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.

If $a \in A$ and $b \in B$ then $(a,b) \in \mathcal{P}(\mathcal{P}(A \cup B))$

Statement

Let $A$ and $B$ be sets. If $a \in A$ and $b \in B$ , then $(a,b) \in \mathcal{P}(\mathcal{P}(A \cup B))$ .

Facts & Assumptions

Given: sets $A$ and $B$ , and elements $a \in A$ and $b \in B$ .

[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]

$z \in a \cup b$ holds if and only if $z \in a$ or $z \in b$ ( $\bigcup \varnothing = \varnothing$ , $\bigcup \{a\} = a$ , $\bigcup \{a,b\} = a \cup b$ , $\bigcap \{a\} = a$ , and $\bigcap \{a,b\} = a \cap b$ ).

[L4]

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

[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]

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

Proof

technique · direct

1.1

$a \in A$ gives $a \in A \cup B$ , and $b \in B$ gives $b \in A \cup B$ .

L3L6given

2.1

The elements of $\{a\}$ are $a$ alone and the elements of $\{a,b\}$ are $a$ and $b$ , so both sets are included in $A \cup B$ and are therefore elements of $\mathcal{P}(A \cup B)$ .

L2L4L5step 1.1

3.1

The elements of $\{\{a\},\{a,b\}\}$ are $\{a\}$ and $\{a,b\}$ , so that set is included in $\mathcal{P}(A \cup B)$ and is therefore an element of $\mathcal{P}(\mathcal{P}(A \cup B))$ ; and that set is $(a,b)$ .

L1L2L4L5step 2.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 16 results over 6 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)
C. Wilson, A Brief Introduction to ZFC (Chicago REU 2016), §2.4 (standard reference, not scraped)
Ordered pair (Wikipedia) (standard reference, not scraped)

If a∈Aa \in Aa∈A and b∈Bb \in Bb∈B then (a,b)∈P(P(A∪B))(a,b) \in \mathcal{P}(\mathcal{P}(A \cup B))(a,b)∈P(P(A∪B))