LemmaStatement: Literature-sourcedProof: 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.

Every relation $R$ satisfies $R \subseteq \operatorname{dom} R \times \operatorname{ran} R$ , and $R$ is a relation from $A$ to $B$ if and only if $\operatorname{dom} R \subseteq A$ and $\operatorname{ran} R \subseteq B$

Statement

Let $R$ be a relation and let $A$ and $B$ be sets. Then

(i) $R \subseteq \operatorname{dom} R \times \operatorname{ran} R$ ;
(ii) $R \subseteq A \times B$ if and only if $\operatorname{dom} R \subseteq A$ and $\operatorname{ran} R \subseteq B$ .

Facts & Assumptions

Given: a relation $R$ and sets $A$ , $B$ .

[L1]

A relation is a set $R$ every element of which is an ordered pair (Relation, $\operatorname{dom} R$ , $\operatorname{ran} R$ , $\operatorname{fld} R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

[L2]

$\operatorname{dom} R := \{\, a : \exists b\ (a,b) \in R \,\}, \qquad \operatorname{ran} R := \{\, b : \exists a\ (a,b) \in R \,\}$ (Relation, $\operatorname{dom} R$ , $\operatorname{ran} R$ , $\operatorname{fld} R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

[L3]

$z \in A \times B$ holds if and only if $z = (a,b)$ for some $a \in A$ and some $b \in B$ (The Cartesian product $A \times B := \{\, z \in \mathcal{P}(\mathcal{P}(A \cup B)) : \exists a \in A\ \exists b \in B\ z = (a,b) \,\}$ ).

[L4]

$(a,b) = (c,d)$ if and only if $a = c$ and $b = d$ ( $(a,b) = (c,d)$ if and only if $a = c$ and $b = d$ ).

[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

Claim (i): let $z \in R$ . Since $R$ is a relation, $z = (a,b)$ for some sets $a$ and $b$ ; then $a \in \operatorname{dom} R$ and $b \in \operatorname{ran} R$ by the defining conditions, so $z \in \operatorname{dom} R \times \operatorname{ran} R$ .

L1L2L3L5

1.2

Claim (ii), from left to right: assume $R \subseteq A \times B$ . If $a \in \operatorname{dom} R$ then $(a,b) \in R$ for some $b$ , so $(a,b) \in A \times B$ , so $(a,b) = (a',b')$ with $a' \in A$ and $b' \in B$ , and the characterising property gives $a = a' \in A$ . The argument for $\operatorname{ran} R \subseteq B$ is the same on the second coordinate.

L2L3L4L5

1.3

Claim (ii), from right to left: assume $\operatorname{dom} R \subseteq A$ and $\operatorname{ran} R \subseteq B$ , and let $z \in R$ . Then $z = (a,b)$ with $a \in \operatorname{dom} R \subseteq A$ and $b \in \operatorname{ran} R \subseteq B$ , so $z \in A \times B$ .

L1L2L3L5

2.1

Claims (i) and (ii) are established, which is the statement.

step 1.1step 1.2step 1.3∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 18 results over 8 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)
Cartesian product (Wikipedia) (standard reference, not scraped)

Every relation RRR satisfies R⊆dom⁡R×ran⁡RR \subseteq \operatorname{dom} R \times \operatorname{ran} RR⊆domR×ranR, and RRR is a relation from AAA to BBB if and only if dom⁡R⊆A\operatorname{dom} R \subseteq AdomR⊆A and ran⁡R⊆B\operatorname{ran} R \subseteq BranR⊆B