CorollaryStatement: 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 $x \cup \{x\} = y \cup \{y\}$ then $x = y$

Statement

For all sets $x$ and $y$ , if $x \cup \{x\} = y \cup \{y\}$ then $x = y$ .

Facts & Assumptions

Given: sets $x$ and $y$ with $x \cup \{x\} = y \cup \{y\}$ ; the Axiom of Foundation, through the result cited as [L3].

[L1]

$z \in a \cup b$ 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 $\bigcap \{a,b\} = a \cap b$ ).

[L2]

$\{x\} := \{x,x\}$ , the singleton of $x$ , is the set whose only element is $x$ (The unordered pair $\{x,y\}$ and the singleton $\{x\} = \{x,x\}$ ).

[L3]

There are no sets $x$ and $y$ with $x \in y$ and $y \in x$ (Under Foundation, $x \notin x$ for every set $x$ , there are no sets with $x \in y \in x$ , and there are no sets with $x \in y \in z \in x$ ).

[L4]

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

Proof

technique · direct

1.1

$x \in \{x\}$ , so $x \in x \cup \{x\}$ ; likewise $y \in y \cup \{y\}$ .

L1L2L4

2.1

By hypothesis the two sets are equal, so $x \in y \cup \{y\}$ and $y \in x \cup \{x\}$ ; hence $x \in y$ or $x = y$ , and $y \in x$ or $y = x$ .

L1L2givenstep 1.1

3.1

If $x \neq y$ then both alternatives $x = y$ and $y = x$ fail, leaving $x \in y$ and $y \in x$ , which is impossible; therefore $x = y$ .

L3step 2.1∎

Remarks

The general statement, and its special case for $\omega$ . This holds for all sets and is proved from Foundation. The corresponding clause for the natural numbers, that the successor is injective on $\omega$ , is part (P2) of The von Neumann naturals form a Peano system ↗, which proves it by a different route that does not use Foundation.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 13 results over 7 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), Lemma 13 (standard reference, not scraped)
Axiom of regularity (Wikipedia) (standard reference, not scraped)
Axiom of infinity (Wikipedia) (standard reference, not scraped)

If x∪{x}=y∪{y}x \cup \{x\} = y \cup \{y\}x∪{x}=y∪{y} then x=yx = yx=y