LemmaStatement: AI-adaptedProof: 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 $b = d$ ) or ( $a = d$ and $b = c$ ); in particular $\{a\} = \{c,d\}$ if and only if $a = c = d$

Statement

For all sets $a$ , $b$ , $c$ , $d$ :

$\{a,b\} = \{c,d\} \ \leftrightarrow\ \bigl((a = c \wedge b = d) \vee (a = d \wedge b = c)\bigr).$

In particular, taking $b := a$ , $\{a\} = \{c,d\}$ if and only if $a = c$ and $a = d$ .

Facts & Assumptions

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

[L1]

$\{x,y\}$ is the set whose elements are exactly $x$ and $y$ , and $\{x\} := \{x,x\}$ (The unordered pair $\{x,y\}$ and the singleton $\{x\} = \{x,x\}$ ).

[L2]

If every $z$ satisfies $z \in x$ if and only if $z \in y$ , then $x = y$ (The Axiom of Extensionality: $\forall x\,\forall y\,(\forall z\,(z \in x \leftrightarrow z \in y) \to x = y)$ ).

Proof

technique · cases

1.1

Right to left: if $a = c$ and $b = d$ , or $a = d$ and $b = c$ , then for every $t$ the conditions " $t = a$ or $t = b$ " and " $t = c$ or $t = d$ " are the same, so the two pairs have the same elements and are equal.

L1L2

1.2

Left to right, setting up: assume $\{a,b\} = \{c,d\}$ . Since $a$ and $b$ are elements of the left-hand set they are elements of the right-hand one, so $a$ is $c$ or $d$ and $b$ is $c$ or $d$ ; likewise $c$ is $a$ or $b$ and $d$ is $a$ or $b$ .

2.1

Suppose $a = b$ . Then $c$ is $a$ or $b$ , so $c = a$ ; and $d$ is $a$ or $b$ , so $d = a$ . Hence $a = c$ and $b = a = d$ , the first disjunct.

assume-case samestep 1.2

2.2

Suppose instead $a \neq b$ . If $a = c$ , then $b$ is $c$ or $d$ , and $b = c = a$ is excluded, so $b = d$ and the first disjunct holds. Otherwise $a = d$ ; then $c$ is $a$ or $b$ , and $c = a$ would give $c = a = d$ , whence $b$ , being $c$ or $d$ , equals $a$ , contradicting $a \neq b$ ; so $c = b$ and the second disjunct holds.

assume-case diffstep 1.2

3.1

The alternatives $a = b$ and $a \neq b$ exhaust the possibilities, so the left-to-right implication holds.

step 2.1step 2.2cases-exhaustive

4.1

Both implications hold, which is the biconditional; putting $b := a$ makes the two disjuncts read " $a = c$ and $a = d$ " and " $a = d$ and $a = c$ ", so $\{a\} = \{c,d\}$ if and only if $a = c$ and $a = d$ .

step 1.1step 3.1L1∎

Depends on

Used by

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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 4 results over 3 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), §1.2 and Lemma 1 (standard reference, not scraped)
Axiom of pairing (Wikipedia) (standard reference, not scraped)
Ordered pair (Wikipedia) (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) or (a=da = da=d and b=cb = cb=c); in particular {a}={c,d}\{a\} = \{c,d\}{a}={c,d} if and only if a=c=da = c = da=c=d