Furthermore, A only bribes legislators who already support x, although some must be bribed in order to raise their level of support high enough to prevent B from trying to buy their votes.
Payments above v(0) would make some bribed members of A's coalition more expensive for B to buy than legislator 0.
A always adopts a leveling strategy because A's bribed coalition must contain more than v(0) members.
To prevent B from invading, A must make offers such that each bribed legislator costs [W.
1](0)] represent the bribed friends of A, that is, legislators who receive payments from A even though they initially support x.
This follows from the fact that if any bribed member of A's coalition costs more than b, then A could decrease the bribe slightly and keep B's cost of invading the same, while strictly decreasing his own costs.
Next, we show that all members of A's bribed coalition (actually, all members with the possible exception of a set of measure zero) cost at least b dollars for B to buy.
Thus, the revised strategy for A is superior, and it follows that for each member of A's bribed coalition v(z) + [a.
It is easily shown that A will not make any member of his bribed coalition cost more than b.
No bribed member of coalition can cost more than b.
Because of this, one can show, as in (3-a), that there exists a way for A to increase the bribe to a positive-measure set of legislators who cost less than b, while decreasing the bribe to all other bribed legislators.
Because A adopts a leveling strategy, each bribed legislator costs b for B to buy.