False statementConstruction: AI-adaptedVerification: 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.

FALSE: $(a \setminus b) \setminus c = a \setminus (b \setminus c)$ for all sets $a$ , $b$ , $c$

Statement

False statement. Set difference is associative: for all sets $a$ , $b$ , $c$ ,

$(a \setminus b) \setminus c = a \setminus (b \setminus c).$

Facts & Assumptions

Given: the claim above, and the sets $a = b = c := \{\varnothing\}$ .

[L1]

$z \in a \setminus b$ holds exactly when $z \in a$ and $z \notin b$ (The difference $a \setminus b$ , the symmetric difference $a \triangle b$ , and the complement $X \setminus a$ relative to a set $X$ ).

[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 is exactly one set with no elements, written $\varnothing$ (There is exactly one set with no elements, written $\varnothing$ ).

[L4]

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

Refutation

technique · direct

1.1

Take $a = b = c := \{\varnothing\}$ .

given

2.1

$a \setminus b$ has no elements, since $z \in a \setminus b$ requires $z \in a$ and $z \notin b$ while $a$ and $b$ are the same set; so $a \setminus b = \varnothing$ , and $(a \setminus b) \setminus c$ likewise has no elements and equals $\varnothing$ .

L1L3L4step 1.1

2.2

By the same computation $b \setminus c = \varnothing$ , and $z \in a \setminus \varnothing$ requires only $z \in a$ , so $a \setminus (b \setminus c) = a = \{\varnothing\}$ .

L1L3L4step 1.1

3.1

The two sides are $\varnothing$ and $\{\varnothing\}$ , which differ because the second has an element and the first has none; the claim is therefore false.

L2L3step 2.1step 2.2∎

Remarks

Where the two sides part company here. On the left, removing $b$ from $a$ already empties it, so the outer difference can only be empty. On the right, $b \setminus c$ is empty, so nothing at all is removed from $a$ . One nonempty set playing all three roles makes both collapses happen at once.

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: 10 results over 5 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

Complement (set theory) (Wikipedia) (standard reference, not scraped)
Algebra of sets (Wikipedia) (standard reference, not scraped)
B. Kaya, MATH 320 Set Theory (METU), Def. 4 (standard reference, not scraped)

FALSE: (a∖b)∖c=a∖(b∖c)(a \setminus b) \setminus c = a \setminus (b \setminus c)(a∖b)∖c=a∖(b∖c) for all sets aaa, bbb, ccc