 Last time we stated a very interesting result, one of the first results in game theory due to Von Neumann and that deals with the game of chess and the result was that one and only one of the following three statements is true. Either white has a winning strategy or black has a winning strategy or each player has a strategy guaranteeing at least a draw because exactly one of them is to be true, therefore the fourth possibility of neither of these three things are true is not a possibility. Now in this module we are going to prove that formally. So because we want to prove it, we will have to develop a little bit of more notation. So each vertex in this game tree, let us start with the game tree of chess is a game situation because this, once you are given a specific node, let us say this node, you exactly know through which path it has been reached and therefore you know the whole game situation. We denote by gamma of x, the subtree rooted at x, including the node x itself. So this whole subtree starting from x, we are going to define that, denote that as gamma of x. Now we also have this notation nx which says that number of vertices in gamma of x because it is a finite game, we can always enumerate and say that there are exactly nx number of vertices in gamma of x. Now consider a node which is a vertex in this subtree of gamma of x and y is not equal to x. So something here and we are clear that because this is a tree, the subtree that is rooted at y and denoted by gamma of y is going to be a subgraph of gamma of x. And therefore this ny if you want to call the number of nodes here, that is certainly is going to be smaller than n of x, the number of nodes. Now we also know that if the nx is equal to 1 that means there exists only one node in this subtree which is x itself, then x itself is a terminal node. There is no future nodes possible, no future actions possible there, the game ends. And when the game ends, it ends in one of the three outcomes, win for white, win for b or draw. Now the proof proceeds via an induction on this number of nodes on nx. So the theorem that we have just mentioned is trivially true for nx equal to 1. And why is that? Because at this terminal node as we said either one of the three outcomes can happen. If the white king is removed then b wins, if the black king is removed then white wins. And if both of them exist and there are no more moves then the game ends in a draw. Alright so we know that for one node system it is definitely true. Now what we are going to look at is suppose we have nx which is strictly greater than 1. And our induction hypothesis is that for all the vertices y in gamma of y such that this in y is less than n of x, then this the statement holds. So for all the subtrees here, so because there is more than one node there must be some y which is not equal to x and that leaves in the sub tree of gamma of x. And in that sub tree every vertex follows three statements, one of those three statements and exactly one of those three statements is true. And as we know that gamma of y is a sub game. So sub tree sub game are used interchangeably in this as we go along in this proof. So I think drawing a figure will make it more clear. So what we know from the induction hypothesis is that if we start from x we do not know how we can conclude which of these three statements or we have not proved that one and exactly one of these three statements is true at x. But we know that at every sub tree of x excluding x this statement is true. That is because in each of these cases the sub tree starting from here and the rest of the network the in y is strictly less than nx and there we already know that one of these three statements and exactly one of these three statements is true. Alright, so let's get back to our definition. So without loss of generality let's assume at this game situation x it's the move for white and a very similar and analogous argument can be used if the it was a move for the black player. All the things can be retraced in a very similar way and I can leave that as an exercise. So let's define this notation c of x by c of x we denote all this type of nodes which are reachable from x in one action of this particular player of the player white. So this is essentially this node so and we know that at each of these nodes in each of the nodes in cx it's a move for player b the black player. Now there can be three possible cases we are going to go over that one by one. So the first case is it may be possible that there exists some y naught so suppose this one is y naught which is in cx and in that place in that node the statement one is true because we know that the one of these three statements is definitely going to hold. Suppose the statement one holds in gamma of y naught then we conclude that this statement one is even true in this gamma of x that is the node starting from here and why is that true because y naught node the white player can guarantee a win the white player who plays here it can just pick this action and reach to that y naught and at y naught it knows that it has a winning strategy. So fair enough so we have found a winning strategy for white player even at the node x if case one holds. So the let's go to the second case so the second case is that there could be a situation that for all y so for all the nodes in this cx the statement two is true that is b has a winning strategy then we know that even two is true in gamma of x because the moment so no matter whatever action the white player picks so suppose y is somewhere here I mean it picks the white player picks this action and reaches this strategy reaches this game situation of y and then player b picks the optimal action or the winning strategy at that y and it will ensure a winning strategy. So the fact that at every node in this cx there exists a winning strategy for b it also ensures that there exists a winning strategy at x. Now the case three is when none of this holds and that is quite interesting so because case one and case two does not hold here then we can conclude both of these things one by one. So because one does not hold then w the white does not have a winning strategy in any of this in any of this nodes in cx and since the induction hypothesis holds for every y in cx therefore either b can have a winning strategy or they both have a draw guaranteeing strategy right this is these are the two possibilities. One of the other two things has to happen. Now we notice that even two does not hold I mean case two also does not hold then you can come back to case two and negate this sentence negate this statement which says that there must exist some y for which this this player b does not have a winning strategy. So what was case two case two was saying that for all nodes in cx player b has a winning strategy the negation of that is that there exists some node at least where b does not have a winning strategy. Now from what we know that there could be two possibilities either b has a winning strategy or both have a draw guaranteeing strategy now we have found some y prime in c of x where b does not have a winning strategy. So player a player white who is making the current move can actually pick that particular action which leads leads it to a specific y prime. So let's say this is the y prime point where b does not have a winning strategy. So this white player can just pick this action and go to a situation where b does not have a winning strategy but both of them has a guaranteed draw guaranteeing strategy and therefore the statement three must be true in this case and therefore we also conclude that if case three holds then you can ensure that three the third statement of that of that proposition is true. All right so that essentially concludes the proof.