LemmaStatement: AI-adaptedProof: AI-generatedprecheck 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.

The order presentation and the positive-cone presentation of an ordered ring determine each other: $P = {x : 0 < x}$ satisfies trichotomy and closure, and $a < b : ⟺ b - a \in P$ recovers the order

Statement

Let $R$ be a ring (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides). Call a subset $P \subseteq R$ a cone when

(C1) trichotomy: for each $x \in R$ exactly one of $x \in P$ , $x = 0$ , $- x \in P$ holds;
(C2) closure: if $x, y \in P$ then $x + y \in P$ and $x y \in P$ .

Then:

If $\leq$ makes $R$ an ordered ring (Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication), then $P : = {x \in R : 0 < x}$ is a cone, and for all $a, b \in R$ , $a < b$ if and only if $b - a \in P$ .
Conversely, let $P \subseteq R$ be a cone and define $a \leq_{P} b$ to mean $b - a \in P$ or $a = b$ . Then $\leq_{P}$ is a total order making $R$ an ordered ring, and its positive cone ${x \in R : 0 <_{P} x}$ is $P$ .
The two constructions are mutually inverse: starting from an ordered ring $(R, \leq)$ , the order $\leq_{P}$ built from its cone $P$ is $\leq$ itself; and starting from a cone $P$ , the cone of $\leq_{P}$ is $P$ .

Facts & Assumptions

Given: A ring $R$ with zero $0$ ; $x - y$ abbreviates $x + (- y)$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L1]

$(R, +, 0)$ is an abelian group: addition is associative and commutative, $x + 0 = x$ , and $x + (- x) = 0$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Group and abelian group).

[L2]

Identities of the abelian group $(R, +, 0)$ , where $x - y : = x + (- y)$ : $x - 0 = x$ ; $x - x = 0$ ; $- (x - y) = y - x$ , since $- (u + v) = (- v) + (- u)$ and $- (- y) = y$ ; $(y - x) + (z - y) = z - x$ , by associativity and commutativity; and $x - y = 0$ exactly when $x = y$ , by cancellation after adding $y$ (Group and abelian group, In a group $e^{- 1} = e$ , $(g^{- 1})^{- 1} = g$ and $(g h)^{- 1} = h^{- 1} g^{- 1}$ , the order of the last product being essential, Cancellation in a group: $g x = g y$ or $x g = y g$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $g x = h$ and $x g = h$ each have exactly one solution).

[L3]

A total order is a reflexive, antisymmetric, transitive relation in which any two elements are comparable, and $a < b$ means $a \leq b$ with $a \neq b$ (Partial order and partially ordered set).

[L4]

An ordered ring is a ring with a total order satisfying (OR1) $a \leq b$ implies $a + c \leq b + c$ , and (OR2) $0 < a$ and $0 < b$ imply $0 < a b$ (Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication).

Proof

technique · direct

1.1

Assume $\leq$ makes $R$ an ordered ring, and put $P = {x : 0 < x}$ . For $a, b \in R$ : $a < b$ if and only if $0 < b - a$ . Indeed, adding $- a$ to $a \leq b$ gives $a - a \leq b - a$ , that is $0 \leq b - a$ , by (OR1); adding $a$ to $0 \leq b - a$ gives $a \leq b$ likewise; and $a = b$ exactly when $b - a = 0$ . So $a < b$ exactly when $b - a \in P$ .

L1L2L4

1.2

Now let $P$ be any cone and define $\leq_{P}$ as in the statement. Reflexivity holds by the clause $a = a$ . Antisymmetry: if $a \neq b$ , $b - a \in P$ and $a - b \in P$ , then $- (b - a) = a - b \in P$ as well, contradicting trichotomy applied to $x = b - a$ ; so $a \leq_{P} b$ and $b \leq_{P} a$ force $a = b$ . Transitivity: if $a \leq_{P} b$ and $b \leq_{P} c$ and the two are not equalities, then $(c - b) + (b - a) = c - a \in P$ by closure; the cases where one of them is an equality are immediate. Comparability: given $a, b$ , trichotomy applied to $b - a$ gives $b - a \in P$ , or $b - a = 0$ and then $a = b$ , or $a - b = - (b - a) \in P$ . So $\leq_{P}$ is a total order.

