Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedprecheck 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.

If gcd⁡(a,b)=1 and a∣bc then a∣c; and if a∣c, b∣c and gcd⁡(a,b)=1 then ab∣c

Statement

Let a,b,c∈Z.

  1. If gcd⁡(a,b)=1 and a∣bc, then a∣c.
  2. If a∣c, b∣c and gcd⁡(a,b)=1, then ab∣c.

Facts & Assumptions

Given: Integers a,b,c.

[L3]

Divisibility is reflexive and transitive; if d∣u and d∣v then d∣ux+vy for all x,y; and d∣u implies d∣uw (Divisibility is reflexive and transitive on Z, and is linear: if d∣a and d∣b then d∣ax+by for all integers x,y; also d∣a implies d∣ac, −d∣a and d∣−a).

[L4]

d∣u means u=dq for some q∈Z (Divisibility in Z: d∣a when a=dq for some integer q).

[L5]

Z is a commutative ring: addition and multiplication are associative and commutative, x⋅1=x, and multiplication distributes over addition (The integers form a commutative ring, Arithmetic on the integers).

Proof

technique · direct
1.1

Claim 1. Assume gcd⁡(a,b)=1 and a∣bc, and fix x,y with ax+by=1.

L1choose
1.2

Claim 2. Assume a∣c, b∣c and gcd⁡(a,b)=1. Write c=as with s∈Z.

L4choose
2.1

Multiplying ax+by=1 by c gives a(xc)+(bc)y=c, using associativity, commutativity and distributivity.

step 1.1L5
3.1

Now a∣a, hence a∣a(xc) by [L3]; and a∣bc, hence a∣(bc)y by [L3]. Applying linearity to these two with coefficients 1 and 1 gives a∣a(xc)+(bc)y=c.

step 1.1step 2.1L3L5
4.1

Then b∣c=as, and gcd⁡(b,a)=gcd⁡(a,b)=1 by [L2], so claim 1 applied with b, a, s in place of a, b, c gives b∣s.

step 3.1step 1.2L2
5.1

Write s=bt; then c=as=a(bt)=(ab)t, so ab∣c.

step 1.2step 4.1L4L5∎

Remarks

Depends on

Used by

Dependency tree · two levels

30 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