Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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.

The elementary recurrences for the real Beta function

Statement

For p,q>0,

B(p+1,q)=pp+qB(p,q),B(p,q+1)=qp+qB(p,q),

and consequently B(p,q)=B(p+1,q)+B(p,q+1).

Facts & Assumptions

Given: Positive real parameters p,q.

[F1]

For p,q>0, B(p,q)=Γ(p)Γ(q)/Γ(p+q) (The real Beta--Gamma identity).

[F2]

For every s>0, Γ(s+1)=sΓ(s) (The real Gamma functional equation Γ(s+1)=sΓ(s)).

Proof

technique · direct
1.1

By [F1] and [F2], B(p+1,q)=Γ(p+1)Γ(q)/Γ(p+q+1)=pB(p,q)/(p+q).

F1F2algebra
1.2

Similarly, B(p,q+1)=qB(p,q)/(p+q).

F1F2algebra
2.1

Adding steps 1.1 and 1.2 and using p+q>0 gives B(p+1,q)+B(p,q+1)=B(p,q).

step 1.1step 1.2algebra

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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