Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)
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.

A two-cell partition of a three-element set, the equivalence relation it induces listed pair by pair, and the quotient set recovered from it

Example

Write u:=∅, v:={∅} and w:={∅,{∅}}, which are pairwise distinct, and put A:={u,v}∪{w}. The collection {{u,v},{w}} has nonempty cells, covers A, and its two cells are disjoint. The equivalence relation it induces on A, namely "a and b lie in the same cell", is

∼  =  {(u,u),(u,v)}∪{(v,u),(v,v)}∪{(w,w)},

with classes [u]=[v]={u,v} and [w]={w}, so that A/∼={{u,v},{w}} is the original collection again.

Facts & Assumptions

Given: u:=∅, v:={∅}, w:={∅,{∅}}, A:={u,v}∪{w}, and ∼ as displayed.

[L1]
[L2]

symmetric: a∼b implies b∼a, for all a,b∈A (Equivalence relation, equivalence class, and the quotient set A/∼).

[L3]

transitive: a∼b and b∼c imply a∼c, for all a,b,c∈A (Equivalence relation, equivalence class, and the quotient set A/∼).

[L4]

[a]  :=  { b∈A  :  a∼b }  ⊆  A (Equivalence relation, equivalence class, and the quotient set A/∼).

[L5]

A/∼  :=  { [a]  :  a∈A } (Equivalence relation, equivalence class, and the quotient set A/∼).

[L7]

{x,y} is the set whose elements are exactly x and y, and {x}:={x,x} (The unordered pair {x,y} and the singleton {x}={x,x}).

[L8]

There is exactly one set with no elements, written ∅ (There is exactly one set with no elements, written ∅).

[L10]

z∈A×B holds if and only if z=(a,b) for some a∈A and some b∈B (The Cartesian product A×B:={ z∈P(P(A∪B)):∃a∈A ∃b∈B z=(a,b) }).

[L11]

(a,b)=(c,d) if and only if a=c and b=d ((a,b)=(c,d) if and only if a=c and b=d).

[L12]
[L13]

If every z satisfies z∈x if and only if z∈y, then x=y (The Axiom of Extensionality: ∀x ∀y (∀z (z∈x↔z∈y)→x=y)).

Verification

technique · direct
1.1L7L8L9L10L12L14L15

u, v and w are pairwise distinct: u has no element, v has exactly u, and w has v as an element while v does not. So A has exactly those three elements, and ∼ is a relation on A, each of its five listed pairs having both coordinates in A.

2.1L1L2L3L11L13step 1.1

∼ is an equivalence relation on A: it contains (u,u), (v,v) and (w,w), so it is reflexive on A; its pairs with distinct coordinates are (u,v) and (v,u), and each of those has its reverse present, so it is symmetric; and every composable pair of its members has its composite present, since chains through u and v stay inside {u,v} and w relates only to itself.

3.1L4L5L7L11L13step 1.1step 2.1

The classes are as stated: [u]={b∈A:u∼b}={u,v}, [v]={u,v} by the pairs (v,u) and (v,v), and [w]={w}. Hence A/∼ has exactly the two elements {u,v} and {w}, which is the collection we started from.

4.1L6step 2.1step 3.1∎

The collection is therefore a partition of A in the sense of the classes being nonempty, covering A and pairwise equal or disjoint, and it is recovered as the quotient set of the equivalence relation it induces.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

24 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