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.

The empty-input initial configuration

Example

Let Q={s,A,R}, Σ={a}, Γ={a,}, with three distinct states and a. The start is s, the accept state A, and the reject state R. Give the transition by its two entries δ(s,a)=(A,a,R),δ(s,)=(A,,R). Here R is a state name and R is the right-movement tag. For this raw tuple M, let tblank(i)= for every iN. Then IM(ε)=(s,0,tblank). Its support is empty and this initial configuration is nonhalting.

Facts & Assumptions

Given: The finite sets, distinct states, two transition entries, and blank tape displayed above.

[F1]

The raw tuple requires totality on the nonhalting state/tape-symbol pairs. The initial tape writes the length-m word at indices below m and blank elsewhere; configurations have a natural head coordinate and finite-support tape, and halting is determined by the two designated states (Initial tapes and machine-relative halting configurations).

Verification

1.1

Removing A,R from Q leaves exactly {s}, so the transition domain is {(s,a),(s,)}. The two entries give one value to each pair, both in Q×Γ×{L,R}. All sets are finite, Σ, and the three designated states are distinct. Thus the data meet every raw-tuple requirement.

givenF1
2.1

The empty word has length zero. For any natural i, the condition i<0 is false, so its initial-tape formula gives tε(i)==tblank(i). Its support is {i:}=, a finite set. Hence it is a tape and IM(ε)=(s,0,tblank) is a configuration; zero is a head coordinate even though it is not an index of an input letter.

step 1.1F1
3.1

Its state is s, unequal to A and R. The accepting and rejecting equalities are both false, so their disjunction is false and the configuration is nonhalting. This conclusion concerns the initial triple and does not apply the transition function.

step 2.1givenF1

Source

This original tuple illustrates the boundary convention fixed in the local definition. Compare the blank empty-input tape in Watrous, §12.1, p. 122; head placement is as specified locally.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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