Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)
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.

Conditional entropy of finite memory given side information

Statement

Let M and R take values in finite sets under a joint probability law p(m,r)≥0 with ∑m,rp(m,r)=1. For any finite random variable Y with law pY, define its Shannon entropy by H(Y):=−∑y:pY(y)>0pY(y)log⁡pY(y); for a pair, H(M,R) means this definition applied to its joint law. Put ar:=∑mp(m,r) and, when ar>0, q(m∣r):=p(m,r)/ar. Thus H(R)=−∑r:ar>0arlog⁡ar. Define H(M∣R):=∑r:ar>0ar[−∑m:q(m∣r)>0q(m∣r)log⁡q(m∣r)]. Then H(M,R)=H(R)+H(M∣R) and H(M∣R)≥0.

Facts & Assumptions

Given: As stated in the item.

[A1]

M,R take values in finite sets under the joint probability law p(m,r).

[F1]

Finite probability spaces, outcome weights, events, and event probabilities and Finite sums and finite products, by recursion: joint weights are nonnegative and sum to one, and all displayed sums are finite.

[F2]

Proof

Given: A finite joint probability law for M and R.

1.1A1F1F2

For every r with ar>0, the numbers q(m∣r) are nonnegative and sum to one. The joint identity p(m,r)=arq(m∣r) on positive entries gives log⁡p(m,r)=log⁡ar+log⁡q(m∣r); summing −plog⁡p over each fixed r yields −arlog⁡ar+ar[−∑m:q(m∣r)>0q(m∣r)log⁡q(m∣r)]. Terms with ar=0 have all p(m,r)=0 and contribute nothing. Summing over finite r proves the chain rule.

2.1A1F1F2step 1.1∎

For each positive conditional probability q(m∣r)≤1, [F2] gives −q(m∣r)log⁡q(m∣r)≥0. Multiplying by ar≥0 and summing proves H(M∣R)≥0.

Depends on

Dependency tree · 2 levels

Sources