Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: 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.

∅ is a relation on every set, is the unique equivalence relation on ∅, is a function ∅→B for every B, is a bijection ∅→∅, and is not a surjection ∅→{∅}

Example

The empty set does the work of five different objects at once.

  • ∅ is a relation, and a relation on A for every set A; its domain, range and field are all ∅.
  • ∅ is the only relation on ∅, and it is an equivalence relation on ∅; so ∅ carries exactly one equivalence relation.
  • ∅ is a function ∅→B for every set B, and it is the only one.
  • ∅ is a bijection ∅→∅.
  • ∅ is not a surjection ∅→{∅}, even though it is an injective function ∅→{∅}.

The last two together are the reason a codomain belongs to the declaration f:A→B rather than to the set f: one and the same set is a bijection under one declaration and a non-surjection under another. The empty function is also the unique element of the empty product.

Facts & Assumptions

Given: the set ∅ and arbitrary sets A and B.

[L2]

dom⁡R:={ a:∃b (a,b)∈R },ran⁡R:={ b:∃a (a,b)∈R } (Relation, dom⁡R, ran⁡R, fld⁡R, and the specialisations "relation from A to B" and "relation on A").

[L3]
[L4]

We write f:A→B, and say f is a function from A to B, when f is a function with dom⁡f=A and ran⁡f⊆B (A function is a relation f with (a,b)∈f and (a,c)∈f implying b=c; f:A→B, the value f(a), domain and codomain).

[L5]

f is injective (one-to-one) if f(x)=f(y) implies x=y, for all x,y∈A (Injection, surjection, bijection).

[L6]

f is surjective (onto) if for every b∈B there is some x∈A with f(x)=b (Injection, surjection, bijection).

[L7]

A binary relation ∼ on A is an equivalence relation when it is reflexive, symmetric and transitive (Equivalence relation, equivalence class, and the quotient set A/∼).

[L8]

R is reflexive on A when (a,a)∈R for every a∈A (Reflexive, irreflexive, symmetric, asymmetric, antisymmetric, transitive, and connex relations on a set).

[L9]

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

[L10]

{x}:={x,x}, the singleton of x, is the set whose only element is x (The unordered pair {x,y} and the singleton {x}={x,x}).

[L12]
[L13]

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) }).

[L15]

an element of ∏i∈IAi is a function with domain I that takes its value at each index inside the member carried by that index (The product ∏i∈IAi:={ f:I→⋃i∈IAi ∣ f(i)∈Ai for every i∈I }).

Verification

technique · direct
1.1

∅ has no elements, so "every element is an ordered pair" holds vacuously and ∅ is a relation; ∅⊆A×A for every A, so it is a relation on every set; and no set satisfies the defining conditions for its domain or its range, so both are ∅, hence so is its field.

L1L2L9L12
1.2

A relation on ∅ is a subset of ∅×∅, which is ∅, so ∅ is the only one. It is reflexive on ∅, symmetric and transitive, since each condition quantifies over elements of ∅; hence it is the unique equivalence relation on ∅.

L7L8L9L11L12L13
2.1

∅ is single valued vacuously, has domain ∅ and range ∅⊆B, so ∅:∅→B for every B; and any function with domain ∅ has no elements, so it is ∅.

L3L4L9L12step 1.1
3.1

As a function ∅→∅ it is injective, since the injectivity condition quantifies over elements of the domain, and surjective, since the surjectivity condition quantifies over elements of the codomain and ∅ has none; so it is a bijection.

L5L6L9step 2.1
3.2

As a function ∅→{∅} it is still injective, for the same reason, but not surjective: ∅ is an element of {∅} and no element of the domain is sent to it.

L5L6L9L10step 2.1
3.3

The empty function is the unique element of the empty product: ∏i∈∅Ai={∅}, and its one element is a function with domain ∅.

L14L15step 2.1
4.1

All five descriptions hold of the single set ∅, and the last two differ only in the declared codomain.

step 1.1step 1.2step 2.1step 3.1step 3.2step 3.3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

31 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