ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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.

$\mathbb{Z}$ sits inside $\mathbb{Q}$ as a subring that is not a subfield, so the inverse-closure clause of the subfield definition is doing work

Example

The integers are not literally a subset of the rationals in this library: $\mathbb{Q}$ is a set of equivalence classes of pairs of integers, so " $\mathbb{Z}$ inside $\mathbb{Q}$ " means the image of the embedding $j : \mathbb{Z} \to \mathbb{Q}$ , $j(k) = [(k,1)]$ , of The integers embed in the rationals. Write $K := j(\mathbb{Z}) \subseteq \mathbb{Q}$ and let $2 := 1 + 1$ in $\mathbb{Z}$ . Then:

$K$ is a subring of the ring $\mathbb{Q}$ (Subring: a subset containing $1_R$ and closed under addition, additive inverses and multiplication);
$K$ is not a subfield of $\mathbb{Q}$ (Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations): the element $j(2)$ is a nonzero member of $K$ whose inverse in $\mathbb{Q}$ does not lie in $K$ ;
so the inverse-closure clause (K2) of Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations is not implied by being a subring.

Facts & Assumptions

Given: The embedding $j : \mathbb{Z} \to \mathbb{Q}$ , $j(k) = [(k,1)]$ , and $K = j(\mathbb{Z})$ ; the numeral $2 = 1 + 1$ in $\mathbb{Z}$ (The integers as equivalence classes of pairs of naturals).

[L1]

$j$ is injective and preserves addition and multiplication; composing with the embedding of $\mathbb{N}$ it also preserves order (The integers embed in the rationals).

[L2]

$\mathbb{Q}$ is a field, hence a commutative ring, with $1_{\mathbb{Q}} \ne 0_{\mathbb{Q}}$ ( $\mathbb{Q}$ and $\mathbb{R}$ are fields, hence commutative rings, integral domains and ordered rings, all of characteristic $0$ , The rationals form a field, Field, Every field is a commutative ring with $1 \ne 0$ ; it is an integral domain, and it is a commutative division ring, Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Commutative ring).

[L3]

$\mathbb{Z}$ is a commutative ring; its order is total and compatible with addition ( $\mathbb{Z}$ is a commutative ring and an ordered ring, the published construction being an instance of the general definitions, The integers form a commutative ring, The integers form a totally ordered ring, Order on the integers).

[L4]

Cancellation in the additive group of $\mathbb{Q}$ ; and, in a field, $x \cdot x = x$ with $x \ne 0$ forces $x = 1$ , since multiplying by $x^{-1}$ gives $x = 1$ (Field, Every field is a commutative ring with $1 \ne 0$ ; it is an integral domain, and it is a commutative division ring, Cancellation in a group: $gx = gy$ or $xg = yg$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $gx = h$ and $xg = h$ each have exactly one solution).

[L5]

Subring criterion: $S \subseteq R$ is a subring exactly when $1_R \in S$ and $u - v \in S$ and $uv \in S$ for all $u, v \in S$ (Subring criterion: $S \subseteq R$ is a subring if and only if $1_R \in S$ and $a - b \in S$ and $ab \in S$ for all $a, b \in S$ ; and an intersection of subrings is a subring, Subring: a subset containing $1_R$ and closed under addition, additive inverses and multiplication).

[L6]

A subfield of a field $F$ is a subring closed under the inverses of its nonzero elements (Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations).

[L7]

The group of units of $\mathbb{Z}$ is $\{1,-1\}$ ( $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ , The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring, Left inverse, right inverse, and invertible element of a monoid).

[L8]

$\iota : \mathbb{N} \to \mathbb{Z}$ is injective and order preserving with $\iota(0) = 0$ , $\iota(1) = 1$ (The naturals embed in the integers, Arithmetic on the integers).

Verification

technique · direct

1.1

$j(0_{\mathbb{Z}}) = 0_{\mathbb{Q}}$ and $j(1_{\mathbb{Z}}) = 1_{\mathbb{Q}}$ . The first: $j(0) + j(0) = j(0+0) = j(0) = j(0) + 0_{\mathbb{Q}}$ , and cancelling $j(0)$ gives $j(0) = 0_{\mathbb{Q}}$ . The second: $j(1)j(1) = j(1 \cdot 1) = j(1)$ , and $j(1) \ne 0_{\mathbb{Q}}$ because $1_{\mathbb{Z}} \ne 0_{\mathbb{Z}}$ and $j$ is injective with $j(0_{\mathbb{Z}}) = 0_{\mathbb{Q}}$ ; so $j(1) = 1_{\mathbb{Q}}$ . Consequently $j(-b) = -j(b)$ for every $b$ , since $j(b) + j(-b) = j(0) = 0_{\mathbb{Q}}$ .

