Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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.

For every set xx there is exactly one set whose elements are precisely the subsets of xx

Statement

For every set xx there is exactly one set pp such that, for every zz, z∈pz \in p holds if and only if z⊆xz \subseteq x.

Facts & Assumptions

Given: a set xx.

[L1]

For every set xx there is a set yy that contains every zz all of whose elements belong to xx (The Axiom of Power Set: ∀x ∃y ∀z (∀t (t∈z→t∈x)→z∈y)\forall x\,\exists y\,\forall z\,(\forall t\,(t \in z \to t \in x) \to z \in y)).

[L2]

For any parameters pˉ\bar p and any set xx, there is a set yy whose elements are exactly the elements zz of xx for which φ(z,pˉ)\varphi(z,\bar p) holds (The Axiom Schema of Separation: for each formula φ\varphi, ∀pˉ ∀x ∃y ∀z (z∈y↔(z∈x∧φ(z,pˉ)))\forall \bar p\,\forall x\,\exists y\,\forall z\,(z \in y \leftrightarrow (z \in x \wedge \varphi(z,\bar p)))).

[L3]

x⊆yx \subseteq y means that every element of xx is an element of yy (Subset x⊆yx \subseteq y, proper subset x⊊yx \subsetneq y, and the separation notation { z∈x:φ(z) }\{\, z \in x : \varphi(z) \,\}).

[L4]

If every zz satisfies z∈xz \in x if and only if z∈yz \in y, then x=yx = y (The Axiom of Extensionality: ∀x ∀y (∀z (z∈x↔z∈y)→x=y)\forall x\,\forall y\,(\forall z\,(z \in x \leftrightarrow z \in y) \to x = y)).

Proof

technique · direct
1.1

Power Set supplies a set yy that contains every zz all of whose elements belong to xx; by the meaning of inclusion, yy contains every zz with z⊆xz \subseteq x.

L1L3
2.1

Apply Separation to yy with the formula φ(z,x):=∀t (t∈z→t∈x)\varphi(z,x) := \forall t\,(t \in z \to t \in x) and the parameter xx: the set p:={ z∈y:z⊆x }p := \{\, z \in y : z \subseteq x \,\} exists, and for every zz, z∈pz \in p holds if and only if z∈yz \in y and z⊆xz \subseteq x.

L2L3step 1.1
3.1

If z⊆xz \subseteq x then z∈yz \in y by step 1.1, so the condition "z∈yz \in y and z⊆xz \subseteq x" reduces to z⊆xz \subseteq x; hence z∈pz \in p holds if and only if z⊆xz \subseteq x, which proves existence.

step 1.1step 2.1
4.1

If p′p' also satisfies "z∈p′z \in p' if and only if z⊆xz \subseteq x" for every zz, then z∈pz \in p holds if and only if z∈p′z \in p' for every zz, so p=p′p = p'; existence and uniqueness together give the statement.

L4step 3.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 5 results over 3 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources