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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24
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  • 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.
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The integer-defining relation is an equivalence relation

Statement

The relation (a,b)∼(c,d)  ⟺  a+d=b+c on N×N (The integers as equivalence classes of pairs of naturals) is an equivalence relation.

Facts & Assumptions

Given: The relation (a,b)∼(c,d)  ⟺  a+d=b+c on N×N.

[A1]

Addition on N is commutative and associative.

[A2]

Cancellation in N: if x+z=y+z then x=y.

Proof

technique · direct
1.1

Reflexivity: for any (a,b) we have a+b=b+a, so (a,b)∼(a,b).

A1
1.2

Symmetry: suppose (a,b)∼(c,d), i.e. a+d=b+c. Then c+b=d+a, which is the defining equation for (c,d)∼(a,b).

A1
1.3

Suppose (a,b)∼(c,d) and (c,d)∼(e,f), i.e. a+d=b+c and c+f=d+e.

given
2.1

Adding the two equations: (a+d)+(c+f)=(b+c)+(d+e).

step 1.3algebra
3.1

Regrouping both sides: (a+f)+(c+d)=(b+e)+(c+d).

step 2.1A1
4.1

Cancelling (c+d): a+f=b+e, so (a,b)∼(e,f); the relation is transitive.

step 3.1A2
5.1

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

step 1.1step 1.2step 4.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Cited to discharge well-definedness by The integers as equivalence classes of pairs of naturals.

Dependency tree · two levels

10 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