TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck 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.

$a \cap (b \cup c) = (a \cap b) \cup (a \cap c)$ and $a \cup (b \cap c) = (a \cup b) \cap (a \cup c)$ ; and for a nonempty set $F$ , $a \cap ⋃ F = ⋃ {a \cap b : b \in F}$ and $a \cup ⋂ F = ⋂ {a \cup b : b \in F}$

Statement

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

$a \cap (b \cup c) = (a \cap b) \cup (a \cap c), a \cup (b \cap c) = (a \cup b) \cap (a \cup c) .$

Let $F$ be a set with $F \neq \emptyset$ . Then ${a \cap b : b \in F}$ and ${a \cup b : b \in F}$ are nonempty sets, and

$a \cap ⋃ F = ⋃ {a \cap b : b \in F}, a \cup ⋂ F = ⋂ {a \cup b : b \in F} .$

Facts & Assumptions

Given: sets $a$ , $b$ , $c$ and a set $F$ with $F \neq \emptyset$ where the family forms are concerned.

[L1]

$z \in a \cup b$ holds if and only if $z \in a$ or $z \in b$ ( $⋃ \emptyset = \emptyset$ , $⋃ {a} = a$ , $⋃ {a, b} = a \cup b$ , $⋂ {a} = a$ , and $⋂ {a, b} = a \cap b$ ).

[L2]

$z \in a \cap b$ holds if and only if $z \in a$ and $z \in b$ ( $⋃ \emptyset = \emptyset$ , $⋃ {a} = a$ , $⋃ {a, b} = a \cup b$ , $⋂ {a} = a$ , and $⋂ {a, b} = a \cap b$ ).

[L3]

$⋃ x$ is the set whose elements are exactly the elements of the elements of $x$ (The union $⋃ x$ of a set, and the binary union $a \cup b : = ⋃ {a, b}$ ).

[L4]

For $x \neq \emptyset$ , $⋂ x$ is the set whose elements are exactly the sets belonging to every element of $x$ (The intersection $⋂ x$ of a nonempty set, the binary intersection $a \cap b : = ⋂ {a, b}$ , and disjointness).

[L5]

For any parameters $\overset{ˉ}{p}$ and any set $x$ , there is a set $y$ whose elements are exactly the elements $z$ of $x$ for which $φ (z, \overset{ˉ}{p})$ holds (The Axiom Schema of Separation: for each formula $φ$ , $\forall \overset{ˉ}{p} \forall x \exists y \forall z (z \in y \leftrightarrow (z \in x \land φ (z, \overset{ˉ}{p})))$ ).

[L6]

$z \in P (x)$ holds if and only if $z \subseteq x$ (The power set $P (x) = {z : z \subseteq x}$ ).

[L7]

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

[L8]

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 · direct

1.1

Every $a \cap b$ is a subset of $a$ , so the class ${a \cap b : b \in F}$ is the separated set ${w \in P (a) : \exists b (b \in F \land w = a \cap b)}$ , hence a set; and it has a member because $F$ does.

L2L5L6L7

1.2

Every $a \cup b$ with $b \in F$ is a subset of $a \cup ⋃ F$ , since an element of $b$ lies in $⋃ F$ ; so ${a \cup b : b \in F}$ is the separated set ${w \in P (a \cup ⋃ F) : \exists b (b \in F \land w = a \cup b)}$ , hence a set, and it has a member because $F$ does.

L1L3L5L6L7

1.3

$z \in a \cap (b \cup c)$ holds exactly when $z \in a$ , and $z \in b$ or $z \in c$ ; that is exactly when $z \in a$ and $z \in b$ , or $z \in a$ and $z \in c$ , which is $z \in (a \cap b) \cup (a \cap c)$ .

L1L2L8

1.4

$z \in a \cup (b \cap c)$ holds exactly when $z \in a$ , or $z \in b$ and $z \in c$ ; that is exactly when $z \in a$ or $z \in b$ , and $z \in a$ or $z \in c$ , which is $z \in (a \cup b) \cap (a \cup c)$ .

L1L2L8

2.1

$z \in a \cap ⋃ F$ holds exactly when $z \in a$ and $z \in s$ for some $s \in F$ , that is, exactly when $z \in a \cap s$ for some $s \in F$ ; the sets $a \cap s$ with $s \in F$ are precisely the elements of ${a \cap b : b \in F}$ , so this holds exactly when $z$ lies in some element of that set, which is $z \in ⋃ {a \cap b : b \in F}$ .

L2L3L8step 1.1

2.2

$z \in ⋂ {a \cup b : b \in F}$ holds exactly when, for every $b \in F$ , $z \in a$ or $z \in b$ . If $z \in a \cup ⋂ F$ then either $z \in a$ , and the condition holds outright, or $z$ lies in every element of $F$ , and it holds again. Conversely, if the condition holds and $z \notin a$ , then $z \in b$ for every $b \in F$ , so $z \in ⋂ F$ ; hence $z \in a \cup ⋂ F$ in both cases.

L1L4L8step 1.2

3.1

The two finite laws and the two family laws are established, which is the statement.

step 1.3step 1.4step 2.1step 2.2∎

Depends on

Used by

For a nonempty index set I: B ∩ ⋃_i ∈ I Aᵢ = ⋃_i ∈ I (B ∩ Aᵢ), B ∪ ⋂_i ∈ I Aᵢ = ⋂_i ∈ I (B ∪ Aᵢ), X ∖ ⋃_i ∈ I Aᵢ = ⋂_i ∈ I (X ∖ Aᵢ), and X ∖ ⋂_i ∈ I Aᵢ = ⋃_i ∈ I (X ∖ Aᵢ) Corollary

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on 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)
Union (set theory) (Wikipedia) (standard reference, not scraped)

a∩(b∪c)=(a∩b)∪(a∩c)a∩(b∪c)=(a∩b)∪(a∩c) and a∪(b∩c)=(a∪b)∩(a∪c)a∪(b∩c)=(a∪b)∩(a∪c); and for a nonempty set FF, a∩⋃F=⋃{ a∩b:b∈F }a∩⋃F=⋃{a∩b:b∈F} and a∪⋂F=⋂{ a∪b:b∈F }a∪⋂F=⋂{a∪b:b∈F}