False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-02 (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.

FALSE: for all sets $A$ and $B$ with $B$ having at least two elements, $A \times B$ is strictly larger than $A$

Statement

FALSE. The statement

for all sets $A$ and $B$ with $B$ having at least two elements, $A ≺ A \times B$

that is, $A$ injects into $A \times B$ and is not equinumerous with it (Equinumerous sets, $A \approx B$ and $A ⪯ B$ ).

The claim generalises the finite product rule in the shape a reader expects: if $∣ A ∣$ is multiplied by at least $2$ , surely the product is bigger. It fails at both ends of the range, trivially when $A$ is empty and substantially when $A$ is infinite.

Facts & Assumptions

Given: The sets $N$ and $\emptyset$ , and $2 = {0, 1}$ (The natural numbers $N$ (von Neumann)).

[L1]

$N \times N \approx N$ : the map $J (m, n) = 2^{m} (2 n + 1)$ is a bijection of $N \times N$ onto $N ∖ {0}$ , and composing with the inverse of the successor gives a bijection onto $N$ ( $N \times N \approx N$ , Finite, countably infinite, countable, uncountable).

[L2]

$X ≺ Y$ means $X ⪯ Y$ and $X \approx̸ Y$ (Equinumerous sets, $A \approx B$ and $A ⪯ B$ ).

[L3]

$∣ A \times B ∣ = ∣ A ∣ ∣ B ∣$ for finite $A$ , $B$ (The product rule: $∣ A \times B ∣ = ∣ A ∣ ∣ B ∣$ , and $∣ \prod_{i < m} A_{i} ∣ = \prod_{i < m} ∣ A_{i} ∣$ ), and $∣ X ∣ = 0$ exactly when $X = \emptyset$ (The cardinality $∣ A ∣$ of a finite set).

[L4]

Order arithmetic of $N$ : $p \leq q$ implies $p k \leq q k$ ; $x \leq y$ implies $x + c \leq y + c$ ; $a \neq 0$ is the same as $1 \leq a$ ; $2 \cdot a = σ (1) \cdot a = 1 \cdot a + a = a + a$ by the successor-left law; $1 \cdot a = a$ ; and multiplication is commutative (Order is compatible with multiplication, Order is compatible with addition, Discreteness: $σ (n)$ is the immediate successor, Zero and one under multiplication, Distributivity and the successor law for multiplication, Multiplication is commutative, Order on the natural numbers).

[L5]

$0 \neq 1$ in $N$ , so $N$ has at least two elements (The von Neumann naturals form a Peano system).

Refutation

technique · direct

1.1

The substantial witness: $A = B = N$ . The set $N$ has at least two elements by [L5], so the hypothesis on $B$ holds. By [L1] there is a bijection $N \times N \to N$ , so $A \approx A \times B$ , and therefore $A ≺ A \times B$ is false by [L2].

givenL1L2L5

1.2

A degenerate witness, which shows the claim fails even for finite $A$ : take $A = \emptyset$ and $B = 2$ . Then $A \times B = \emptyset$ , since a pair in it would have a first coordinate in $\emptyset$ ; so $A \approx A \times B$ and again $A ≺ A \times B$ fails.

givenL2L3

2.1

The corrected finite statement is true. Let $A$ be finite and nonempty and let $B$ be finite with $∣ B ∣ \geq 2$ ; write $a : = ∣ A ∣ \geq 1$ and $b : = ∣ B ∣$ . Then $∣ A \times B ∣ = a b \geq 2 a = a + a \geq a + 1 > a$ by [L3] and [L4], using $1 \leq a$ . So a finite nonempty $A$ is strictly smaller than $A \times B$ in cardinality. Both hypotheses are needed, by step 1.1 and step 1.2 respectively.

step 1.1step 1.2L3L4

3.1

So the displayed statement is false, and what fails is not the product rule but its extension beyond the finite nonempty case: finiteness and nonemptiness of $A$ are exactly the hypotheses under which multiplying by a factor of at least $2$ increases the count.

step 1.1step 1.2step 2.1L2L3∎

Remarks

The contrast with Cantor's theorem is the point. $X ≺ P (X)$ holds for every set whatsoever (Cantor's theorem: $A ≺ P (A)$ ), finite or infinite; strict increase survives to the infinite case there and not here. Passing to the power set is a genuinely different operation from multiplying by a fixed set.
Read the cited theorem before using it. $N \times N \approx N$ states that $J (m, n) = 2^{m} (2 n + 1)$ is a bijection onto the nonzero naturals, and the bijection onto $N$ is obtained by composing with the inverse of the successor. The statement used above is the one that item actually proves.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

47 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

Rule of product (Wikipedia) (standard reference, not scraped)
Cardinality (Wikipedia) (standard reference, not scraped)
Cantor's theorem (Wikipedia) (standard reference, not scraped)
Countable set (Wikipedia) (standard reference, not scraped)

FALSE: for all sets AA and BB with BB having at least two elements, A×BA×B is strictly larger than AA