CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

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

Statement refuted

Refuted claim: the sieve identity for three sets with its triple term deleted, that is FALSE: the real-valued three-set inclusion-exclusion identity remains true after deleting the triple-intersection term:

$ι ∣ A \cup B \cup C ∣ = ι ∣ A ∣ + ι ∣ B ∣ + ι ∣ C ∣ - ι ∣ A \cap B ∣ - ι ∣ A \cap C ∣ - ι ∣ B \cap C ∣ .$

The witness is a family of overlapping but unequal sets, so that no cardinality in the computation is degenerate. Inside $X : = 5 = {0, 1, 2, 3, 4}$ take

$A : = {0, 1, 2}, B : = {1, 2, 3}, C : = {2, 3, 4} .$

The truncated right-hand side returns $4$ , while the union has $5$ elements; the missing triple term is $∣ A \cap B \cap C ∣ = 1$ , and restoring it gives $5$ .

Facts & Assumptions

Given: $X = 5$ , $A = {0, 1, 2}$ , $B = {1, 2, 3}$ , $C = {2, 3, 4}$ , and the canonical natural $ι$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L1]

A listed set with distinct entries has as many elements as entries (The cardinality $∣ A ∣$ of a finite set, clauses (a) and (c), Injection, surjection, bijection).

[L2]

The sieve identity for a sieve family with ambient set $X$ and index set $I = 3$ , whose terms carry the sign $(- 1)^{∣ J ∣ + 1}$ (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, Integer powers $a^{m}$ ).

[L3]

$ι$ is additive and injective, so the arithmetic below may be carried out on natural numbers and read in $R$ , where subtraction is available (Laws of finite sums and products in $N$ , and $ι (\sum_{k < n} a_{k}) = \sum_{k < n} ι (a_{k})$ , clauses 0 and 7, Ordered field, Field).

Counterexample

technique · constructive

1.1

The three sets are subsets of $X$ with $∣ A ∣ = ∣ B ∣ = ∣ C ∣ = 3$ by [L1], their entry lists being strictly increasing and so having distinct entries.

givenL1construct

1.2

The intersections. Comparing the listed sets entry by entry gives $A \cap B = {1, 2}$ , $A \cap C = {2}$ , $B \cap C = {2, 3}$ and $A \cap B \cap C = {2}$ , of sizes $2$ , $1$ , $2$ and $1$ by [L1].

givenL1

1.3

The union. Every element of $X$ lies in one of the three sets: $0, 1, 2 \in A$ , $3 \in B$ and $4 \in C$ . So $A \cup B \cup C = X$ , of size $5$ by [L1].

givenL1

2.1

The truncated right-hand side. By steps 1.1 and 1.2 it is $3 + 3 + 3 - 2 - 1 - 2 = 4$ , while the left-hand side is $5$ by step 1.3. Since $4 \neq 5$ in $R$ by [L3], the refuted claim fails on this family.

step 1.1step 1.2step 1.3L3

3.1

The correct computation. By [L2] the sieve sum carries the triple term with sign $(- 1)^{4} = + 1$ , so it reads $3 + 3 + 3 - 2 - 1 - 2 + 1 = 5$ , which is $ι ∣ A \cup B \cup C ∣$ by step 1.3. The discrepancy of step 2.1 is exactly $∣ A \cap B \cap C ∣ = 1$ , the element $2$ lying in all three sets.

step 1.2step 2.1L2L3discharge-construct∎

Remarks

Every set in the witness is a proper subset of $X$ and no two of them are equal, so the failure is not an artefact of a degenerate family. What makes the truncated formula wrong is only that the triple intersection is nonempty.
The direction of the error. The truncation at depth $2$ under-counts, by exactly the size of the triple intersection. That is the direction the Bonferroni inequalities predict for an even truncation, and the size of the gap here is the single term that the next level of the sieve would add.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

54 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)