TheoremStatement: 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.

$(a,b) = (c,d)$ 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]

$\{a,b\} = \{c,d\} \ \leftrightarrow\ \bigl((a = c \wedge b = d) \vee (a = d \wedge b = c)\bigr)$ ( $\{a,b\} = \{c,d\}$ 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 6 results over 4 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

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