LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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.

In any ring $0 \cdot a = a \cdot 0 = 0$ , $(- a) b = a (- b) = - (a b)$ , $(- a) (- b) = a b$ , $(- 1) a = - a$ and $a (b - c) = a b - a c$

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) and let $a, b, c \in R$ . Then

$0 \cdot a = 0$ and $a \cdot 0 = 0$ ;
$(- a) b = - (a b)$ and $a (- b) = - (a b)$ ;
$(- a) (- b) = a b$ ;
$(- 1) a = - a$ and $a (- 1) = - a$ ;
$a (b - c) = a b - a c$ and $(b - c) a = b a - c a$ .

In particular, if $1 = 0$ in $R$ then $R = {0}$ .

No commutativity is assumed, which is why each claim is stated on both sides.

Facts & Assumptions

Given: A ring $R$ with zero $0$ , identity $1$ , addition $+$ and multiplication $\cdot$ , and elements $a, b, c \in R$ ; $x - y$ 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, $0 + x = x = x + 0$ , and each $x$ has an additive inverse $- x$ with $x + (- x) = 0 = (- x) + x$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Group and abelian group, Left identity, right identity, and two-sided identity for a binary operation).

[L2]

Both distributive laws hold: $x (y + z) = x y + x z$ and $(y + z) x = y x + z x$ for all $x, y, z \in R$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L3]

$1 \cdot x = x = x \cdot 1$ for every $x \in R$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L4]

Cancellation in the additive group: $x + z = y + z$ implies $x = y$ , and $z + x = z + y$ implies $x = y$ (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, Group and abelian group).

[L5]

In a group, $(g^{- 1})^{- 1} = g$ ; written additively, $- (- x) = x$ (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, Group and abelian group).

Proof

technique · direct

1.1L1L2

$0 \cdot a = (0 + 0) \cdot a = 0 \cdot a + 0 \cdot a$ , using $0 + 0 = 0$ and the right distributive law.

1.2L1L2

$a \cdot 0 = a \cdot (0 + 0) = a \cdot 0 + a \cdot 0$ , using $0 + 0 = 0$ and the left distributive law.

1.3L1L2

$(- a) b + a b = ((- a) + a) b = 0 \cdot b$ , by the right distributive law and $(- a) + a = 0$ .

1.4L1L2

$a (- b) + a b = a ((- b) + b) = a \cdot 0$ , by the left distributive law and $(- b) + b = 0$ .

2.1step 1.1L1L4

$0 \cdot a = 0$ : step 1.1 gives $0 + 0 \cdot a = 0 \cdot a = 0 \cdot a + 0 \cdot a$ , and cancelling $0 \cdot a$ on the right gives $0 = 0 \cdot a$ .

2.2step 1.2L1L4

$a \cdot 0 = 0$ : step 1.2 gives $0 + a \cdot 0 = a \cdot 0 = a \cdot 0 + a \cdot 0$ , and cancelling $a \cdot 0$ on the right gives $0 = a \cdot 0$ . This proves claim 1.

3.1step 1.3step 2.1L1L4

$(- a) b = - (a b)$ : by step 1.3 and step 2.1, $(- a) b + a b = 0 \cdot b = 0 = - (a b) + a b$ , and cancelling $a b$ on the right gives $(- a) b = - (a b)$ .

3.2step 1.4step 2.2L1L4

$a (- b) = - (a b)$ : by step 1.4 and step 2.2, $a (- b) + a b = a \cdot 0 = 0 = - (a b) + a b$ , and cancelling $a b$ on the right gives $a (- b) = - (a b)$ . This proves claim 2.

4.1step 3.1step 3.2L5

$(- a) (- b) = - (a (- b)) = - (- (a b)) = a b$ : the first equality is step 3.1 applied with $b$ replaced by $- b$ , the second is step 3.2, and the third is $- (- x) = x$ . This proves claim 3.

4.2step 3.1step 3.2L3

$(- 1) a = - (1 \cdot a) = - a$ and $a (- 1) = - (a \cdot 1) = - a$ , by step 3.1 and step 3.2 applied with $a$ or $b$ equal to $1$ , together with the identity law. This proves claim 4.

4.3step 3.1step 3.2L2

$a (b - c) = a (b + (- c)) = a b + a (- c) = a b + (- (a c)) = a b - a c$ , and $(b - c) a = (b + (- c)) a = b a + (- c) a = b a + (- (c a)) = b a - c a$ , using the two distributive laws and step 3.1 and step 3.2. This proves claim 5.

5.1step 2.1L3∎

If $1 = 0$ then for every $x \in R$ we have $x = 1 \cdot x = 0 \cdot x = 0$ by the identity law and step 2.1, so $R = {0}$ .

Remarks

These are the general statements; the field versions already in the library are instances of them. For a field $F$ , claim 1 is Multiplication by zero: $0 \cdot a = 0$ and claims 2 and 3 are Sign rules for products: $(- a) b = - (a b)$ and $(- a) (- b) = a b$ , both proved earlier from the field axioms alone. Once Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring identifies a field as a commutative ring, those two lemmas are exactly the present ones read in $F$ . The direction of generality matters for citation: a proof about an arbitrary ring must use this lemma, never the field lemmas, since a ring need not be a field.
Claim 1 is where the additive group is doing the work. The identity $0 \cdot a = 0$ is not an axiom and does not follow from the multiplicative structure; it follows from distributivity together with the fact that addition cancels, which is available because $(R, +, 0)$ is a group and not merely a monoid.
The last sentence of the statement is why the zero ring exists at all. A ring with $1 = 0$ collapses to one element, so requiring $1 \neq 0$ , as Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors and Division ring: a ring with $1 \neq 0$ in which every nonzero element is a unit do, is exactly the requirement that the ring is not that one-element ring.

In any ring $0 \cdot a = a \cdot 0 = 0$ , $(- a) b = a (- b) = - (a b)$ , $(- a) (- b) = a b$ , $(- 1) a = - a$ and $a (b - c) = a b - a c$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

In any ring 0⋅a=a⋅0=00⋅a=a⋅0=0, (−a)b=a(−b)=−(ab)(−a)b=a(−b)=−(ab), (−a)(−b)=ab(−a)(−b)=ab, (−1)a=−a(−1)a=−a and a(b−c)=ab−aca(b−c)=ab−ac

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

In any ring $0 \cdot a = a \cdot 0 = 0$ , $(- a) b = a (- b) = - (a b)$ , $(- a) (- b) = a b$ , $(- 1) a = - a$ and $a (b - c) = a b - a c$