,r} and E({r})={r}.

givenF1
2.1

The empty-word value is δ^(p,ε)={p,q}. For a, the successor union from this set is δ(p,a)δ(q,a)={q,r}={q,r}, already closed; hence δ^(p,a)={q,r}. For b, the corresponding union is =, so δ^(p,b)=.

step 1.1givenF1
3.1

After a the set is {q,r}. Its a-successor union is {q,r}={q,r}, and its b-successor union is {r}={r}. Both are closed, giving δ^(p,aa)={q,r} and δ^(p,ab)={r}. From {r} the a-successor union is empty, so δ^(p,aba)=.

step 1.1step 2.1givenF1
4.1

The sets for a,aa,ab each contain r, whereas the sets for ε,b,aba do not. Intersecting them with F={r} gives {r} in the first three cases and in the other three, proving every asserted acceptance and rejection.

step 2.1step 3.1F1given

Source

The witness is original to this example. The calculation uses the closure and word-transition rules of Gallier, Definitions 3.6–3.7, pp. 30–32, via the local interface.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources