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.

For $f : A \to B$ and $T, U \subseteq B$ : $f^{- 1} [T \cup U] = f^{- 1} [T] \cup f^{- 1} [U]$ , $f^{- 1} [T \cap U] = f^{- 1} [T] \cap f^{- 1} [U]$ , $f^{- 1} [T ∖ U] = f^{- 1} [T] ∖ f^{- 1} [U]$ , and $f^{- 1} [B ∖ T] = A ∖ f^{- 1} [T]$

Statement

Let $f : A \to B$ and let $T, U \subseteq B$ . Then

$f^{- 1} [T \cup U] = f^{- 1} [T] \cup f^{- 1} [U], f^{- 1} [T \cap U] = f^{- 1} [T] \cap f^{- 1} [U],$

$f^{- 1} [T ∖ U] = f^{- 1} [T] ∖ f^{- 1} [U], f^{- 1} [B ∖ T] = A ∖ f^{- 1} [T] .$

Facts & Assumptions

Given: a function $f : A \to B$ and subsets $T, U$ of $B$ .

[L1]

$a \in R^{- 1} [B]$ holds if and only if $(a, b) \in R$ for some $b \in B$ (The image $R [A]$ and the preimage $R^{- 1} [B]$ of a set under a relation).

[L2]

We write $f : A \to B$ , and say $f$ is a function from $A$ to $B$ , when $f$ is a function with $dom f = A$ and $ran f \subseteq B$ (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).

[L3]

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

[L4]

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

[L5]

$z \in a ∖ b$ 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 $X$ ).

[L6]

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

[L7]

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

[L8]

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

Proof

technique · direct

1.1

Membership criterion: for any $S \subseteq B$ , $a \in f^{- 1} [S]$ holds exactly when $a \in A$ and $f (a) \in S$ . Indeed $a \in f^{- 1} [S]$ says $(a, b) \in f$ for some $b \in S$ , and $(a, b) \in f$ says $a \in dom f = A$ and $b = f (a)$ .

L1L2L7L8

2.1

First identity: $a \in f^{- 1} [T \cup U]$ exactly when $a \in A$ and $f (a) \in T$ or $f (a) \in U$ , which is exactly $a \in f^{- 1} [T]$ or $a \in f^{- 1} [U]$ .

L3L6step 1.1

2.2

Second identity: $a \in f^{- 1} [T \cap U]$ exactly when $a \in A$ and $f (a)$ lies in both $T$ and $U$ , which is exactly $a \in f^{- 1} [T]$ and $a \in f^{- 1} [U]$ .

L4L6step 1.1

2.3

Third identity: $a \in f^{- 1} [T ∖ U]$ exactly when $a \in A$ , $f (a) \in T$ and $f (a) \notin U$ ; the last two conditions say $a \in f^{- 1} [T]$ and $a \notin f^{- 1} [U]$ , since for $a \in A$ membership in $f^{- 1} [U]$ is exactly $f (a) \in U$ .

L5L6step 1.1

2.4

Fourth identity: $a \in f^{- 1} [B ∖ T]$ exactly when $a \in A$ , $f (a) \in B$ and $f (a) \notin T$ ; the condition $f (a) \in B$ is automatic because $ran f \subseteq B$ , so this says $a \in A$ and $a \notin f^{- 1} [T]$ .

L2L5L6step 1.1

3.1

The four identities are established, which is the statement.

step 2.1step 2.2step 2.3step 2.4∎

Depends on

Used by

The image-preimage adjunctions explain the set-operation preservation laws Remark

Dependency tree · two levels

18 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

Image (mathematics) (Wikipedia) (standard reference, not scraped)
B. Kaya, MATH 320 Set Theory (METU), §2.2 (standard reference, not scraped)
Algebra of sets (Wikipedia) (standard reference, not scraped)

For f:A→Bf:A→B and T,U⊆BT,U⊆B: f−1[T∪U]=f−1[T]∪f−1[U]f−1[T∪U]=f−1[T]∪f−1[U], f−1[T∩U]=f−1[T]∩f−1[U]f−1[T∩U]=f−1[T]∩f−1[U], f−1[T∖U]=f−1[T]∖f−1[U]f−1[T∖U]=f−1[T]∖f−1[U], and f−1[B∖T]=A∖f−1[T]f−1[B∖T]=A∖f−1[T]