CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck 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 $A$ , $B$ , $C$ with $(A \times B) \times C \neq A \times (B \times C)$

Statement refuted

Refuted claim: $(A \times B) \times C = A \times (B \times C)$ for all sets $A$ , $B$ , $C$ . The witness is $A = B = C = {\emptyset}$ : the left-hand side has the element $((\emptyset, \emptyset), \emptyset)$ , whose first coordinate is an ordered pair, and every element of the right-hand side has first coordinate $\emptyset$ .

This is why the convention $(a, b, c) : = ((a, b), c)$ of The ordered triple $(a, b, c) : = ((a, b), c)$ and the iterated products $A \times B \times C : = (A \times B) \times C$ has to be fixed rather than assumed harmless.

Facts & Assumptions

Given: $A = B = C = {\emptyset}$ .

[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 $A \times B : = {z \in P (P (A \cup B)) : \exists a \in A \exists b \in B z = (a, b)}$ ).

[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]

$(a, b) : = {{a}, {a, b}}$ (The Kuratowski ordered pair $(a, b) : = {{a}, {a, b}}$ ).

[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 $\emptyset$ ).

[L6]

If every $z$ satisfies $z \in x$ if and only if $z \in y$ , then $x = y$ (The Axiom of Extensionality: $\forall x \forall y (\forall z (z \in x \leftrightarrow z \in y) \to x = y)$ ).

Counterexample

technique · direct

1.1

$(\emptyset, \emptyset) = {{\emptyset}, {\emptyset, \emptyset}} = {{\emptyset}}$ , which has ${\emptyset}$ as an element; $\emptyset$ has no element, so $(\emptyset, \emptyset) \neq \emptyset$ .

L3L4L5L6

2.1

Since $\emptyset$ is the only element of each of $A$ , $B$ , $C$ , the product $A \times B$ has $(\emptyset, \emptyset)$ as its only element, so $((\emptyset, \emptyset), \emptyset)$ is an element of $(A \times B) \times C$ .

L1L4step 1.1

2.2

Every element of $A \times (B \times C)$ has the form $(x, y)$ with $x \in A$ , hence with $x = \emptyset$ ; if $((\emptyset, \emptyset), \emptyset)$ were such an element then the characterising property would give $(\emptyset, \emptyset) = \emptyset$ , which step 1.1 refutes.

L1L2L4step 1.1

3.1

The set $(A \times B) \times C$ therefore has an element that $A \times (B \times C)$ does not, so the two products are different.

step 2.1step 2.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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

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

Sets AA, BB, CC with (A×B)×C≠A×(B×C)(A×B)×C=A×(B×C)