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, Non-triviality holds because for p 2 S the set W p contains values other than 1, and because S 6 = ? (as X 6 ` 6 = ? by violation of negation-connectedness). Finally, let Cr be any credence function, and let us show that f (Cr) is consistent and deductively closed

. Consistency, For a contradiction, let f (Cr) be inconsistent. Then f (Cr) has a minimal inconsistent subset Y. By Lemma 16, Y contains at most one proposition from S. So there is a q 2 Y such that Y \{q} ? X\S

Y. \{q} and C. , 1 as p 2 f (Cr) and W p = {1}. So, as Y \{q} entails q, Cr(q) = 1, i.e., Cr(q) = 0. Hence q 6 2 f (Cr) as 0 6 2 W q

, Assume f (Cr) entails q 2 X. We must show that q 2 f (Cr)

, Let Z be a minimal subset of f (Cr) that entails q. Note that Z

?. Case and ;. , Then for all p 2 Z, Cr(p) = 1, as p 2 f (Cr) and W p = {1}. So Cr(q) = 1, as Z entails q. Hence q 2 f (Cr), vol.1

, By Lemma 16 and the minimal inconsistency of Z [ {q}, we can infer three things. First, Z \ S is singleton, say Z \ S = {z}. Second, q 6 2 S. Third, (Z [ {q}) \ X ` = ?, i.e., Z [ {q} ? X 6 `. As q 2 X 6 ` \S, we have q 2 S. For all p 2 Z\{z}, Cr(p) = 1, as p 2 f (Cr) and W p = {1}. So, as Z entails q, Cr(q) Cr(z). Meanwhile, Cr(z) > 0

, Cr(q) > 0. Hence, q 2 f (Cr)