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.

If every element of $R$ is an ordered pair, then ${a : \exists b (a, b) \in R}$ and ${b : \exists a (a, b) \in R}$ are sets, both included in $⋃ ⋃ R$

Statement

Let $R$ be a set every element of which is an ordered pair. Then the classes

${a : \exists b (a, b) \in R} and {b : \exists a (a, b) \in R}$

are sets, and both are included in $⋃ ⋃ R$ .

Facts & Assumptions

Given: a set $R$ every element of which is an ordered pair.

[L1]

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

[L2]

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

[L3]

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

[L4]

For any parameters $\overset{ˉ}{p}$ and any set $x$ , there is a set $y$ whose elements are exactly the elements $z$ of $x$ for which $φ (z, \overset{ˉ}{p})$ holds (The Axiom Schema of Separation: for each formula $φ$ , $\forall \overset{ˉ}{p} \forall x \exists y \forall z (z \in y \leftrightarrow (z \in x \land φ (z, \overset{ˉ}{p})))$ ).

[L5]

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

Proof

technique · direct

1.1

Let $(a, b) \in R$ . The elements of $(a, b)$ are ${a}$ and ${a, b}$ , and $(a, b)$ is an element of $R$ , so ${a}$ and ${a, b}$ are elements of $⋃ R$ .

L1L2L3

2.1

The elements of ${a}$ and of ${a, b}$ are elements of $⋃ ⋃ R$ ; hence $a \in ⋃ ⋃ R$ and $b \in ⋃ ⋃ R$ whenever $(a, b) \in R$ .

L2L3step 1.1

3.1

Separation applied to $⋃ ⋃ R$ with the formula $\exists b ((a, b) \in R)$ and the parameter $R$ gives the set ${a \in ⋃ ⋃ R : \exists b (a, b) \in R}$ ; by step 2.1 every $a$ satisfying that formula already lies in $⋃ ⋃ R$ , so this set has exactly the elements of the first class. The second class is obtained the same way from the formula $\exists a ((a, b) \in R)$ .

L4L5step 2.1∎

Depends on

Used by

Relation, dom R, ran R, fld R, and the specialisations "relation from A to B" and "relation on A" Definition

Dependency tree · two levels

9 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), §2.1 (standard reference, not scraped)
Binary relation (Wikipedia) (standard reference, not scraped)
C. Wilson, A Brief Introduction to ZFC (Chicago REU 2016), Def. 2.9 (standard reference, not scraped)

If every element of RR is an ordered pair, then { a:∃b (a,b)∈R }{a:∃b (a,b)∈R} and { b:∃a (a,b)∈R }{b:∃a (a,b)∈R} are sets, both included in ⋃⋃R⋃⋃R