Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (openai/gpt-5.4)audited 2026-07-24
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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 integers have no zero divisors; multiplicative cancellation

Statement

If x,yZx, y \in \mathbb{Z} are nonzero then xy0xy \ne 0. Consequently, if xz=yzxz = yz and z0z \ne 0, then x=yx = y.

Facts & Assumptions

Given: Nonzero integers x,yx, y.

[L1]

Z\mathbb{Z} is a totally ordered ring; in particular trichotomy holds (each integer is negative, zero, or positive) and positives are closed under multiplication (The integers form a totally ordered ring).

[L2]

Ring arithmetic in Z\mathbb{Z}: (u)v=(uv)(-u)v = -(uv), (u)=u-(-u) = u, and distributivity (The integers form a commutative ring).

Proof

technique · cases
1.1

By trichotomy, each of xx and yy is positive or negative; split on the sign pattern.

L1cases-exhaustive
1.2

Case both positive: 0<xy0 < xy by closure of positives, so xy0xy \ne 0.

assume-case bothL1
1.3

Case exactly one negative, say y<0y < 0 (the other case is identical by commutativity): 0<y0 < -y, so 0<x(y)=(xy)0 < x(-y) = -(xy), hence xy<0xy < 0 and xy0xy \ne 0.

assume-case mixedL1L2
1.4

Case both negative: 0<x0 < -x and 0<y0 < -y, so 0<(x)(y)=xy0 < (-x)(-y) = xy, and xy0xy \ne 0.

assume-case negnegL1L2
2.1

In every case xy0xy \ne 0: the product of nonzero integers is nonzero.

step 1.2step 1.3step 1.4cases
3.1

Cancellation: if xz=yzxz = yz with z0z \ne 0, then (xy)z=xzyz=0(x-y)z = xz - yz = 0; were xy0x - y \ne 0, the product would be nonzero, so xy=0x - y = 0, i.e. x=yx = y.

step 2.1L2

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 31 results over 12 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