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

1.1L1L4

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 $\emptyset$ .

1.2L1L3

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 $z \in a$ .

1.3L1L3

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 $z \in b$ .

1.4L2L3

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 $z \in a$ .

1.5L2L3

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 $z \in b$ .

2.1L5step 1.1step 1.2step 1.3step 1.4step 1.5∎

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.

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

⋃∅=∅⋃∅=∅, ⋃{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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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