False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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FALSE: $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ for all finite $A$ and $B$

Statement

FALSE. The statement

$∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$

for all finite sets $A$ and $B$ .

This is The sum rule: a finite disjoint union is finite with $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ and $∣ ⋃_{i \in I} A_{i} ∣ = \sum_{i \in I} ∣ A_{i} ∣$ , and a sum over a finite index set splits along a partition clause 1 with the hypothesis $A \cap B = \emptyset$ deleted. It is the single most common way a count goes wrong, and the smallest witness is as small as a witness can be.

Facts & Assumptions

Given: The sets ${0}$ , ${0, 1}$ and ${1, 2}$ , with $0$ , $1$ , $2$ the von Neumann naturals (The natural numbers $N$ (von Neumann)).

[L1]

$∣ n ∣ = n$ for a natural $n$ , and a bijection transports cardinality (The cardinality $∣ A ∣$ of a finite set, Injection, surjection, bijection).

[L2]

$∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ when $A \cap B = \emptyset$ (The sum rule: a finite disjoint union is finite with $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ and $∣ ⋃_{i \in I} A_{i} ∣ = \sum_{i \in I} ∣ A_{i} ∣$ , and a sum over a finite index set splits along a partition, clause 1).

[L3]

Distinct naturals are distinct, so $1 \neq 2$ and $3 \neq 4$ (Trichotomy of the order on $N$ ).

Refutation

technique · direct

1.1

The smallest witness. Take $A = B = {0}$ . Then $A \cup B = {0}$ , which is $1$ as a von Neumann natural, so $∣ A \cup B ∣ = 1$ by [L1]; while $∣ A ∣ + ∣ B ∣ = 1 + 1 = 2$ . Since $1 \neq 2$ by [L3], the displayed statement fails.

givenL1L3

2.1

A witness with partial overlap, to show that the failure is not an artefact of taking $A$ and $B$ equal. Take $A = {0, 1} = 2$ and $B = {1, 2}$ . Then $∣ A ∣ = 2$ by [L1], and $∣ B ∣ = 2$ because $0 \mapsto 1$ , $1 \mapsto 2$ is a bijection of $2$ onto $B$ . But $A \cup B = {0, 1, 2} = 3$ , so $∣ A \cup B ∣ = 3$ , whereas $∣ A ∣ + ∣ B ∣ = 4$ . Again the two differ.

step 1.1L1L3

3.1

The missing hypothesis is disjointness. [L2] proves the identity whenever $A \cap B = \emptyset$ , and in step 1.1 the intersection is ${0}$ while in step 2.1 it is ${1}$ ; in each case the excess of the right-hand side over the left is the cardinality of that intersection.

step 1.1step 2.1L2∎

Remarks

What the general correction is. Adding the counts and then subtracting the count of the overlap is the two-set case of inclusion and exclusion. That principle is the next page of this track and is not available here, so no formula for the general case is stated: what is established above is only that the identity as displayed is false, and where its hypothesis went.
The same failure at the level of a family is exhibited concretely in A count that overcounts because the blocks are not disjoint, and exactly where the sum rule's hypothesis is spent, where twelve subsets are counted as sixteen.

Depends on

Used by

Nothing in the library uses this result yet.

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

Rule of sum (Wikipedia) (standard reference, not scraped)
Inclusion-exclusion principle (Wikipedia) (standard reference, not scraped)
Cardinality (Wikipedia) (standard reference, not scraped)
J. Sylvestre, Elementary Foundations 12.02, Properties of finite sets and their cardinality (LibreTexts) (standard reference, not scraped)

FALSE: ∣A∪B∣=∣A∣+∣B∣∣A∪B∣=∣A∣+∣B∣ for all finite AA and BB