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LemmaStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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.

Integer multiplication is well defined

Statement

The operation [(a,b)]⋅[(c,d)]=[(ac+bd, ad+bc)][(a,b)] \cdot [(c,d)] = [(ac+bd,\, ad+bc)] on Z\mathbb{Z} (Arithmetic on the integers) is independent of the chosen representatives.

Facts & Assumptions

Given: Pairs with (a,b)∼(a′,b′)(a,b) \sim (a',b') and (c,d)∼(c′,d′)(c,d) \sim (c',d') in the sense of The integers as equivalence classes of pairs of naturals.

[A1]

Addition on N\mathbb{N} is commutative and associative.

[A2]

Multiplication on N\mathbb{N} is commutative.

[A3]

Distributivity in N\mathbb{N}: x(y+z)=xy+xzx(y+z) = xy + xz.

Proof

technique · direct
1.1

By hypothesis a+b′=b+a′a + b' = b + a'; write ss for this common value.

given
1.2

Regrouping and factoring: (ac+bd)+(a′d+b′c)=c(a+b′)+d(b+a′)(ac+bd) + (a'd+b'c) = c(a+b') + d(b+a').

A1A2A3
1.3

Regrouping and factoring: (ad+bc)+(a′c+b′d)=d(a+b′)+c(b+a′)(ad+bc) + (a'c+b'd) = d(a+b') + c(b+a').

A1A2A3
2.1

By step 1.1 both right-hand sides equal cs+dscs + ds, so (ac+bd)+(a′d+b′c)=(ad+bc)+(a′c+b′d)(ac+bd) + (a'd+b'c) = (ad+bc) + (a'c+b'd).

step 1.1step 1.2step 1.3
3.1

That equation is precisely the defining relation (ac+bd, ad+bc)∼(a′c+b′d, a′d+b′c)(ac+bd,\, ad+bc) \sim (a'c+b'd,\, a'd+b'c): the product class is unchanged when the first factor's representative changes.

step 2.1
4.1

The product formula is symmetric in its two arguments: swapping (a,b)↔(c,d)(a,b) \leftrightarrow (c,d) sends (ac+bd, ad+bc)(ac+bd,\, ad+bc) to (ca+db, cb+da)(ca+db,\, cb+da), the same pair. Hence, by the argument of steps 1.1–3.1 applied to the second factor, the product class is also unchanged when (c,d)(c,d) is replaced by (c′,d′)(c',d').

step 3.1A1A2
5.1

Replacing first (a,b)(a,b) by (a′,b′)(a',b') and then (c,d)(c,d) by (c′,d′)(c',d'): [(a,b)]⋅[(c,d)]=[(a′,b′)]⋅[(c,d)]=[(a′,b′)]⋅[(c′,d′)][(a,b)] \cdot [(c,d)] = [(a',b')] \cdot [(c,d)] = [(a',b')] \cdot [(c',d')]; multiplication is well defined.

step 3.1step 4.1∎

Depends on

Used by

Cited to discharge well-definedness by Arithmetic on the integers.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 21 results over 11 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