(b) If the minimizer is on play in the given position, the successors are obtained by making playing any card c that is legal in any element of X  and then restricting Z to those situations for which c is in fact a legal play.

4. Terminal nodes are nodes where all cards have been played, and therefore correspond to single situations s. For such a terminal position, if the declarer has made his contract, the value is ({s},{S}).  If the declarer has failed to make his contract, the value is
( { s }, { S = { s }}).
5. The maximization and minimization functions are given by (8) and (9) respectively.

Theorem 5.0.7 Suppose that the value of the imperfect information game for G is T.

Then a set of positions T is a subset of an element of T if and only if the maximizer has a strategy that wins in every element of T, assuming that the minimizer plays with perfect in for mation. Proof. Once again, the proof proceeds by induction on the depth of the game tree. And once again, the case where p is a terminal position is handled easily by the definition. For the inductive case, we consider the maximizer and minimizer separately.

For the maximizer, suppose that there is some set T of situations that satisfies the conditions of the theorem, so that the maximizer has a strategy that caters to all of the elements of T . Then the first move of that strategy will be some single move to a position pi that is a successor of p and that caters to the elements of T. Thus if the value of the successful child is F , T is a subset of some F ∈ F by the inductive hypothesis. Thus if the value of the original game is G, T is a subset of an element of G by virtue of (8).

Alternatively, if T is a set for which the maximizer has no such strategy, then clearly the maximizer cannot have a strategy after making any of the moves to the successor positions p i . This means that no superset U D T in any ev c ( p i ), and thus no superset of T in ev c ( p )
either.

The minimizing case is not really any harder. Suppose first that the maximizer has no strategy for succeeding in every situation in T . Then the minimizer (playing with perfect information) must have some move to a position p i with value F i such that T is not a subset of any element of Fi. Now if Fi = {Ti}, recall that

$${\mathfrak{s}}\ T\ \mathbb{Z}\ T_{i}\cap U_{j}\ {\mathfrak{l}}$$

min( {Ti}, {Ui}) = {Ti ∩ Uj},

$$\mathbb{Z}\ T_{i}\ \mathrm{for\each}\ i.\ \ \mathrm{Thus}$$

and T Q T i for each i . Thus T Q T i N U i for each i and j , and there is no V D T with

$$\{U_{i}\}=\{I_{i}\cap U_{j}\}$$

$${\mathrm{~o~}}V\supseteq T{\mathrm{~with~}}$$
$\mathbb{L}_{\infty}\subset\mathbb{R}$
$(\mathbb{T},\mathbb{I})$ ... 

## V E Min({Ti}, {Ui})

For the last case, suppose that the maximizer does have a strategy for succeeding in ery situation in T . That means that after any move for the minimizer, the maximizer will still have a strategy that succeeds in T i so that if pi are the successors of p and ev c ( pi ) = Ti, then there is a T i ∈ T i with T ⊆ T i . Now T ⊆ ∩ T i ∈ min( T i ) = ev c ( p ). Thus ev c ( p ) contains an element that is a superset of T.

Using this result, we can in theory compute exactly the set of things we might play for given a single-dummy bridge problem. Before we turn to the issues involved in doing so in practice, however, let me repeat the example of this section using the imperfect information technique. Here is the game tree again with values assigned to the terminal nodes: