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 an indexed family (Ai)iI(A_i)_{i \in I} the collection of functions ff with domain II and f(i)Aif(i) \in A_i for every iIi \in I is a set

Statement

Let (Ai)iI(A_i)_{i \in I} be an indexed family. Then there is a set whose elements are exactly the functions ff with domf=I\operatorname{dom} f = I and f(i)Aif(i) \in A_i for every iIi \in I, and it is a subset of CIC^{I} where C:=iIAiC := \bigcup_{i \in I} A_i.

Facts & Assumptions

Given: an indexed family (Ai)iI(A_i)_{i \in I}, and C:=iIAiC := \bigcup_{i \in I} A_i.

[L1]

An indexed family with index set II is a function AA with domA=I\operatorname{dom} A = I (An indexed family (Ai)iI(A_i)_{i \in I} is a function with domain II; {Ai:iI}\{A_i : i \in I\} is its range).

[L3]

fBAf \in B^{A} holds if and only if f:ABf : A \to B (The set BAB^{A} of all functions ABA \to B).

[L4]

We write f:ABf : A \to B, and say ff is a function from AA to BB, when ff is a function with domf=A\operatorname{dom} f = A and ranfB\operatorname{ran} f \subseteq B (A function is a relation ff with (a,b)f(a,b) \in f and (a,c)f(a,c) \in f implying b=cb = c; f:ABf : A \to B, the value f(a)f(a), domain and codomain).

[L7]

ranR:={b:a (a,b)R}\operatorname{ran} R := \{\, b : \exists a\ (a,b) \in R \,\} (Relation, domR\operatorname{dom} R, ranR\operatorname{ran} R, fldR\operatorname{fld} R, and the specialisations "relation from AA to BB" and "relation on AA").

Proof

technique · direct
1.1

Let ff be a function with domf=I\operatorname{dom} f = I and f(i)Aif(i) \in A_i for every iIi \in I. Every element of ranf\operatorname{ran} f is f(i)f(i) for some iIi \in I, hence lies in AiA_i and therefore in CC; so ranfC\operatorname{ran} f \subseteq C and f:ICf : I \to C, that is, fCIf \in C^{I}.

L1L2L3L4L6L7
2.1

Separating inside CIC^{I} with the formula saying that z(i)Aiz(i) \in A_i for every iIi \in I, with parameters II and the family, gives a set whose elements are exactly the members of CIC^{I} with that property; by step 1.1 every function of the kind described already lies in CIC^{I}, so this set has exactly the intended elements and is included in CIC^{I}.

L3L5L6step 1.1

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 25 results over 10 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