LemmaStatement: AI-adaptedProof: 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.

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$P (α (G (n, p)) \geq s) \leq (\binom{n}{s}) (1 - p)^{(\binom{s}{2})} \leq n^{s} \exp (- p (\binom{s}{2}))$ for $s \leq n$

Statement

For $0 \leq s \leq n$ and $p \in [0, 1]$ , $P (α (G (n, p)) \geq s) \leq (\binom{n}{s}) (1 - p)^{(\binom{s}{2})} \leq n^{s} \exp ⁣ (- p (\binom{s}{2})) .$ For $s > n$ , the event is empty. At $s = 0$ every displayed quantity equals $1$ . At $s = 1$ both bounds equal $n$ , so the second inequality is an equality while the first is strict whenever $n \geq 2$ .

Facts & Assumptions

Given: Naturals $n, s$ and $p \in [0, 1]$ .

[L1]

$α (G)$ is the greatest size of an independent vertex set (Cliques, independent sets, clique number and independence number).

[L2]

A prescribed set of absent edges has probability $(1 - p)^{m}$ (A prescribed set of present and absent edges in $G (n, p)$ has product probability).

[L3]

Probability of a finite union is at most the sum of the event probabilities (The finite union bound).

[L4]

There are $(\binom{n}{s})$ subsets of size $s$ (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

[L5]

For $s \leq n$ , $(\binom{n}{s}) s! = n^{\underline{s}}$ ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ ).

[L6]

$(1 - p)^{m} \leq \exp (- p m)$ for $0 \leq p \leq 1$ ( $1 + x \leq \exp (x)$ for every real $x$ , hence $(1 - p)^{m} \leq \exp (- m p)$ ).

Proof

technique · cases

1.1

Suppose $s \leq n$ . If $α (G (n, p)) \geq s$ , some $s$ -set has all of its $(\binom{s}{2})$ possible internal edges absent.

assume-case inrangeL1

1.2

Suppose $s > n$ . Then no $s$ -subset exists and the event $α (G (n, p)) \geq s$ is empty.

assume-case outrangeL1L4

2.1

A fixed $s$ -set is independent with probability $(1 - p)^{(\binom{s}{2})}$ , so [L3] and [L4] give the first bound.

step 1.1L2L3L4

3.1

By [L5], $(\binom{n}{s}) \leq n^{\underline{s}} \leq n^{s}$ , including $s = 0$ . Applying [L6] with $m = (\binom{s}{2})$ gives the second bound.

step 2.1L5L6algebra

4.1

The cases are exhaustive. At $s = 0$ or $1$ , $(\binom{s}{2}) = 0$ and the displayed expressions have their stated boundary values.

step 3.1step 1.2cases-exhaustive∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 109 results over 25 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, proof of Theorem 4.2.1 (standard reference, not scraped)
Y. Zhao, MIT 18.218 Probabilistic Method in Combinatorics, proof of Theorem 6.3 (standard reference, not scraped)

P(α(G(n,p))≥s)≤(ns)(1−p)(s2)≤nsexp⁡(−p(s2))P(α(G(n,p))≥s)≤(sn​)(1−p)(2s​)≤nsexp(−p(2s​)) for s≤ns≤n