Manuscript II@2 · II.tex · proposition · 50d64def

Proposition: A complete finite separator excludes global membership

Coverage and the evaluation contract › Finite membership and complete separation · lines 1013–1031 · proof included or linked

II@2#prop:finite-separation

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If the catalogue is complete and

z N_{S} = 0, z m_{S} \neq = 0,

then $m \in / C$ . Moreover,

z r_{S} (d, θ) = q (d) z m_{S}

for every allowed constant contact vector.

Proof. A global identity $m = \sum_{ℓ} α_{ℓ} n_{ℓ}$ would imply $m_{S} = N_{S} α$ , contradicting $z m_{S} \neq = 0$ . Multiply the evaluated factorization by $z$ for the second claim.

Displayed formulas in this passage

unlabeled · line 1016
$z N_{S} = 0, z m_{S} \neq = 0,$
unlabeled · line 1021
$z r_{S} (d, θ) = q (d) z m_{S}$

Nearby in this section

section: Finite membership and complete separation
section: Finite membership and complete separation