ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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.

$(Z, +)$ is an abelian group, $(Z, \cdot)$ is a commutative monoid that is not a group, and its group of units is ${1, - 1}$

Example

Let $Z$ be the integers with the operations of Arithmetic on the integers. Then

$(Z, +, 0)$ is an abelian group (Group and abelian group);
$(Z, \cdot, 1)$ is a commutative monoid (Semigroup and monoid) which is not a group, because $0$ has no multiplicative inverse;
its group of units (The invertible elements of a monoid form a group under the restricted operation) is $Z^{\times} = {1, - 1}$ , and $1 \neq - 1$ .

Facts & Assumptions

Given: The integers $Z$ with $+$ , $\cdot$ , $0$ and $1$ (The integers as equivalence classes of pairs of naturals, Arithmetic on the integers), and the embedding $ι : N \to Z$ of The naturals embed in the integers.

[L1]

In $Z$ : addition is associative and commutative, $x + 0 = x$ , and every $x = [(a, b)]$ has the additive inverse $- x = [(b, a)]$ with $x + (- x) = 0$ ; multiplication is associative and commutative with $x \cdot 1 = x$ , and it distributes over addition. These are the ring axioms, verified one by one in the proof of The integers form a commutative ring (The integers form a commutative ring, Arithmetic on the integers).

[L2]

The order on $Z$ is total, antisymmetric and transitive, is compatible with addition, and positives are closed under multiplication (The integers form a totally ordered ring, Order on the integers).

[L3]

$ι$ is injective, preserves addition, multiplication and order, and its image is exactly the nonnegative integers; $ι (0) = 0$ and $ι (1) = 1$ (The naturals embed in the integers).

[L4]

On $N$ : every $k \neq 0$ is a successor $σ (j) = 1 + j$ , so $k \neq 0$ implies $1 \leq k$ (Every nonzero natural number is a successor, Addition is commutative, Order on the natural numbers, The natural numbers $N$ (von Neumann)).

[L5]

A group is a monoid in which every element is invertible; the units of a monoid form a group (Group and abelian group, Semigroup and monoid, Left inverse, right inverse, and invertible element of a monoid, Left identity, right identity, and two-sided identity for a binary operation, The invertible elements of a monoid form a group under the restricted operation).

Verification

technique · direct

1.1L1L5

Addition on $Z$ is associative and commutative, $x + 0 = x$ and hence $0 + x = x$ , and every $x$ has a two-sided additive inverse $- x$ ; so $(Z, +, 0)$ is an abelian group.

1.2L1L5

Multiplication on $Z$ is associative and commutative and $x \cdot 1 = x = 1 \cdot x$ ; so $(Z, \cdot, 1)$ is a commutative monoid.

1.3L1

For every $y \in Z$ , $0 \cdot y = 0$ : by distributivity $0 \cdot y = (0 + 0) \cdot y = 0 \cdot y + 0 \cdot y$ , and adding $- (0 \cdot y)$ gives $0 = 0 \cdot y$ .

1.4L3L4

$1 \neq 0$ in $Z$ : $ι$ is injective with $ι (1) = 1$ and $ι (0) = 0$ , and $1 \neq 0$ in $N$ since $1 = σ (0)$ contains $0$ as an element while $0 = \emptyset$ has none.

1.5L3L4

Discreteness in $Z$ : if $0 < x$ then $1 \leq x$ . Indeed $x \geq 0$ gives $x = ι (k)$ with $k \in N$ , and $k \neq 0$ since $ι (0) = 0 \neq x$ ; so $1 \leq k$ in $N$ and, $ι$ preserving the order, $1 = ι (1) \leq ι (k) = x$ .

2.1step 1.3step 1.4L5

$0$ is not invertible in $(Z, \cdot, 1)$ : $0 \cdot y = 0 \neq 1$ for every $y$ . Hence $(Z, \cdot, 1)$ is not a group.

2.2step 1.3L1

$1$ and $- 1$ are units: $1 \cdot 1 = 1$ , and $(- 1) (- 1) = 1$ because $(- 1) (- 1) + (- 1) = (- 1) ((- 1) + 1) = (- 1) \cdot 0 = 0$ , so $(- 1) (- 1)$ is the additive inverse of $- 1$ , which is $1$ .

2.3step 1.5L1L2

If $0 < u$ and $1 \leq v$ then $u \leq u v$ : from $1 \leq v$ we get $0 \leq v - 1$ , so $u (v - 1) \geq 0$ , either factor being possibly zero, and $u v - u = u (v - 1) \geq 0$ gives $u \leq u v$ .

2.4step 1.3step 1.4L1L2L3

Let $x y = 1$ . Then $x \neq 0$ and $y \neq 0$ by step 1.3 and step 1.4. If $x > 0$ and $y < 0$ then $0 < x (- y) = - (x y) = - 1$ , so $1 < 0$ , contradicting $0 < 1$ , which holds by step 1.4 and totality since $1 \geq 0$ ; the case $x < 0$ , $y > 0$ is the same with the names interchanged. So $x$ and $y$ are both positive or both negative.

3.1step 1.5step 2.3L2

Both positive: $1 \leq x$ and $1 \leq y$ by step 1.5, so $x \leq x y = 1$ by step 2.3, and with $1 \leq x$ antisymmetry gives $x = 1$ .

4.1step 3.1L1L2

Both negative: then $0 < - x$ and $0 < - y$ and $(- x) (- y) = x y = 1$ by ring arithmetic, so $- x = 1$ by step 3.1, that is $x = - 1$ .

5.1step 2.2step 2.4step 3.1step 4.1L1L2L5∎

By steps 2.2, 2.4, 3.1 and 4.1 the units of $(Z, \cdot, 1)$ are exactly $1$ and $- 1$ ; and $1 \neq - 1$ , since $1 = - 1$ would give $1 + 1 = 0$ , while $0 < 1$ gives $1 < 1 + 1$ and hence $0 < 1 + 1$ . So $Z^{\times} = {1, - 1}$ , a group under multiplication with two elements.

Remarks

The Statement of The integers form a commutative ring is quoted here by its content, not by its name. That theorem says $Z$ is "a commutative ring with multiplicative identity", a phrase not defined at this point in the reading order; what is used above is the list of equations its proof verifies one at a time. Nothing here presupposes a definition of a ring.
Cancellation without invertibility. $(Z, \cdot)$ satisfies cancellation by nonzero elements yet is not a group, and $(N, +)$ is a sharper example still, being cancellative outright (A commutative monoid in which cancellation holds need not be a group: $(N, +)$ ).
$Z^{\times} = {1, - 1}$ is the first finite group in the library that is not trivial. It is cyclic of order $2$ , generated by $- 1$ .

$(Z, +)$ is an abelian group, $(Z, \cdot)$ is a commutative monoid that is not a group, and its group of units is ${1, - 1}$

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

(Z,+)(Z,+) is an abelian group, (Z,⋅)(Z,⋅) is a commutative monoid that is not a group, and its group of units is {1,−1}{1,−1}

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

$(Z, +)$ is an abelian group, $(Z, \cdot)$ is a commutative monoid that is not a group, and its group of units is ${1, - 1}$