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.

Finite word-calculus rearrangement

Statement

For every positive integer d,

∀a1⋯∀ad∃x1⋯∃xd ⊢d ∃A1⋯∃Ad∀x1⋯∀xd.

Facts & Assumptions

Given: A positive integer d and the two displayed endpoint words.

[F1]

The language Ld, elementary commutations, matched-pair replacement, finite block permutation, and ⊢d are exactly the finite calculus in the preceding definition. The finite word calculus for the Halpern–Läuchli argument

Proof

technique · induction on $d$
1.1F1

Write Um=∀a1⋯∀am, Xm=∃x1⋯∃xm, Em=∃A1⋯∃Am, and Vm=∀x1⋯∀xm. For d>1 there is a bridge ∀adEd−1Vd−1∃xd⊢d∃AdUd−1Xd−1∀xd: Rule 3 moves Ed−1 left of ∀ad; Rule 1 rearranges the middle as (∃Ai∀xi)i=1d−1∀ad∃xd; Rule 2 changes all d matched pairs to (∀ai∃xi)i=1d−1∃Ad∀xd; Rule 1 moves each earlier ∃xi past later universal a-symbols and commutes the existential symbols to give Ud−1∃AdXd−1∀xd; and Rule 3 moves ∃Ad left of Ud−1.

1.2F1

Derivations in Ld−1 lift through either outside matched pair: W⊢d−1W′ implies both ∀adW∃xd⊢d∀adW′∃xd and ∃AdW∀xd⊢d∃AdW′∀xd. It suffices to lift one rule step. Rules 1 and 2 apply unchanged inside the context. For Rule 3, Rule 1 first rearranges its adjacent prefix of a- and A-symbols into a universal block followed by an existential block, Rule 3 exchanges the blocks, and Rule 1 restores the required order. The outside coordinate remains a complete ordered pair; induction on the finite derivation length proves both implications.

1.3F1base

If d=1, the asserted derivation is exactly ∀a1∃x1⊢1∃A1∀x1, an instance of Rule 2.

2.1F1step 1.1step 1.2ih

Suppose d>1 and the result holds in dimension d−1. Rule 1 gives UdXd⊢d∀adUd−1Xd−1∃xd. Lift the induction hypothesis by the first implication in step 1.2, apply the bridge from step 1.1, and lift the induction hypothesis by the second implication in step 1.2; thus ∀adUd−1Xd−1∃xd⊢d∀adEd−1Vd−1∃xd⊢d∃AdUd−1Xd−1∀xd⊢d∃AdEd−1Vd−1∀xd. Rule 1 finally commutes the A-symbols and the x-symbols to obtain EdVd.

3.1F1step 1.3step 2.1discharge-induction∎

Every intermediate word lies in Ld: elementary commutation changes no coordinate's selected pair, Rule 2 replaces one legal ordered pair by the other, and Rule 3 is invoked only with its result in Ld. Steps 1.3 and 2.1 therefore prove the assertion for every positive d.

Depends on

Used by

Dependency tree · two levels

2 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