L1L2L3

2.1

Trichotomy for the positive cone in claim 1. Return here to the ordered-ring order and its set $P = {x : 0 < x}$ from step 1.1. Let $x \in R$ . By totality and antisymmetry exactly one of $0 < x$ , $0 = x$ , $x < 0$ holds. By step 1.1 applied with $a = x$ , $b = 0$ , the last is equivalent to $0 - x = - x \in P$ . So exactly one of $x \in P$ , $x = 0$ , $- x \in P$ holds.

step 1.1L2L3

2.2

$\leq_{P}$ makes $R$ an ordered ring. (OR1): $(b + c) - (a + c) = b - a$ , so $a \leq_{P} b$ implies $a + c \leq_{P} b + c$ . (OR2): $0 <_{P} a$ means $a - 0 = a \in P$ , so if $0 <_{P} a$ and $0 <_{P} b$ then $a b \in P$ by closure, that is $0 <_{P} a b$ .

step 1.2L1L2L4

3.1

Closure for $P$ . Let $x, y \in P$ . Then $0 < x$ and, adding $x$ to $0 \leq y$ , $x \leq x + y$ ; with $0 < x$ and transitivity this gives $0 < x + y$ , so $x + y \in P$ . And $x y \in P$ is (OR2) verbatim. So $P$ is a cone, which with step 2.1 and step 1.1 proves claim 1.

step 1.1step 2.1L1L3L4

3.2

The cone of $\leq_{P}$ is $P$ : $0 <_{P} x$ means $x - 0 = x \in P$ and $x \neq 0$ , and $0 \notin P$ by trichotomy, so ${x : 0 <_{P} x} = P$ . With steps 1.2 and 2.2 this proves claim 2.

step 1.2step 2.2L2

4.1

Claim 3. Starting from an ordered ring $(R, \leq)$ with cone $P$ , step 1.1 says $a < b$ exactly when $b - a \in P$ , hence $a \leq b$ exactly when $b - a \in P$ or $a = b$ , which is $a \leq_{P} b$ ; so $\leq_{P}$ and $\leq$ are the same relation. Starting from a cone $P$ , step 3.2 says the cone of $\leq_{P}$ is $P$ .

step 1.1step 2.2step 3.2∎

Remarks

This is what keeps one notion of "ordered" in the library rather than two. The published The integers form a totally ordered ring presents the order on $Z$ as a relation; the published Ordered field presents the order on a field by its positive cone. Without this lemma the two would be different-looking hypotheses and every later statement would have to choose one. With it, Every ordered field is an ordered ring, and its order is the one its positive cone induces is a two-line consequence.
Trichotomy is doing the work in both directions. In claim 1 it comes from totality plus antisymmetry of the order; in claim 2 it is what supplies comparability and antisymmetry. Closure under addition, by contrast, is a strict statement in one direction and needs transitivity to recover in the other, which is why step 3.1 argues through $x \leq x + y$ rather than quoting (OR1) directly.
Nothing here uses multiplication except (OR2) and (C2), which correspond to each other verbatim. That is why the lemma holds for rings that are not commutative as well.

Depends on

Used by

Dependency tree · two levels

17 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

Ordered ring (Wikipedia) (standard reference, not scraped)
Positive cone (Wikipedia) (standard reference, not scraped)

The order presentation and the positive-cone presentation of an ordered ring determine each other: P={ x:0<x }P={x:0<x} satisfies trichotomy and closure, and a<b: ⟺ b−a∈Pa<b:⟺b−a∈P recovers the order