LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24

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 rational-defining relation is an equivalence relation

The relation $(a, b) \sim (c, d) ⟺ a d = c b$ on pairs of integers with nonzero second component (The rationals as equivalence classes of pairs of integers) is an equivalence relation.

Given: Pairs $(a, b), (c, d), (e, f)$ of integers with $b, d, f \neq 0$ .

[L1]

[L2]

Multiplicative cancellation in $Z$ : $u d = v d$ with $d \neq 0$ implies $u = v$ (The integers have no zero divisors; multiplicative cancellation).

technique · direct

1.1L1

Reflexivity: $a b = b a$ , so $(a, b) \sim (a, b)$ .

1.2L1

Symmetry: if $a d = c b$ then $c b = a d$ , which is the defining equation for $(c, d) \sim (a, b)$ .

1.3given

Suppose $(a, b) \sim (c, d)$ and $(c, d) \sim (e, f)$ , i.e. $a d = c b$ and $c f = e d$ .

2.1step 1.3L1

Multiplying the first equation by $f$ and the second by $b$ : $a d f = c b f$ and $c f b = e d b$ .

3.1step 2.1L1

Chaining: $(a f) d = a d f = c b f = c f b = e d b = (e b) d$ .

4.1step 3.1L2

Cancelling the nonzero $d$ : $a f = e b$ , so $(a, b) \sim (e, f)$ ; the relation is transitive.

5.1step 1.1step 1.2step 4.1∎

The relation is reflexive, symmetric, and transitive, hence an equivalence relation.

Depends on