PropositionStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 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 deletion-alteration method converts an expected defect count into a deterministic lower bound

Statement

Suppose a finite random object has integer size $Y$ and comes with a finite listed collection of $X$ defects. If a deterministic repair deletes at most one unit for every listed defect and produces an admissible object, then some repaired outcome has size at least $E [Y] - E [X] .$ Repeated or redundant listed defects are allowed; they can only weaken the lower bound.

Facts & Assumptions

Given: Integer-valued random variables $X, Y$ and a repair as in the Statement.

[L1]

Expectation is linear without independence, so $E [Y - X] = E [Y] - E [X]$ (Expectation is linear for every finite family of random variables, without any independence hypothesis).

[L2]

Some outcome of a finite random variable has value at least its expectation (Expectation preserves pointwise order and lies between the minimum and maximum attained values).

Proof

technique · direct

1.1

On every outcome, deleting at most one unit per listed defect leaves an admissible object of size at least $Y - X$ .

given

1.2

By [L2], there is an outcome with $Y - X \geq E [Y - X]$ .

2.1

Repair that outcome. Its size is at least $E [Y - X] = E [Y] - E [X]$ by [L1]. Overlisting defects increases $X$ and therefore cannot invalidate the lower bound.

step 1.1step 1.2L1∎

Depends on

Used by

For all positive k,ℓ, some finite graph has girth greater than ℓ and chromatic number greater than k Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 32 results over 11 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, Chapter 4 (standard reference, not scraped)
Y. Zhao, MIT 18.218 Probabilistic Method in Combinatorics, Section 6.3 (standard reference, not scraped)