CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-06 (claude-opus-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.

Sets with $(A \times C) \cup (B \times D) \subsetneq (A \cup B) \times (C \cup D)$

Statement refuted

Refuted claim: $(A \times C) \cup (B \times D) = (A \cup B) \times (C \cup D)$ for all sets $A$ , $B$ , $C$ , $D$ . The witness is $A = C := \{\varnothing\}$ and $B = D := \{\{\varnothing\}\}$ : the pair $(\varnothing,\{\varnothing\})$ mixes a first coordinate from $A$ with a second coordinate from $D$ , so it lies in the right-hand side and in neither product on the left.

Facts & Assumptions

Given: $A = C := \{\varnothing\}$ and $B = D := \{\{\varnothing\}\}$ .

[L1]

$z \in A \times B$ holds if and only if $z = (a,b)$ for some $a \in A$ and some $b \in B$ (The Cartesian product $A \times B := \{\, z \in \mathcal{P}(\mathcal{P}(A \cup B)) : \exists a \in A\ \exists b \in B\ z = (a,b) \,\}$ ).

[L2]

$(a,b) = (c,d)$ if and only if $a = c$ and $b = d$ ( $(a,b) = (c,d)$ if and only if $a = c$ and $b = d$ ).

[L3]

$z \in a \cup b$ holds if and only if $z \in a$ or $z \in b$ ( $\bigcup \varnothing = \varnothing$ , $\bigcup \{a\} = a$ , $\bigcup \{a,b\} = a \cup b$ , $\bigcap \{a\} = a$ , and $\bigcap \{a,b\} = a \cap b$ ).

[L4]

$\{x,y\}$ is the set whose elements are exactly $x$ and $y$ , and $\{x\} := \{x,x\}$ (The unordered pair $\{x,y\}$ and the singleton $\{x\} = \{x,x\}$ ).

[L5]

There is exactly one set with no elements (There is exactly one set with no elements, written $\varnothing$ ).

[L6]

$x \subseteq y$ means that every element of $x$ is an element of $y$ (Subset $x \subseteq y$ , proper subset $x \subsetneq y$ , and the separation notation $\{\, z \in x : \varphi(z) \,\}$ ).

[L7]

$a \cup b := \bigcup\{a,b\}$ (The union $\bigcup x$ of a set, and the binary union $a \cup b := \bigcup \{a,b\}$ ).

Counterexample

technique · direct

1.1

$\varnothing \neq \{\varnothing\}$ , since the second has an element and the first has none; so $\varnothing \notin B$ and $\{\varnothing\} \notin C$ , while $\varnothing \in A$ and $\{\varnothing\} \in D$ .

L4L5

1.2

The inclusion from left to right always holds: an element of $A \times C$ is a pair $(x,y)$ with $x \in A$ and $y \in C$ , hence with $x \in A \cup B$ and $y \in C \cup D$ , so it lies in $(A \cup B) \times (C \cup D)$ ; the same argument applies to $B \times D$ .

L1L3L6L7

2.1

The pair $(\varnothing,\{\varnothing\})$ lies in $(A \cup B) \times (C \cup D)$ , because $\varnothing \in A$ gives $\varnothing \in A \cup B$ and $\{\varnothing\} \in D$ gives $\{\varnothing\} \in C \cup D$ .

L1L3L7step 1.1

2.2

It lies in neither product on the left. Membership in $A \times C$ would give $\{\varnothing\} \in C$ , and membership in $B \times D$ would give $\varnothing \in B$ ; the characterising property makes the coordinates unambiguous, and step 1.1 rules out both.

L1L2L3step 1.1

3.1

The inclusion of step 1.2 therefore omits the element exhibited at step 2.1, so it is proper.

step 1.2step 2.1step 2.2∎

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: 19 results over 8 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

Cartesian product (Wikipedia) (standard reference, not scraped)
Algebra of sets (Wikipedia) (standard reference, not scraped)
B. Kaya, MATH 320 Set Theory (METU), §2.1 (standard reference, not scraped)

Sets with (A×C)∪(B×D)⊊(A∪B)×(C∪D)(A \times C) \cup (B \times D) \subsetneq (A \cup B) \times (C \cup D)(A×C)∪(B×D)⊊(A∪B)×(C∪D)