For the first talk, see here. The second talk took place on September 21.

We want to prove of Theorem 1, that if , then II has a winning coding strategy in .

The argument makes essential use of the following:

Coding Lemma.Let be a poset such that for all ,

Suppose that . Then there is a map such that

**Proof.** Note that is infinite. We may then identify it with some infinite cardinal . It suffices to show that for any partial ordering on as in the hypothesis, there is a map such that for any , there is a with such that .

Well-order in type , and call this ordering. We define by transfinite recursion through . Given , let be the set of its -predecessors,

.

Our inductive assumption is that for any pair , we have chosen some with , and defined . Let us denote by the domain of the partial function we have defined so far. Note that . Since has size , it must meet . Take to be least in this intersection, and set , thus completing the stage of this recursion.

At the end, the resulting map can be extended to a map with domain in an arbitrary way, and this function clearly is as required.

Back to the proof of . Fix a perfect information winning strategy for II in , and a set cofinal in of least possible size. Pick a such that for all we have .

Given , let . Now we consider two cases, depending on whether for some we have or not.

Suppose first that for all . Then the Coding Lemma applies with in the role of , and as chosen. Let be as in the lemma.

We define as follows:

- Given , let be such that , and set .
- Given with , let be such that , and set .

Clearly, is winning: In any run of the game with II following , player II’s moves cover their responses following , and we are done since is winning.

The second case, when there is some with , will be dealt with in the next talk.