Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-12
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.

PDA fragment variables compose

Statement

Let M be a PDA that accepts by empty stack.

  1. If x is in the language of [p,Z,r] and y is in the language of [r,Y,q], then xy removes the stack prefix ZY while changing state from p to q and leaving the remaining stack untouched.
  2. Conversely, consider a computation starting at (p,w,ZYα), and let (q,v,α) be its first configuration whose stack is exactly α. Then there are words x,yΣ and a state rQ such that w=xyv, x[p,Z,r], and y[r,Y,q]. The two split points are the first moments when Yα and then α are exposed.

Facts & Assumptions

Given: A PDA M that accepts by empty stack.

[L1]

By The PDA fragment variable [p,Z,q], x[p,Z,r] means that for every stack suffix α one has (p,x,Zα)M(r,ε,α).

Proof

technique · direct
1.1

Assume x[p,Z,r] and y[r,Y,q], and let α be any stack suffix. Then [L1] gives (p,x,ZYα)M(r,ε,Yα) and also (r,y,Yα)M(q,ε,α). Concatenating these two computations yields (p,xy,ZYα)M(q,ε,α).

L1given
1.2

Conversely, let (q,v,α) be the first configuration with stack α in a computation starting at (p,w,ZYα). Before that point there is a first configuration whose stack is Yα; write it as (r,yv,Yα) and write w=xyv, where x is the input prefix consumed before it is reached and y is consumed between the two first-exposure configurations. The computation therefore contains (p,xyv,ZYα)M(r,yv,Yα)M(q,v,α). Neither segment inspects the suffix α before its endpoint. Hence the same transition sequences work above every stack suffix β, giving (p,x,Zβ)M(r,ε,β) and (r,y,Yβ)M(q,ε,β). By [L1], x[p,Z,r] and y[r,Y,q].

L1construct
2.1

Therefore fragment computations compose exactly by concatenating the consumed words and, conversely, by splitting at the first moment when the lower stacked symbol is exposed.

step 1.1step 1.2

Depends on

Used by

Dependency tree · two levels

3 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