
## 6.4 Application To Imperfect Information

In order to apply these ideas to games of imperfect information treated as in Section 5, we need to show that the value set introduced there is a (hopefully distributive) lattice.

To do this, recall that there is redundant information in an arbitrary element F of 22 5 since if F contains both T and U with T ⊆ U (in other words, the maximizer can play for either T or for U but U is properly better), the set T can be removed from F without affecting the maximizer's options in any interesting way. This suggests the following:
Definition 6.4.1 Let F ∈ 225 for an arbitrary set S.  We will say that F is reduced if there are no T, U E F with T ⊆ U. We will say that F 1 is a reduction of F 2 if F 1 is reduced and F1 G F2. Lemma 6.4.2 Every F E 225 has a unique reduction.

Proof. This is immediate; just remove the subsumed elements from F .

p We will denote the reduction of F by r(F).

Armed with this definition, we can now modify Definition 5.0.6 in the obvious way, replacing the value set V with the set of reduced elements of V and the maximizing and minimizing functions (8) and (9) with the reduced versions thereof, so that

$$\operatorname*{max}({\mathcal{F}},{\mathcal{G}})=r({\mathcal{F}}\cup{\mathcal{G}})$$

(19)
and

$$\operatorname*{min}(\,\{F_{i}\,\},\,\{G_{j}\,\}\,)=r(\,\{F_{i}\,\cap G_{j}\,\}\,)$$
$\left(20\right)$ . 
Remember that we typically write V for max and A for min. Proposition 6.4.3 Given the above definitions, (V, V, A) is a distributive lattice. Proof. We need to show that max and min as defined above are commutative, associative, and idempotent, that they distribute with respect to one another, and that the absorption identity (12) is satisfied. Since the reduction operator clearly commutes with the initial definitions of max and min, commutativity, associativity and distributivity are obvious, as is the fact that V is idempotent. To see that A is idempotent, we have F ∧ F = r({min( F i, F j )}) = r({F; ∩ F j })
but each element of the set on the righthand side is a subset of Fi NFi so F ∧ F = r({F;}) = r(F) = F
For the absorption identity, we need to show that

$\bigcap F_j\}$). 
FV(F AG) = F
But

$${\mathcal{F}}\wedge{\mathcal{G}}=r\left\{F_{i}\cap G_{j}\right\}$$
