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.

P(∅)={∅}, P({∅})={∅,{∅}}, and P({∅,{∅}}) listed in full

Example

Write A:={∅,{∅}}, and abbreviate {x,y,z,w}:={x,y}∪{z,w}. Then

P(∅)={∅},P({∅})={∅,{∅}},P(A)={∅,{∅},{{∅}},A},

and the four sets listed in P(A) are pairwise distinct.

Facts & Assumptions

Given: A:={∅,{∅}} and the abbreviation {x,y,z,w}:={x,y}∪{z,w}.

[L2]

z∈P(x) holds if and only if z⊆x (The power set P(x)={ z:z⊆x }).

[L4]

{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}).

[L5]

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)).

[L6]

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

Verification

technique · direct
1.1

P(∅)={∅} is the general identity applied at ∅.

L1
1.2

A set z satisfies z⊆{∅} exactly when every element of z equals ∅, that is, exactly when z has no element or ∅ is its only element; those two sets are ∅ and {∅}, and both are indeed included in {∅}. Hence P({∅})={∅,{∅}}.

L2L3L4L5L6
1.3

A set z satisfies z⊆A exactly when every element of z is ∅ or {∅}, so z is determined by whether it has ∅ as an element and whether it has {∅} as an element; the four resulting sets are ∅, {∅}, {{∅}} and A, and each is included in A. Hence P(A)={∅,{∅},{{∅}},A}.

L2L3L4L5L6L7L8
1.4

The four are pairwise distinct: ∅ has no element, {∅} has ∅ as its only element, {{∅}} has {∅} as its only element, and A has both; moreover ∅≠{∅}, since the second has an element and the first does not.

L4L5L6
2.1

The three power sets are computed and the four members of P(A) are distinct, which is the example.

step 1.1step 1.2step 1.3step 1.4∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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