CounterexampleConstruction: AI-adaptedVerification: 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 function $f$ and sets $S, T$ with $f [S \cap T] ⊊ f [S] \cap f [T]$

Statement refuted

Refuted claim: $f [S \cap T] = f [S] \cap f [T]$ for every function $f : A \to B$ and all $S, T \subseteq A$ . Write $u : = \emptyset$ and $v : = {\emptyset}$ , and take

$A : = {u, v}, B : = {u}, f : = {(u, u), (v, u)}, S : = {u}, T : = {v} .$

Then $S \cap T = \emptyset$ , so $f [S \cap T] = \emptyset$ , while $f [S] = f [T] = {u}$ and therefore $f [S] \cap f [T] = {u}$ . The inclusion $f [S \cap T] \subseteq f [S] \cap f [T]$ of For $f : A \to B$ and $S, T \subseteq A$ : $f [S \cup T] = f [S] \cup f [T]$ , $f [S \cap T] \subseteq f [S] \cap f [T]$ and $f [S] ∖ f [T] \subseteq f [S ∖ T]$ ; both inclusions are equalities for all $S$ and $T$ if and only if $f$ is injective is proper here, and it is proper exactly because $f$ is not injective.

Facts & Assumptions

Given: $u : = \emptyset$ , $v : = {\emptyset}$ , $A : = {u, v}$ , $B : = {u}$ , $f : = {(u, u), (v, u)}$ , $S : = {u}$ and $T : = {v}$ .

[L1]

$f [S \cap T] \subseteq f [S] \cap f [T]$ (For $f : A \to B$ and $S, T \subseteq A$ : $f [S \cup T] = f [S] \cup f [T]$ , $f [S \cap T] \subseteq f [S] \cap f [T]$ and $f [S] ∖ f [T] \subseteq f [S ∖ T]$ ; both inclusions are equalities for all $S$ and $T$ if and only if $f$ is injective).

[L2]

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

[L3]

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).

[L4]

$f$ is injective (one-to-one) if $f (x) = f (y)$ implies $x = y$ , for all $x, y \in A$ (Injection, surjection, bijection).

[L5]

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

[L6]

${x, y}$ is the set whose elements are exactly $x$ and $y$ , and ${x} : = {x, x}$ (The unordered pair ${x, y}$ and the singleton ${x} = {x, x}$ ).

[L7]

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

[L8]

$(a, b) : = {{a}, {a, b}}$ (The Kuratowski ordered pair $(a, b) : = {{a}, {a, b}}$ ).

[L9]

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

[L10]

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

[L11]

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

Counterexample

technique · direct

1.1

$u \neq v$ , since $v$ has an element and $u$ has none; so $A$ has exactly the two elements $u$ and $v$ , and $S$ and $T$ are the two singletons inside it.

L6L7

2.1

$f$ is a function $A \to B$ : its two pairs have distinct first coordinates, so it is single valued; its domain is $A$ and its range is ${u} = B$ . It is not injective, since $f (u) = u = f (v)$ while $u \neq v$ .

L3L4L8L9step 1.1

2.2

$S \cap T$ has no element, since an element would have to equal both $u$ and $v$ ; so $S \cap T = \emptyset$ and $f [S \cap T] = \emptyset$ , no element of $\emptyset$ being related to anything.

L2L5L6L7step 1.1

3.1

$f [S] = {u}$ and $f [T] = {u}$ , since $f (u) = u$ and $f (v) = u$ ; hence $f [S] \cap f [T] = {u}$ .

L2L5L6step 2.1

4.1

The general inclusion holds and is proper here: $u$ lies in $f [S] \cap f [T]$ and not in $f [S \cap T]$ , which has no element at all.

L1L7L10L11step 2.2step 3.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

22 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)
Bijection, injection and surjection (Wikipedia) (standard reference, not scraped)
B. Kaya, MATH 320 Set Theory (METU), §2.2 (standard reference, not scraped)

A function ff and sets S,TS,T with f[S∩T]⊊f[S]∩f[T]f[S∩T]⊊f[S]∩f[T]