CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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.

The symmetric Lovász Local Lemma under $e p (d + 1) \leq 1$

Statement

Let $d \in N$ and $p \geq 0$ . Let $(A_{i})_{i \in I}$ have a dependency digraph of maximum out-degree at most $d$ . If $P (A_{i}) \leq p$ for every $i$ and $e p (d + 1) \leq 1,$ then $P (⋂_{i} A_{i}^{c}) > 0$ .

Facts & Assumptions

Given: A finite event family, its dependency digraph, and $p, d$ satisfying the Statement.

[L1]

The asymmetric Local Lemma applies when $P (A_{i}) \leq x_{i} \prod_{i \to j} (1 - x_{j})$ with $0 \leq x_{i} < 1$ (The asymmetric Lovász Local Lemma for finitely many events).

[L2]

$1 + y \leq \exp (y)$ for every real $y$ ( $1 + x \leq \exp (x)$ for every real $x$ , hence $(1 - p)^{m} \leq \exp (- m p)$ ).

[L3]

$\exp (u + v) = \exp (u) \exp (v)$ ; natural powers preserve order on nonnegative bases; and positive inequalities may be multiplied and inverted using the ordered-field laws (The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ , Integer powers $a^{m}$ , Laws of integer exponents, Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , The reals form a totally ordered field).

Proof

technique · cases

1.1

Suppose $d = 0$ and set every $x_{i} = 1 / e$ . Applying [L2] at $y = 1$ gives $2 \leq e$ , so $0 < x_{i} < 1$ . The hypothesis gives $p \leq 1 / e = x_{i}$ , and the empty neighbour product is $1$ , so [L1] applies.

assume-case zeroL1L2algebra

1.2

Suppose $d \geq 1$ and set every $x_{i} = 1 / (d + 1)$ . From [L2] at $y = 1 / d$ and [L3], $(1 + 1 / d)^{d} \leq e$ , hence $(1 - 1 / (d + 1))^{d} = (d / (d + 1))^{d} \geq 1 / e$ .

assume-case positiveL2L3algebra

2.1

Each vertex has at most $d$ out-neighbours, so $x_{i} \prod_{i \to j} (1 - x_{j}) \geq 1 / (e (d + 1)) \geq p$ by the hypothesis. Thus [L1] applies.

step 1.2L1L3algebra

3.1

The cases $d = 0$ and $d \geq 1$ are exhaustive and both give positive probability that no bad event occurs.

step 1.1step 2.1cases-exhaustive∎

Depends on

Used by

A k-uniform hypergraph is 2-colourable when every edge meets at most d other edges and e(d+1)≤2ᵏ⁻¹ Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 119 results over 31 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

J. Matousek and J. Vondrak, The Probabilistic Method, Corollary 5.1.2 (standard reference, not scraped)
Y. Zhao, MIT 18.218 Probabilistic Method in Combinatorics, Corollary 5.3 (standard reference, not scraped)

The symmetric Lovász Local Lemma under ep(d+1)≤1ep(d+1)≤1