LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 $\bigcup\bigcup R$

Statement

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

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

are sets, and both are included in $\bigcup\bigcup R$ .

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]

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

[L4]

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

[L5]

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

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 $\bigcup R$ .

L1L2L3

2.1

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

L2L3step 1.1

3.1

Separation applied to $\bigcup\bigcup R$ with the formula $\exists b\,((a,b) \in R)$ and the parameter $R$ gives the set $\{\, a \in \bigcup\bigcup R : \exists b\ (a,b) \in R \,\}$ ; by step 2.1 every $a$ satisfying that formula already lies in $\bigcup\bigcup R$ , 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 9 results over 4 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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 RRR is an ordered pair, then { a:∃b (a,b)∈R }\{\, a : \exists b\ (a,b) \in R \,\}{a:∃b (a,b)∈R} and { b:∃a (a,b)∈R }\{\, b : \exists a\ (a,b) \in R \,\}{b:∃a (a,b)∈R} are sets, both included in ⋃⋃R\bigcup\bigcup R⋃⋃R