LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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.

Regularization preserves value quantitatively

Statement

Let $G$ have $m = ∣ E ∣ > 0$ ordinary edges, with the fixed nonempty alphabet and paired-loop convention. Its cloud graph $G_{1}$ is degree $129$ , has $2 m$ vertices and $129 m$ ordinary edges, and is constructible in polynomial time without changing the alphabet. Put $K = \max (1, 2 / h_{0}) = 20 / 7$ and $c = 1 / (129 K)$ . Then $c UNSAT (G) \leq UNSAT (G_{1}) \leq UNSAT (G) / 129.$ For every labeling $τ$ of $G_{1}$ , plurality decoding $D τ$ satisfies ${UNSAT}_{D τ} (G) \leq 129 K {UNSAT}_{τ} (G_{1})$ . For an edgeless input use the empty output convention and UNSAT zero.

Facts & Assumptions

Given: the objects and hypotheses in the statement above.

[F1]

For any labeling of the cloud graph $G_{1}$ of a nonempty-edge constraint graph $G$ , decode each original vertex by its cloud's plurality label, using fixed tie breaking. Let $S$ count the ports disagreeing with that label, and let $U_{i n t}, U_{e x t}$ count violated internal equality and external edges. Then $U_{i n t} \geq \frac{h_{0}}{2} S, U_{G} \leq U_{e x t} + S, h_{0} = 7 / 10,$ where $U_{G}$ is the decoded violation count in $G$ . (Cloud plurality rounding).

Proof

1.1

Each original edge has two ports even if it is a loop. Every port has $128$ internal slots and one external slot, so there are $2 m$ vertices and $129 (2 m) / 2 = 129 m$ edges. Listing each all-size cloud expander and copying the original relation tables takes polynomial time in the explicit input size; the sum of polynomial cloud costs is polynomial since their total size is $2 m$ .

2.1

For an arbitrary output labeling, the rounding inequalities give $U_{G} \leq U_{e x t} + (2 / h_{0}) U_{i n t} \leq K (U_{e x t} + U_{i n t})$ . Divide by $m$ and use the $129 m$ output edge count to get the assignment-level inequality. Minimizing the output violation fraction then gives the lower bound with $c$ .

F1step 1.1

3.1

An optimal original labeling exists because the alphabet and vertex set are finite nonempty (the empty vertex set has its one labeling). Extend it constantly on each cloud. No equality edge fails and exactly the original bad external edges fail, giving the upper bound after division by $129 m$ . If $m = 0$ , both UNSAT values are zero by the stipulated empty-output convention, without these divisions.

step 1.1algebra∎

Depends on

Cloud plurality rounding

Used by

Constraint cloud rounding and loop counts Example
Constraint expander overlay Lemma

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

Irit Dinur, The PCP theorem by gap amplification; §4 Lemma4.1, PDF pages13–14. (standard reference, not scraped)