TheoremStatement: 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.

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

Statement

For all sets $a$ , $b$ , $c$ , $d$ : $(a, b) = (c, d)$ if and only if $a = c$ and $b = d$ .

Facts & Assumptions

Given: sets $a$ , $b$ , $c$ , $d$ .

[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]

${a, b} = {c, d} \leftrightarrow ((a = c \land b = d) \lor (a = d \land b = c))$ ( ${a, b} = {c, d}$ if and only if ( $a = c$ and $b = d$ ) or ( $a = d$ and $b = c$ ); in particular ${a} = {c, d}$ if and only if $a = c = d$ ).

Proof

technique · cases

1.1

Right to left: if $a = c$ and $b = d$ then ${a} = {c}$ and ${a, b} = {c, d}$ , so ${{a}, {a, b}}$ and ${{c}, {c, d}}$ are the same set, that is $(a, b) = (c, d)$ .

L1L2

1.2

Left to right, setting up: assume ${{a}, {a, b}} = {{c}, {c, d}}$ . Applying the equality criterion for unordered pairs to these two sets, either ${a} = {c}$ and ${a, b} = {c, d}$ , or ${a} = {c, d}$ and ${a, b} = {c}$ .

L1L3

2.1

In the first alternative, ${a} = {c}$ reads ${a, a} = {c, c}$ and gives $a = c$ . Then ${a, b} = {c, d}$ gives $a = c$ and $b = d$ , or $a = d$ and $b = c$ ; in the second of these $b = c = a = d$ , so $b = d$ holds as well. Either way $a = c$ and $b = d$ .

assume-case firststep 1.2L2L3

2.2

In the second alternative, ${a, b} = {c}$ reads ${a, b} = {c, c}$ and gives $a = c$ and $b = c$ , while ${a} = {c, d}$ reads ${a, a} = {c, d}$ and gives $a = c$ and $a = d$ . Hence $a = c$ and $b = c = a = d$ .

assume-case secondstep 1.2L2L3

3.1

The two alternatives supplied by step 1.2 are exhaustive, so the left-to-right implication holds.

step 2.1step 2.2cases-exhaustive

4.1

Both implications hold, which is the statement.

step 1.1step 3.1∎

Depends on

Used by

Dependency tree · two levels

5 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

Ordered pair (Wikipedia) (standard reference, not scraped)
B. Kaya, MATH 320 Set Theory (METU), Lemma 1 (standard reference, not scraped)
C. Wilson, A Brief Introduction to ZFC (Chicago REU 2016), Thm. 2.7 (standard reference, not scraped)

(a,b)=(c,d)(a,b)=(c,d) if and only if a=ca=c and b=db=d