Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-09-12
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The start, accept, and transition constraints have polynomial size

Statement

For a fixed machine and a polynomially bounded tableau, the Boolean formulas forcing the correct start row, forcing an accepting state somewhere, and forcing legal row-to-row transitions all have polynomial total size.

Facts & Assumptions

Given: A tableau of side length T+1 for a fixed machine N on input w.

[L1]

A local 2×3 window is legal exactly when it matches one allowed local update of the machine, by Local 2x3 tableau windows and legality.

[L2]

Each tableau cell ranges over a constant-size alphabet for fixed N, by For a fixed machine, each tableau cell ranges over a constant-size extended alphabet.

Proof

technique · direct
1.1

The start constraint fixes each cell of the first row to its predetermined initial symbol, so it contributes one constant-size literal condition per column and therefore O(T) total size. The accept constraint is the disjunction over all cells that one of them carries an accepting-state tag, so it has O(T2) size.

givenconstruct
1.2

For each adjacent-row position and each illegal 2×3 pattern, add a clause forbidding that pattern. By [L1] and [L2], there are O(T2) window positions and only constantly many patterns to inspect at each position, each producing a constant-size clause.

L1L2givenconstruct
2.1

Combining steps 1.1 and 1.2 shows that the start, accept, and transition families together have polynomial total size.

step 1.1step 1.2

Depends on

Used by

Dependency tree · two levels

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Sources