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
1.1

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.

L7L8L9L10L12L14L15
2.1

∼ 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.

L1L2L3L11L13step 1.1
3.1

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.

L4L5L7L11L13step 1.1step 2.1
4.1

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.

L6step 2.1step 3.1∎

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