False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)verified 2026-07-29 (claude-fable-5)

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.

FALSE: the real-valued three-set inclusion-exclusion identity remains true after deleting the triple-intersection term

Statement

FALSE. The statement

for all finite sets $A$ , $B$ , $C$ , in $R$ , $ι ∣ A \cup B \cup C ∣ = ι ∣ A ∣ + ι ∣ B ∣ + ι ∣ C ∣ - ι ∣ A \cap B ∣ - ι ∣ A \cap C ∣ - ι ∣ B \cap C ∣ .$

This is the sieve identity of Inclusion and exclusion: $ι ∣ ⋃_{i \in I} A_{i} ∣ = \sum_{\emptyset \neq J \subseteq I} (- 1)^{∣ J ∣ + 1} ι ∣ A_{J} ∣$ , together with the complementary form counting the elements in none of the $A_{i}$ for a family of three sets with the term at the triple intersection deleted. The identity is correct only when the triple term $+ ι ∣ A \cap B \cap C ∣$ is present, and the claim above is refuted by a family in which that term is not $0$ .

Facts & Assumptions

Given: The one-element set $A : = B : = C : = {0}$ , taken inside the ambient set $X : = {0}$ , and the canonical natural $ι$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L1]

$∣ {0} ∣ = 1$ and $ι (1) = 1$ (The cardinality $∣ A ∣$ of a finite set, clause (a), The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L2]

The sieve identity for a finite family of subsets of a finite ambient set, in the form $ι ∣ ⋃_{i \in I} A_{i} ∣ = \sum_{J \neq \emptyset} (- 1)^{∣ J ∣ + 1} ι ∣ A_{J} ∣$ (Inclusion and exclusion: $ι ∣ ⋃_{i \in I} A_{i} ∣ = \sum_{\emptyset \neq J \subseteq I} (- 1)^{∣ J ∣ + 1} ι ∣ A_{J} ∣$ , together with the complementary form counting the elements in none of the $A_{i}$ , clause 1, A finite family $(A_{i})_{i \in I}$ of subsets of a finite set $X$ , the intersections $A_{J}$ for $J \subseteq I$ , and the convention $A_{\emptyset} = X$ , The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form).

[L3]

$(- 1)^{0} = 1$ and $(- 1)^{p + 1} = - (- 1)^{p}$ , so $(- 1)^{2} = 1$ , $(- 1)^{3} = - 1$ and $(- 1)^{4} = 1$ (Integer powers $a^{m}$ ).

[L4]

$R$ is an ordered field, so $1 \neq 0$ and its arithmetic is available (Ordered field, Field).

Refutation

technique · constructive

1.1

Take $I : = 3 = {0, 1, 2}$ , $X : = {0}$ and $A_{0} : = A_{1} : = A_{2} : = {0}$ , so that the family of the displayed claim is $A = A_{0}$ , $B = A_{1}$ , $C = A_{2}$ . Every intersection of a nonempty subfamily is ${0}$ , and the union is ${0}$ .

givenconstruct

1.2

Every set occurring in the computation is ${0}$ , so by [L1] each of $ι ∣ A ∣$ , $ι ∣ B ∣$ , $ι ∣ C ∣$ , $ι ∣ A \cap B ∣$ , $ι ∣ A \cap C ∣$ , $ι ∣ B \cap C ∣$ , $ι ∣ A \cap B \cap C ∣$ and $ι ∣ A \cup B \cup C ∣$ equals $1$ .

2.1

The right-hand side of the displayed claim is therefore $1 + 1 + 1 - 1 - 1 - 1 = 0$ , while its left-hand side is $1$ . Since $1 \neq 0$ in $R$ by [L4], the claim is false at this family.

step 1.1step 1.2L4

3.1

The true identity at the same family. By [L2] the sieve sum has the three singleton terms with sign $(- 1)^{2} = 1$ , the three two-element terms with sign $(- 1)^{3} = - 1$ and the one three-element term with sign $(- 1)^{4} = 1$ , so it reads $1 + 1 + 1 - 1 - 1 - 1 + 1 = 1$ , which is $ι ∣ A \cup B \cup C ∣$ . The deleted triple term is exactly the discrepancy found in step 2.1.

step 1.2step 2.1L2L3discharge-construct∎

Remarks

The claim is the sieve truncated at depth $2$ , and the Bonferroni inequalities say what such a truncation does in general: an even truncation under-estimates. Here it under-estimates by $1$ , and the claim asserts equality, so the failure is in the direction the inequality predicts.
The witness is as small as it can be. Three sets are needed for a triple intersection to exist, and the discrepancy is the size of that intersection, so any family with a nonempty triple intersection refutes the claim. Taking all three sets equal to a single point makes every cardinality in the computation equal to $1$ .

Depends on

Used by

A three-set count that drops the triple intersection and returns the wrong answer Counterexample

Dependency tree · two levels

34 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

Inclusion-exclusion principle (Wikipedia) (standard reference, not scraped)
Cardinality (Wikipedia) (standard reference, not scraped)
Guichard, The Inclusion-Exclusion Formula (LibreTexts) (standard reference, not scraped)