Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13
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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.

  • 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.

Every k-uniform hypergraph with fewer than 2k1 edges is 2-colourable

Statement

Let k1. Every finite k-uniform hypergraph with fewer than 2k1 edges admits a vertex two-colouring with no monochromatic edge.

Facts & Assumptions

Given: A finite k-uniform hypergraph H=(V,E) with k1 and E<2k1.

[L1]

An edge in a k-uniform hypergraph has exactly k vertices (r-uniform hypergraphs and complete balanced r-partite r-graphs Ks,,s(r)).

[L2]

Independent coordinate events in a finite product space have product probability (Product weights normalize, and coordinate events are mutually independent).

[L3]

A sum of indicators counts the corresponding events, each indicator has expectation equal to its event probability, and expectation is linear without independence (Indicators turn event probabilities, intersections, and finite counts into expectations and products, Expectation is linear for every finite family of random variables, without any independence hypothesis).

[L4]

A nonnegative integer-valued variable with expectation less than 1 vanishes at some outcome (The first-moment method for avoiding or forcing a finite count of bad events).

Proof

technique · direct
1.1

Colour every vertex independently and uniformly red or blue. For a fixed edge, its k colours are all red or all blue with probability 22k=21k.

L1L2
2.1

Let X count monochromatic edges. Then E[X]=E21k<1.

step 1.1L3algebra
3.1

By [L4], some colouring has X=0 and is proper. If k=1, the edge hypothesis forces E=, and the same proof applies.

step 2.1L4

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 54 results over 16 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources