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.

For sets $A$ and $B$ the collection of all functions $A \to B$ is a set, being a subset of $\mathcal{P}(A \times B)$

Statement

Let $A$ and $B$ be sets. Then there is a set whose elements are exactly the functions $f : A \to B$ , and it is a subset of $\mathcal{P}(A \times B)$ .

Facts & Assumptions

Given: sets $A$ and $B$ .

[L1]

We write $f : A \to B$ , and say $f$ is a function from $A$ to $B$ , when $f$ is a function with $\operatorname{dom} f = A$ and $\operatorname{ran} f \subseteq B$ (A function is a relation $f$ with $(a,b) \in f$ and $(a,c) \in f$ implying $b = c$ ; $f : A \to B$ , the value $f(a)$ , domain and codomain).

[L2]

For a relation $R$ , $R \subseteq A \times B$ if and only if $\operatorname{dom} R \subseteq A$ and $\operatorname{ran} R \subseteq B$ (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 $\operatorname{ran} R \subseteq B$ ).

[L3]

$z \in \mathcal{P}(x)$ holds if and only if $z \subseteq x$ (The power set $\mathcal{P}(x) = \{\, z : z \subseteq x \,\}$ ).

[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 $\{\, z \in x : \varphi(z) \,\}$ ).

[L6]

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

Proof

technique · direct

1.1

A function $f : A \to B$ is a relation with $\operatorname{dom} f = A$ and $\operatorname{ran} f \subseteq B$ , so $f \subseteq A \times B$ and therefore $f \in \mathcal{P}(A \times B)$ .

L1L2L3L5L6

2.1

Separating inside $\mathcal{P}(A \times B)$ with the formula saying that $z$ is a function and $\operatorname{dom} z = A$ , with parameters $A$ and $B$ , gives a set whose elements are exactly those elements of $\mathcal{P}(A \times B)$ that are functions $A \to B$ ; by step 1.1 every function $A \to B$ is such an element, so that set has exactly the intended elements and is included in $\mathcal{P}(A \times B)$ .

L1L3L4L5step 1.1∎

Depends on

Used by

The set B^A of all functions A → B Definition

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 21 results over 9 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), §2.2 (standard reference, not scraped)
Function (mathematics) (Wikipedia) (standard reference, not scraped)
Power set (Wikipedia) (standard reference, not scraped)

For sets AAA and BBB the collection of all functions A→BA \to BA→B is a set, being a subset of P(A×B)\mathcal{P}(A \times B)P(A×B)