If p is nonterminal, suppose first that it is a maximizing node. Now let s ∈ S be some particular situation. If the maximizer can win in s, then there is some successor ( p ′ , S ′ ) to ( p, S ) where the maximizer wins, and hence by the inductive hypothesis, the value of
(p', S') is a set U with s E U. But since the maximizer moves in p, the value assigned to (p, S) is a superset of the value assigned to any subnode, so that s E evc(p, S) = T.

If, on the other hand, the maximizer cannot win in s, then he cannot win in any child of s. If ( p i , S i ) are the successors of ( p, S ) in the game tree, then again by the inductive hypothesis, we must have s & evc(pi, Si) for each i. But

$$\mathbf{ev}_{c}(p,\,S)=\,\sqcup_{i}\mathbf{ev}_{c}(p_{i},\,S_{i}\,)$$
$$\mathbb{Z}_{c}\left(p_{i},S_{i}\right)$$

## So That S & Evc(P, S) = T,

For the minimizing case, suppose that the maximizer wins in s. Then the maximizer must win in every successor of s, so that s E evc(pi, Si) for each such successor and therefore s E evc ( p, S ).  Alternatively, if the minimizer wins in s, he must have a legal winning option so that s & evc (pi, Si) for some i and therefore s & evc (p, S).

Unfortunately, Proposition 5.0.5 is in some sense exactly what we wanted not to prove:
it says that our modified game computes the set of situations in which it is possible for the maximizer to make his contract if he has perfect information about the opponents' cards, not the set of situations in which it is possible for him to make his contract given his actual state of incomplete information.

Before we go on to deal with this, however, let me look at an example in some detail.

The example we will use is similar to that of Section 3 and involves a situation where the maximizer can make his contract if either West has the ♠Q or East has three hearts. I will denote by S the set of situations where West has the ♠Q, and by T the set where East has three hearts. It's possible to tie in the "defer the guess" example from Section 3 as well, so I will do that also. Here is the game tree for the game in question:

![0_image_0.png](0_image_0.png)

At the root node, the maximizer has four choices. If he makes the move on the left
(playing for S , as it turns out), the minimizer then moves in a situation where the maximizer wins if S holds and loses if T holds. For the second move, where the maximizer is essentially playing for T, the reverse is true.

In the third case, the maximizer defers the guess. We suppose that he is on play again immediately, forced to commit between playing for S and playing for T . In the last case, he wins independent of whether T or S obtains.