CorollaryStatement: AI-adaptedProof: 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.

For a nonempty index set $I$ : $B \cap ⋃_{i \in I} A_{i} = ⋃_{i \in I} (B \cap A_{i})$ , $B \cup ⋂_{i \in I} A_{i} = ⋂_{i \in I} (B \cup A_{i})$ , $X ∖ ⋃_{i \in I} A_{i} = ⋂_{i \in I} (X ∖ A_{i})$ , and $X ∖ ⋂_{i \in I} A_{i} = ⋃_{i \in I} (X ∖ A_{i})$

Statement

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

$B \cap ⋃_{i \in I} A_{i} = ⋃_{i \in I} (B \cap A_{i}), B \cup ⋂_{i \in I} A_{i} = ⋂_{i \in I} (B \cup A_{i}),$

$X ∖ ⋃_{i \in I} A_{i} = ⋂_{i \in I} (X ∖ A_{i}), X ∖ ⋂_{i \in I} A_{i} = ⋃_{i \in I} (X ∖ A_{i}) .$

Facts & Assumptions

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

[L1]

$⋃_{i \in I} A_{i} : = ⋃ {A_{i} : i \in I}$ ( $⋃_{i \in I} A_{i} : = ⋃ {A_{i} : i \in I}$ , and $⋂_{i \in I} A_{i} : = ⋂ {A_{i} : i \in I}$ for $I \neq \emptyset$ ).

[L2]

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

[L3]

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

[L4]

$a \cap ⋃ F = ⋃ {a \cap b : b \in F}, a \cup ⋂ F = ⋂ {a \cup b : b \in F}$ for a set $F$ with $F \neq \emptyset$ ( $a \cap (b \cup c) = (a \cap b) \cup (a \cap c)$ 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 $a \cup ⋂ F = ⋂ {a \cup b : b \in F}$ ).

[L5]

$X ∖ ⋃ F = ⋂ {X ∖ a : a \in F}, X ∖ ⋂ F = ⋃ {X ∖ a : a \in F}$ for a set $F$ with $F \neq \emptyset$ ( $X ∖ (a \cup b) = (X ∖ a) \cap (X ∖ b)$ and $X ∖ (a \cap b) = (X ∖ a) \cup (X ∖ b)$ ; and for a nonempty set $F$ , $X ∖ ⋃ F = ⋂ {X ∖ a : a \in F}$ and $X ∖ ⋂ F = ⋃ {X ∖ a : a \in F}$ ).

[L6]

$z \in a ∖ b$ holds exactly when $z \in a$ and $z \notin b$ (The difference $a ∖ b$ , the symmetric difference $a △ b$ , and the complement $X ∖ a$ 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]

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

[L9]

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

[L10]

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 $φ$ , $\forall \overset{ˉ}{p} \forall x \exists y \forall z (z \in y \leftrightarrow (z \in x \land φ (z, \overset{ˉ}{p})))$ ).

[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 $A \times B : = {z \in P (P (A \cup B)) : \exists a \in A \exists b \in B z = (a, b)}$ ).

[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

1.1L2L9L10L11L12

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

2.1L2L3L8step 1.1

Write $F : = {A_{i} : i \in I}$ , which is $ran A$ 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.

3.1L1L4L5L6L7step 2.1

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

4.1step 1.1step 2.1step 3.1∎

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

For a nonempty index set $I$ : $B \cap ⋃_{i \in I} A_{i} = ⋃_{i \in I} (B \cap A_{i})$ , $B \cup ⋂_{i \in I} A_{i} = ⋂_{i \in I} (B \cup A_{i})$ , $X ∖ ⋃_{i \in I} A_{i} = ⋂_{i \in I} (X ∖ A_{i})$ , and $X ∖ ⋂_{i \in I} A_{i} = ⋃_{i \in I} (X ∖ A_{i})$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources