False statementConstruction: 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.

FALSE: $⋂ \emptyset = \emptyset$

Statement

False statement. The condition defining $⋂ x$ determines a set at $x = \emptyset$ , and that set is $\emptyset$ :

$⋂ \emptyset = \emptyset .$

The claim is tempting because $⋃ \emptyset = \emptyset$ is true and the two operations look symmetric. They are not. The condition defining $⋃ x$ asks for a witness inside $x$ , so it fails for every $z$ when $x$ has no elements; the condition defining $⋂ x$ is a universal statement about the elements of $x$ , so it holds for every $z$ when $x$ has no elements.

Facts & Assumptions

Given: the claim above.

[L1]

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

[L2]

For $x \neq \emptyset$ , $⋂ x$ is the set whose elements are exactly the sets belonging to every element of $x$ (The intersection $⋂ x$ of a nonempty set, the binary intersection $a \cap b : = ⋂ {a, b}$ , and disjointness).

[L3]

There is no set $y$ such that, for every set $x$ , $x \in y$ holds if and only if $x$ belongs to every element of $\emptyset$ (There is no set $y$ with $x \in y \leftrightarrow \forall s (s \in \emptyset \to x \in s)$ , so $⋂ \emptyset$ is undefined).

[L4]

There is no set $U$ such that $y \in U$ for every set $y$ (There is no set $U$ with $y \in U$ for every set $y$ ).

Refutation

technique · contradiction

1.1

Suppose the condition defining the intersection determines a set at $\emptyset$ , and call it $c$ .

assume-contra

2.1

$\emptyset$ has no elements, so every set $x$ satisfies " $x$ belongs to every element of $\emptyset$ " vacuously; hence every set is an element of $c$ .

L1L2step 1.1

3.1

No set has every set as an element, so no such $c$ exists; in particular the equation of the claim asserts something of an object that is not there.

L3L4step 2.1

4.1

The claim also fails on its own terms: were $c$ equal to $\emptyset$ it would have no elements, while step 2.1 puts $\emptyset$ itself among its elements. Both readings collapse, so $⋂ \emptyset$ is left undefined rather than assigned the value $\emptyset$ .

L1step 2.1step 3.1discharge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 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

Intersection (set theory) (Wikipedia) (standard reference, not scraped)
C. Wilson, A Brief Introduction to ZFC (Chicago REU 2016), §2.3 (standard reference, not scraped)
B. Kaya, MATH 320 Set Theory (METU), §1.2 (standard reference, not scraped)

FALSE: ⋂∅=∅⋂∅=∅