LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck pass

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.

Private-message conditioning lowers output entropy

Statement

Under the hypotheses and decomposition of Conditioning on a private message separates average acceptance, the weights and posteriors can be chosen so that Σ_jq_j H(D_jA)≤H(DA)−1/2.

Facts & Assumptions

Given: Matrices and distribution as in Conditioning on a private message separates average acceptance, with the resulting $q_{j}, D^{(j)}, X_{j}$ .

[F1]

Conditioning on a private message separates average acceptance: For rows $X_{j}$ equal to the transposes of the columns of $M$ , $\sum_{j} q_{j} (D A) X_{j}^{T} \leq 0.02$ and $\sum_{j} q_{j} (D^{(j)} A) X_{j}^{T} \geq 0.99$ , with $D = \sum_{j} q_{j} D^{(j)}$ .

[F2]

A separating event forces relative entropy: If $P X^{T} \leq 0.06$ and $Q X^{T} \geq 0.94$ , then $D (Q ∥ P) \geq 1$ .

Proof

technique · direct

1.1givenF1algebra

Put $P = D A$ and $Q_{j} = D^{(j)} A$ . These are probability vectors and $P = \sum_{j} q_{j} Q_{j}$ . The weighted acceptance gap in [F1] is at least $0.97$ . Since every difference $β_{j} = (Q_{j} - P) X_{j}^{T}$ lies in $[- 1, 1]$ , the set $U = {j : β_{j} \geq 0.94}$ has weight at least $1 / 2$ : otherwise the weighted average is less than $0.94 + 0.06 \cdot (1 / 2) = 0.97$ . For $j \in U$ , $Q_{j} X_{j}^{T} \geq 0.94$ and $P X_{j}^{T} = Q_{j} X_{j}^{T} - β_{j} \leq 0.06$ .

2.1step 1.1F1F2algebra∎

By [F2], $D (Q_{j} ∥ P) \geq 1$ for each $j \in U$ . Direct expansion of the finite sums, using $P_{i} = \sum_{j} q_{j} (Q_{j})_{i}$ , gives $H (P) - \sum_{j} q_{j} H (Q_{j}) = \sum_{j} q_{j} D (Q_{j} ∥ P) .$ If $P_{i} = 0$ , every positive-weight $(Q_{j})_{i}$ is zero, so no undefined logarithm enters. The right side is at least the weight of $U$ , hence at least $1 / 2$ . Substituting $P = D A$ and $Q_{j} = D^{(j)} A$ proves the entropy drop.

Depends on

Used by

Packing bound for separated randomized message rows Lemma

Dependency tree · two levels

5 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

Ilan Newman and Mario Szegedy, Public vs. Private Coin Flips in One Round Communication Games (standard reference, not scraped)