L1L2L3L4L8

1.2

$0 < 1 < 2$ in $\mathbb{Z}$ , and $-1 < 0$ : the first because $1 = \iota(1)$ is nonnegative and $1 \ne 0$ by injectivity of $\iota$ , the second by adding $1$ to $0 < 1$ , and the third by adding $-1$ to $0 < 1$ . Hence $2 \ne 0$ , $2 \ne 1$ and $2 \ne -1$ , so $2 \notin \mathbb{Z}^{\times}$ by [L7].

L3L7L8

2.1

$1_{\mathbb{Q}} = j(1_{\mathbb{Z}}) \in K$ by step 1.1; and for $u = j(a)$ , $v = j(b)$ in $K$ we have $u - v = j(a) + (-j(b)) = j(a) + j(-b) = j(a-b) \in K$ and $uv = j(a)j(b) = j(ab) \in K$ . So $K$ is a subring of $\mathbb{Q}$ by [L5]. This is claim 1.

step 1.1L1L2L5

2.2

$j(2) \ne 0_{\mathbb{Q}}$ : by step 1.2, $2 \ne 0_{\mathbb{Z}}$ , and $j$ is injective with $j(0_{\mathbb{Z}}) = 0_{\mathbb{Q}}$ by step 1.1. So $j(2)$ has an inverse $j(2)^{-1}$ in the field $\mathbb{Q}$ .

step 1.1step 1.2L1L2

3.1

$j(2)^{-1} \notin K$ . Suppose it were, say $j(2)^{-1} = j(m)$ for some $m \in \mathbb{Z}$ . Then $j(2m) = j(2)j(m) = 1_{\mathbb{Q}} = j(1_{\mathbb{Z}})$ , so $2m = 1_{\mathbb{Z}}$ by injectivity of $j$ ; commutativity also gives $m2=1$ , so $m$ is a two-sided inverse and $2$ is a unit of $\mathbb{Z}$ , contradicting step 1.2.

step 1.1step 1.2step 2.2L1L3L7

4.1

Claims 2 and 3: by claim 1 the set $K$ is a subring of $\mathbb{Q}$ , and by steps 2.2 and 3.1 it contains a nonzero element whose inverse in $\mathbb{Q}$ is not in $K$ ; so (K2) of [L6] fails and $K$ is not a subfield. Since $K$ satisfies (K1), the clause (K2) is not implied by (K1).

step 2.1step 2.2step 3.1L6∎

Remarks

The example is about an image, not a subset. The integers embed in the rationals is what makes " $\mathbb{Z}$ inside $\mathbb{Q}$ " meaningful, since an integer and a rational are different kinds of object in this library. Every claim above is about $K = j(\mathbb{Z})$ , and $j$ being injective and operation-preserving is what lets facts about $\mathbb{Z}$ , in particular $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ , be used about $K$ .
A subring of a field is automatically an integral domain, since it is a commutative ring with $1 \ne 0$ inheriting the absence of zero divisors from the field. So $K$ is a domain and not a field, which is the same phenomenon as $\mathbb{Z}$ is an integral domain of characteristic $0$ whose group of units is $\{1,-1\}$ , so it is not a field: $2$ is nonzero and not invertible seen inside $\mathbb{Q}$ .
One witness is enough. Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations asks that every nonzero element of $K$ have its inverse in $K$ , so a single element failing it settles the matter; the example produces $j(2)$ and nothing more.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 110 results over 27 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Subring (Wikipedia) (standard reference, not scraped)
Field (mathematics) (Wikipedia) (standard reference, not scraped)
Field (mathematics) (Wikipedia) (standard reference, not scraped)
Integral domain (Wikipedia) (standard reference, not scraped)

Z\mathbb{Z}Z sits inside Q\mathbb{Q}Q as a subring that is not a subfield, so the inverse-closure clause of the subfield definition is doing work