False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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 $\mathbb{R}$ , $\iota\lvert A \cup B \cup C\rvert = \iota\lvert A\rvert + \iota\lvert B\rvert + \iota\lvert C\rvert - \iota\lvert A \cap B\rvert - \iota\lvert A \cap C\rvert - \iota\lvert B \cap C\rvert .$

This is the sieve identity of Inclusion and exclusion: $\iota\lvert\bigcup_{i \in I} A_i\rvert = \sum_{\varnothing \ne J \subseteq I}(-1)^{\lvert J\rvert + 1}\,\iota\lvert A_J\rvert$ , 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 $+\,\iota\lvert A \cap B \cap C\rvert$ 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 $\iota$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

[L1]

$\lvert\{0\}\rvert = 1$ and $\iota(1) = 1$ (The cardinality $\lvert A\rvert$ of a finite set, clause (a), The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

[L2]

The sieve identity for a finite family of subsets of a finite ambient set, in the form $\iota\lvert\bigcup_{i \in I}A_i\rvert = \sum_{J \ne \varnothing}(-1)^{\lvert J\rvert+1}\iota\lvert A_J\rvert$ (Inclusion and exclusion: $\iota\lvert\bigcup_{i \in I} A_i\rvert = \sum_{\varnothing \ne J \subseteq I}(-1)^{\lvert J\rvert + 1}\,\iota\lvert A_J\rvert$ , 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 $A_\varnothing = X$ , 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]

$\mathbb{R}$ is an ordered field, so $1 \ne 0$ 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 $\iota\lvert A\rvert$ , $\iota\lvert B\rvert$ , $\iota\lvert C\rvert$ , $\iota\lvert A \cap B\rvert$ , $\iota\lvert A \cap C\rvert$ , $\iota\lvert B \cap C\rvert$ , $\iota\lvert A \cap B \cap C\rvert$ and $\iota\lvert A \cup B \cup C\rvert$ 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 $1 \ne 0$ in $\mathbb{R}$ 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 $\iota\lvert A \cup B \cup C\rvert$ . 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 76 results over 23 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

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