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Suppose that C is a regular terminal class, with lower and upper bounds Then A i cannot have an influential player j ? N \ S * whenever i ? S * ,
This is always possible since necessarily one of the S k 's does not contain j (for if j ? S k for all k = 1, . . . , p, then j ? S * ) This implies 1 > A i (1 N \j ) ? A i, S ? ), which contradicts (*) ,
Again this is always possible since j ? S * implies j ? S k ? K k for some k. Then A i (1 S ? ) ? A i (1 j ) > 0 ,
one can obtain a sufficient condition (B) for forbidding any normal regular terminal class not reduced to the trivial terminal classes. Condition (B): for any, By applying the above lemma to every pair ,
which is in fact Lemma 5: j influential in A i rules out every regular class with lower and upper bounds, Proof of Theorem 3 We consider the following rule R ??arc going into S * ) or i ? S * , j ? S * (arc outgoing from S * ) ,
with 1 ? |S * | < n ? 1 and 2 ? |S * | < n. If S * has an ingoing arc, it is ruled out by R ?? . If it has no ingoing arc, then by hypothesis every node outside S * is linked to S * by a path. Since N \ S * is always nonempty, taking any node i in N \ S * , there is a path from S * to i ,
Then A i (1 S ) ? A i (1 N \S ) < 1, hence S is not a terminal class by Theorem 1 (ii) Now, suppose that S Then 0 < A i (1 S ? ) ? A i (1 S ), which implies that S is not a terminal class. Necessity: Suppose that S is not a terminal state. Then by Theorem 1 (ii), either there is some i ? S such that A i (1 S ) < 1 or some i ? S such that A i (1 S ) > 0. Suppose that for some ,
exhaustivity of A i that A i (1 S ? ) = 1, a contradiction with the definition of S ? . Therefore, A i (1 N \S ? ) > 0. Finally, we have for all S ?? ) = 1, and the claim is proved. Suppose now that for some i ? S, A i (1 S ) > 0. Take the smallest S ? ? S such that A i (1 S ? ) > 0. Then S ? is influential for i. Indeed, A i (1 S ? ) > 0, and by exhaustivity of A i we must have A i ,
Then A i (1 S ) ? A i (1 N \S ? ) < 1, hence [S, S ? K] cannot be a Boolean terminal class, for any K. Now, suppose that ,