LemmaStatement: AI-adaptedProof: 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.

$⋃ \emptyset = \emptyset$ , $⋃ {a} = a$ , $⋃ {a, b} = a \cup b$ , $⋂ {a} = a$ , and $⋂ {a, b} = a \cap b$

Statement

For all sets $a$ and $b$ :

(i) $⋃ \emptyset = \emptyset$ ;
(ii) $⋃ {a} = a$ ;
(iii) $⋃ {a, b} = a \cup b$ , and $z \in a \cup b$ holds if and only if $z \in a$ or $z \in b$ ;
(iv) $⋂ {a} = a$ ;
(v) $⋂ {a, b} = a \cap b$ , and $z \in a \cap b$ holds if and only if $z \in a$ and $z \in b$ .

The equalities in (iii) and (v) are the definitions of $a \cup b$ and $a \cap b$ written out; what is proved about them here is the membership criterion beside each.

Facts & Assumptions

Given: sets $a$ and $b$ .

[L1]

$⋃ x$ is the set whose elements are exactly the elements of the elements of $x$ , and $a \cup b : = ⋃ {a, b}$ (The union $⋃ x$ of a set, and the binary union $a \cup b : = ⋃ {a, b}$ ).

[L2]

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

[L3]

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

[L4]

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

[L5]

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

Proof

technique · direct

1.1

Claim (i): $z \in ⋃ \emptyset$ requires a member $s$ of $\emptyset$ with $z \in s$ , and $\emptyset$ has no members, so $⋃ \emptyset$ has no elements and is therefore $\emptyset$ .

L1L4

1.2

Claim (ii): $z \in ⋃ {a}$ requires a member $s$ of ${a}$ with $z \in s$ , and the only member of ${a}$ is $a$ , so $z \in ⋃ {a}$ holds exactly when $z \in a$ .

L1L3

1.3

Claim (iii): $a \cup b$ is $⋃ {a, b}$ by definition, and $z \in ⋃ {a, b}$ requires a member $s$ of ${a, b}$ with $z \in s$ ; the members of ${a, b}$ are $a$ and $b$ , so $z \in a \cup b$ holds exactly when $z \in a$ or $z \in b$ .

L1L3

1.4

Claim (iv): ${a}$ is nonempty since $a$ is a member, and $z \in ⋂ {a}$ holds exactly when $z$ belongs to every member of ${a}$ , that is, exactly when $z \in a$ .

L2L3

1.5

Claim (v): $a \cap b$ is $⋂ {a, b}$ by definition, ${a, b}$ is nonempty, and $z \in ⋂ {a, b}$ holds exactly when $z$ belongs to every member of ${a, b}$ ; the members are $a$ and $b$ , so $z \in a \cap b$ holds exactly when $z \in a$ and $z \in b$ .

L2L3

2.1

In each of the five claims the elements of the two sides have been shown to satisfy the same condition, so the sets are equal, which is the statement.

L5step 1.1step 1.2step 1.3step 1.4step 1.5∎

Depends on

Used by

Dependency tree · two levels

10 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

B. Kaya, MATH 320 Set Theory (METU), §1.2 (standard reference, not scraped)
Union (set theory) (Wikipedia) (standard reference, not scraped)
Intersection (set theory) (Wikipedia) (standard reference, not scraped)

⋃∅=∅⋃∅=∅, ⋃{a}=a⋃{a}=a, ⋃{a,b}=a∪b⋃{a,b}=a∪b, ⋂{a}=a⋂{a}=a, and ⋂{a,b}=a∩b⋂{a,b}=a∩b