PropositionStatement: 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.

Union and intersection are commutative, associative, idempotent and absorptive; $a \cup \varnothing = a$ , $a \cap \varnothing = \varnothing$ , $a \setminus \varnothing = a$ , $a \setminus a = \varnothing$ ; and $a \subseteq b$ if and only if $a \cup b = b$ , if and only if $a \cap b = a$ , if and only if $a \setminus b = \varnothing$

Statement

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

(i) $a \cup b = b \cup a$ and $a \cap b = b \cap a$ ;
(ii) $(a \cup b) \cup c = a \cup (b \cup c)$ and $(a \cap b) \cap c = a \cap (b \cap c)$ ;
(iii) $a \cup a = a$ and $a \cap a = a$ ;
(iv) $a \cup (a \cap b) = a$ and $a \cap (a \cup b) = a$ ;
(v) $a \cup \varnothing = a$ , $a \cap \varnothing = \varnothing$ , $a \setminus \varnothing = a$ and $a \setminus a = \varnothing$ ;
(vi) $a \subseteq b$ , $a \cup b = b$ , $a \cap b = a$ and $a \setminus b = \varnothing$ are equivalent.

Facts & Assumptions

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

[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]

$z \in a \cap b$ holds if and only if $z \in a$ and $z \in b$ ( $\bigcup \varnothing = \varnothing$ , $\bigcup \{a\} = a$ , $\bigcup \{a,b\} = a \cup b$ , $\bigcap \{a\} = a$ , and $\bigcap \{a,b\} = a \cap b$ ).

[L3]

$z \in a \setminus b$ 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 $X$ ).

[L4]

$x \subseteq y$ means that every element of $x$ is an element of $y$ (Subset $x \subseteq y$ , proper subset $x \subsetneq y$ , and the separation notation $\{\, z \in x : \varphi(z) \,\}$ ).

[L5]

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

[L6]

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

Proof

technique · direct

1.1

Claim (i): $z \in a \cup b$ holds exactly when $z \in a$ or $z \in b$ , which is the same condition as $z \in b$ or $z \in a$ , so $a \cup b$ and $b \cup a$ have the same elements; the same argument with "and" in place of "or" gives $a \cap b = b \cap a$ .

L1L2L5

1.2

Claim (ii): $z \in (a \cup b) \cup c$ holds exactly when $z \in a$ or $z \in b$ or $z \in c$ , in either bracketing, so the two sides have the same elements; replacing "or" by "and" throughout gives the statement for intersection.

L1L2L5

1.3

Claim (iii): $z \in a \cup a$ holds exactly when $z \in a$ or $z \in a$ , and $z \in a \cap a$ exactly when $z \in a$ and $z \in a$ ; both conditions are $z \in a$ .

L1L2L5

1.4

Claim (iv): $z \in a \cup (a \cap b)$ holds exactly when $z \in a$ , or $z \in a$ and $z \in b$ , which holds exactly when $z \in a$ ; and $z \in a \cap (a \cup b)$ exactly when $z \in a$ , and $z \in a$ or $z \in b$ , which again holds exactly when $z \in a$ .

L1L2L5

1.5

Claim (v): $z \in a \cup \varnothing$ exactly when $z \in a$ or $z \in \varnothing$ , and the second alternative never holds, so this is $z \in a$ ; $z \in a \cap \varnothing$ requires $z \in \varnothing$ , so $a \cap \varnothing$ has no elements; $z \in a \setminus \varnothing$ exactly when $z \in a$ and $z \notin \varnothing$ , and the second condition always holds; $z \in a \setminus a$ requires $z \in a$ and $z \notin a$ , so $a \setminus a$ has no elements.

L1L2L3L5L6

1.6

Claim (vi), that $a \subseteq b$ and $a \cup b = b$ are equivalent: if $a \subseteq b$ then $z \in a \cup b$ , which says $z \in a$ or $z \in b$ , holds exactly when $z \in b$ , so $a \cup b = b$ ; conversely if $a \cup b = b$ and $z \in a$ , then $z \in a \cup b = b$ .

L1L4L5

1.7

Claim (vi), that $a \subseteq b$ and $a \cap b = a$ are equivalent: if $a \subseteq b$ then $z \in a \cap b$ , which says $z \in a$ and $z \in b$ , holds exactly when $z \in a$ , so $a \cap b = a$ ; conversely if $a \cap b = a$ and $z \in a$ , then $z \in a \cap b$ , so $z \in b$ .

L2L4L5

1.8

Claim (vi), that $a \subseteq b$ and $a \setminus b = \varnothing$ are equivalent: if $a \subseteq b$ then no $z$ satisfies $z \in a$ and $z \notin b$ , so $a \setminus b$ has no elements; conversely if $a \setminus b = \varnothing$ and $z \in a$ , then $z \notin b$ is impossible, so $z \in b$ .

L3L4L6

2.1

Claims (i) to (vi) are established, and the four conditions of (vi) are equivalent because each of the last three was shown equivalent to the first.

step 1.1step 1.2step 1.3step 1.4step 1.5step 1.6step 1.7step 1.8∎

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 6 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

Algebra of sets (Wikipedia) (standard reference, not scraped)
B. Kaya, MATH 320 Set Theory (METU), §1.2 (standard reference, not scraped)
Complement (set theory) (Wikipedia) (standard reference, not scraped)