LemmaStatement: Literature-sourcedProof: 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 $a \in A$ and $b \in B$ then $(a, b) \in P (P (A \cup B))$

Statement

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

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 $z \in b$ ( $⋃ \emptyset = \emptyset$ , $⋃ {a} = a$ , $⋃ {a, b} = a \cup b$ , $⋂ {a} = a$ , and $⋂ {a, b} = a \cap b$ ).

[L4]

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

[L5]

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

[L6]

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

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 $P (A \cup B)$ .

L2L4L5step 1.1

3.1

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

L1L2L4L5step 2.1∎

Depends on

Used by

Dependency tree · two levels

14 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)
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∈A and b∈Bb∈B then (a,b)∈P(P(A∪B))(a,b)∈P(P(A∪B))