CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-02 (claude-opus-5)

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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.

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A count that overcounts because the blocks are not disjoint, and exactly where the sum rule's hypothesis is spent

Statement refuted

Refuted claim: clause 2 of 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 with its disjointness hypothesis deleted, that is, for a finite index set $I$ and any family $(A_{i})_{i \in I}$ of finite sets,

$∣ ⋃_{i \in I} A_{i} ∣ = \sum_{i \in I} ∣ A_{i} ∣ .$

The witness lives inside $P (X)$ for $X : = 4 = {0, 1, 2, 3}$ . Take $I : = 2$ , let $A_{0}$ be the set of subsets of $X$ containing $0$ and $A_{1}$ the set of subsets containing $1$ . Then $∣ A_{0} ∣ = ∣ A_{1} ∣ = 8$ , so the right-hand side is $16$ , while $∣ A_{0} \cup A_{1} ∣ = 12$ .

Facts & Assumptions

Given: $X = 4 = {0, 1, 2, 3}$ ; $A_{0} : = {S \subseteq X : 0 \in S}$ ; $A_{1} : = {S \subseteq X : 1 \in S}$ ; and $C : = {S \subseteq X : 0 \notin S and 1 \notin S}$ .

[L1]

$∣ P (Y) ∣ = 2^{∣ Y ∣}$ ( $∣ P (A) ∣ = 2^{∣ A ∣}$ for finite $A$ , Exponentiation of natural numbers, $m^{n}$ , and its agreement with the integer power in $R$ ), and $2^{2} = 4$ , $2^{3} = 8$ , $2^{4} = 16$ .

[L2]

The sum rule for two disjoint blocks: $∣ S \cup T ∣ = ∣ S ∣ + ∣ T ∣$ , and its proof, whose only use of disjointness is the injectivity of the splice map $h$ (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, step 1.1 there).

[L3]

Cardinality (The cardinality $∣ A ∣$ of a finite set): transport along a bijection; a subset of a finite set is finite (A subset of a finite set is finite, with $∣ B ∣ \leq ∣ A ∣$ , and equality holds if and only if $B = A$ ).

[L4]

Maps (Injection, surjection, bijection): a map with a two-sided inverse is a bijection.

[L5]

Cancellation in $N$ (Addition is cancellative), and sums over a finite index set (The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form).

Counterexample

technique · direct

1.1

The three sets are subsets of the finite set $P (X)$ , hence finite by [L3], and $∣ P (X) ∣ = 2^{4} = 16$ by [L1].

givenL1L3

2.1

Each block has eight elements. The map $S \mapsto S ∖ {0}$ sends $A_{0}$ into $P ({1, 2, 3})$ and $T \mapsto T \cup {0}$ sends $P ({1, 2, 3})$ into $A_{0}$ ; the two composites are the identity, because $0 \in S$ for $S \in A_{0}$ and $0 \notin T$ for $T \subseteq {1, 2, 3}$ . So $∣ A_{0} ∣ = 2^{3} = 8$ by [L1], [L3] and [L4], and the same argument at the point $1$ gives $∣ A_{1} ∣ = 8$ . Hence $\sum_{i \in 2} ∣ A_{i} ∣ = 8 + 8 = 16$ .

step 1.1L1L3L4L5

2.2

The union has twelve. A subset of $X$ lies in $A_{0} \cup A_{1}$ exactly when it contains $0$ or contains $1$ , so $P (X)$ is the disjoint union of $A_{0} \cup A_{1}$ and $C$ ; and $C$ is in bijection with $P ({2, 3})$ under the identity map, since a subset of $X$ containing neither $0$ nor $1$ is precisely a subset of ${2, 3}$ , giving $∣ C ∣ = 2^{2} = 4$ . By [L2], $16 = ∣ A_{0} \cup A_{1} ∣ + 4$ , so $∣ A_{0} \cup A_{1} ∣ = 12$ by [L5].

step 1.1L1L2L3L4L5

3.1

The claim fails: $12 \neq 16$ . The overcount is exactly $4$ , the number of subsets containing both $0$ and $1$ , and it agrees with $∣ C ∣$ because $S \mapsto S \cup {0, 1}$ is a bijection of $C$ onto $A_{0} \cap A_{1}$ , with inverse $T \mapsto T ∖ {0, 1}$ .

step 2.1step 2.2L4∎

Remarks

Where the proof of 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 breaks. Its step 1.1 splices bijections $f : p \to A$ and $g : q \to B$ into $h : p + q \to A \cup B$ , and the only place disjointness enters is the third case of injectivity: an index below $p$ must give a value different from an index at least $p$ , and that is guaranteed only because $f$ lands in $A$ , $g$ lands in $B$ and $A \cap B = \emptyset$ . Here the four subsets containing both $0$ and $1$ are each hit twice, so $h$ is surjective but not injective, and one gets $∣ A_{0} \cup A_{1} ∣ \leq 16$ rather than equality.
The systematic repair is inclusion and exclusion, which subtracts the count of the overlap. It is the next page of this track and is not available here, so the correction is not stated as a formula: what this item establishes is that some correction is needed.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 81 results over 27 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

Rule of sum (Wikipedia) (standard reference, not scraped)
Inclusion-exclusion principle (Wikipedia) (standard reference, not scraped)
Power set (Wikipedia) (standard reference, not scraped)