CorollaryStatement: 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.

For a nonempty index set $I$ : $B \cap \bigcup_{i \in I} A_i = \bigcup_{i \in I} (B \cap A_i)$ , $B \cup \bigcap_{i \in I} A_i = \bigcap_{i \in I} (B \cup A_i)$ , $X \setminus \bigcup_{i \in I} A_i = \bigcap_{i \in I} (X \setminus A_i)$ , and $X \setminus \bigcap_{i \in I} A_i = \bigcup_{i \in I} (X \setminus A_i)$

Statement

Let $(A_i)_{i \in I}$ be an indexed family with $I \neq \varnothing$ and let $B$ and $X$ be sets. Then $(B \cap A_i)_{i \in I}$ , $(B \cup A_i)_{i \in I}$ and $(X \setminus A_i)_{i \in I}$ are indexed families with index set $I$ , and

$B \cap \bigcup_{i \in I} A_i = \bigcup_{i \in I} (B \cap A_i), \qquad B \cup \bigcap_{i \in I} A_i = \bigcap_{i \in I} (B \cup A_i),$

$X \setminus \bigcup_{i \in I} A_i = \bigcap_{i \in I} (X \setminus A_i), \qquad X \setminus \bigcap_{i \in I} A_i = \bigcup_{i \in I} (X \setminus A_i).$

Facts & Assumptions

Given: an indexed family $(A_i)_{i \in I}$ with $I \neq \varnothing$ , and sets $B$ and $X$ .

[L1]

$\bigcup_{i \in I} A_i \;:=\; \bigcup \{A_i : i \in I\}$ ( $\bigcup_{i \in I} A_i := \bigcup \{A_i : i \in I\}$ , and $\bigcap_{i \in I} A_i := \bigcap \{A_i : i \in I\}$ for $I \neq \varnothing$ ).

[L2]

An indexed family with index set $I$ is a function $A$ with $\operatorname{dom} A = I$ (An indexed family $(A_i)_{i \in I}$ is a function with domain $I$ ; $\{A_i : i \in I\}$ is its range).

[L3]

$\{A_i : i \in I\} \;:=\; \operatorname{ran} A$ (An indexed family $(A_i)_{i \in I}$ is a function with domain $I$ ; $\{A_i : i \in I\}$ is its range).

[L4]

$a \cap \bigcup F = \bigcup \{\, a \cap b : b \in F \,\}, \qquad a \cup \bigcap F = \bigcap \{\, a \cup b : b \in F \,\}$ for a set $F$ with $F \neq \varnothing$ ( $a \cap (b \cup c) = (a \cap b) \cup (a \cap c)$ and $a \cup (b \cap c) = (a \cup b) \cap (a \cup c)$ ; and for a nonempty set $F$ , $a \cap \bigcup F = \bigcup \{\, a \cap b : b \in F \,\}$ and $a \cup \bigcap F = \bigcap \{\, a \cup b : b \in F \,\}$ ).

[L5]

$X \setminus \bigcup F = \bigcap \{\, X \setminus a : a \in F \,\}, \qquad X \setminus \bigcap F = \bigcup \{\, X \setminus a : a \in F \,\}$ for a set $F$ with $F \neq \varnothing$ ( $X \setminus (a \cup b) = (X \setminus a) \cap (X \setminus b)$ and $X \setminus (a \cap b) = (X \setminus a) \cup (X \setminus b)$ ; and for a nonempty set $F$ , $X \setminus \bigcup F = \bigcap \{\, X \setminus a : a \in F \,\}$ and $X \setminus \bigcap F = \bigcup \{\, X \setminus a : a \in F \,\}$ ).

[L6]

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

[L7]

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

[L8]

$\operatorname{ran} R := \{\, b : \exists a\ (a,b) \in R \,\}$ (Relation, $\operatorname{dom} R$ , $\operatorname{ran} R$ , $\operatorname{fld} R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

[L9]

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

[L10]

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

[L11]

$z \in A \times B$ holds if and only if $z = (a,b)$ for some $a \in A$ and some $b \in B$ (The Cartesian product $A \times B := \{\, z \in \mathcal{P}(\mathcal{P}(A \cup B)) : \exists a \in A\ \exists b \in B\ z = (a,b) \,\}$ ).

[L12]

A function is a relation $f$ such that $(a,b) \in f$ and $(a,c) \in f$ imply $b = c$ (A function is a relation $f$ with $(a,b) \in f$ and $(a,c) \in f$ implying $b = c$ ; $f : A \to B$ , the value $f(a)$ , domain and codomain).

Proof

technique · direct

1.1

The three derived families exist. Each $B \cap A_i$ is a subset of $B$ , so separating inside $I \times \mathcal{P}(B)$ with the formula $\exists i\,\exists w\,(z = (i,w) \wedge i \in I \wedge w = B \cap A_i)$ gives a set; it is a function, since the value at each $i$ is determined, and its domain is $I$ . The same construction inside $I \times \mathcal{P}(B \cup \bigcup_{i \in I} A_i)$ and inside $I \times \mathcal{P}(X)$ gives $(B \cup A_i)_{i \in I}$ and $(X \setminus A_i)_{i \in I}$ .

L2L9L10L11L12

2.1

Write $F := \{A_i : i \in I\}$ , which is $\operatorname{ran} A$ and is nonempty because $I$ is. The range of $(B \cap A_i)_{i \in I}$ is exactly $\{\, B \cap b : b \in F \,\}$ , since the values of the derived family are the sets $B \cap A_i$ with $i \in I$ and the elements of $F$ are exactly the $A_i$ with $i \in I$ ; the same holds for the other two derived families.

L2L3L8step 1.1

3.1

Substituting into the family laws for $F$ therefore gives all four identities: the indexed operations are by definition the primitive $\bigcup$ and $\bigcap$ applied to the range of the family concerned, and step 2.1 identifies those ranges with the sets appearing in the laws.

L1L4L5L6L7step 2.1

4.1

The derived families exist and the four identities hold, which is the statement.

step 1.1step 2.1step 3.1∎

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: 28 results over 12 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

De Morgan's laws (Wikipedia) (standard reference, not scraped)
Indexed family (Wikipedia) (standard reference, not scraped)
B. Kaya, MATH 320 Set Theory (METU), §2.3 (standard reference, not scraped)