 Let us now prove the other case of this proof where the median of all these peaks, the agent peaks and the phantom peaks is an agent peak. So one of these here. So what we are going to do here is that we will prove this for two players, only two players. The general case essentially is a little nontrivial extension, but the idea of this proof remains the same. So I will provide the references if you are interested in reading that general proof. So before proceeding, we first make the following claim. Suppose we have these two players, we are now only looking at two players. And suppose there are two preference profiles which agree on their peaks. So maybe the preferences might be, so one preference might have a much sharper fall than the other one. But the point is that their peaks remain the same or the relative ordering between the alternatives on this side versus this side, that might be different because there is no rule which says that what will happen for those alternatives which are on two other sides of the peak, that might change, but their peaks do not change. So for both player one as well as for player two, in these two different preference profiles, their peaks remain the same. And then the claim says that then the outcome should also remain the same. So how can we prove that? So let us assume that player one speak in these two profiles is A and that for player two is B. And suppose that FB, the outcome in the first case is X and the outcome when we change the first player's preference to the second preference ordering which is P1 prime and keep the second player's preference the same, then the outcome is Y. So and quite naturally, we are assuming for contradiction that Y should not be equal to X and let us see what happens. Now since F is strategy proof, we know that under this preference profile P1, X should be more preferred than Y because here it is assumed that this P1 is the true preference and P1 prime is the misreported one. So if player one's true preference was P1, that agent should have preferred X more than Y. And similarly you can say the same for the other preference, now Y will be more preferred than X. Now since the peaks of P1 and P1 prime are the same, then if X and Y remains on the same side of the peak, so imagine a situation where both of them let us say this is the P1 and both X and Y are on this side and in one case you have X is more preferred than Y and also at the same time Y is more preferred than X. This can happen because these preferences are anti-symmetric that is if both these things happen, X is preferred then Y, Y is preferred than X then they must be the same. So there cannot be any kind of indifference here and because of that fact then if they are on the same side then they must be the same. So we have already we already get the result. The only possibility is when X and Y are on the different side then this is possible but now we are going to show that it is impossible under this setting to have X and Y on two different sides. And why is that? Because remember X and Y are outcome of a social choice function and that social choice function has one additional property than strategy-proofness. So it is strategy-proof and on-to. So we have also seen that under this situation on-to-ness and paired efficiency are equivalent when the social choice function is strategy-proof. We have already seen a result like this. Now we are going to use this paired efficiency and we have also seen for paired efficiency what should happen. So let us assume without loss of generality. Now X is on the left-hand side of this peak and Y is on the right-hand side of the peak. And again without loss of generality A is less than P. So the peak of the first player is on the left of the peak of the second player. Now what should happen for paired to efficiency is that the outcome should lie between A and B and this is a very crucial point. You can already observe that X is living on the left of it. So that is already a contradiction because X is an outcome of a social choice function of the same social choice function which is paired to efficient then this cannot happen. I mean we have this interval A is here and B is here. So these are the two peaks of these two players. The any outcome of the social choice function must live inside it. But we have this X which is living outside so it cannot be paired to efficient. So for all paired to efficient social choice function this cannot happen. So therefore what we have assumed was wrong. And you can look at the other cases as well I mean if B was less than A and something else was true I mean this condition again there should always be some outcome of the social choice function which is living outside that interval and we will have a contradiction to paired to efficiency. Now that is for one transition so we have done the transition from P to P1 prime P2. Now we can complete this transition from P1 prime P2 that we know that the outcome is the same and then we go to P1 prime P2 prime and there also the outcome should be same. So that with that small result let us get back to the actual proof. Now we have a profile P1 P2 which is given by B and we know that the peak of the first player is A, peak of the second player is B. And let us denote Y1 is the phantom peak. Now by assumption because we are in this case 2 the median of these three things, these three numbers here is going to be an agent peak so either A or B and we can without loss of generality assume the median is A for B again a very similar argument will follow. So let us assume for contradiction the outcome is C which is not equal to A. We will have to show that this median has to be the same. This outcome should be exactly equal to that median but let us assume that is not true. Then by paired to efficiency we know that C must be within this interval, within the interval between A and B. So that is something that we have seen. So there are two possible cases in which A can be a median either B is less than A less than Y1 in that case A is a median or Y1 is less than A less than B in that case also A is a median. So these are the two exhaustive cases. So let us look at the first sub case that is B is less than A less than Y1 and by paired to efficiency we know that C has to be somewhere in between. In particular C has to be less than A. Now we are going to construct and show a count example if that this cannot happen. So we construct a P1 prime where the peak remains the same as the previous one as the as P1 itself but Y1 and C I mean Y1 is more preferred under this preference than C. Why is that possible to do? Because now we know that A is living somewhere here, B is here, C is somewhere here because of the paired to efficiency and Y1 is here because Y1 and C are on two different sides of this peak I mean peak is this A then this is possible. You can always construct a preference profile which has Y1 more preferred than C. Now by the earlier claim whenever the peaks remain the same if an outcome I mean if C is the outcome in this preference profile then it better be the outcome in the new preference profile as well. So in P1 prime and P2 under that preference profile the outcome should also be C. So this is let us say point number one and now we are going to consider a very special profile of player 1 but this is P11 remember this is the peak is on the right most position. So this is something like this the right most position is the peak and then it is falling down towards the left. So if that is the preference profile we already know that that is the extreme right. So therefore the P2 or the for player 2 the peak is B that is by the choice that we have made and B is actually less than Y I mean that is how we have. So this is this case B is less than Y and Y is less than or equal to this P111 so of course this is the right most point so therefore it must be larger than all these things. So in that case so if you just look at this particular preference profile P11 and P2 there are these three peaks so one peak is on the extreme right the second is the phantom one and you have another agent peak which is B and because of this peaks are essentially because of this three peaks the median happens to be this Y1 which is a phantom peak and if it is a phantom peak we already know the result of case one where the median was the phantom peak. So then by case one we already know that the outcome the social choice outcome should be exactly equal to the median which is Y1 and now the contradiction happens. Now we know that by construction Y1 is more preferred than C and we know that the outcome at that case that preference profile is C then you already see a kind of a manipulation which is possible by player one it prefers Y1 more than C but the outcome is actually a worse outcome while if it manipulates its preference from P1 prime to P11 it gets this outcome which is Y1 which is more preferred than the current outcome which is C. So F cannot be strategy proof so this is the contradiction that we actually lead to and therefore what we have assumed was incorrect so this should also be the outcome should be exactly equal to A which is the agent peak. And a very similar complementary analysis you can do when Y1 is less than A less than B and the pattern efficiency will imply that A must be so C should be between A and B and you can just replicate the same arguments the same proof I just skip that for you to observe and redo all the steps are done here but you just have to understand that them appropriately. So there also you will have a contradiction to the strategy proofness and essentially then you can conclude that the outcome in both these cases has to be the median which is the agent peak in this case. And now to complete this proof so this is this completes the proof of case 2 for 2 agents for the n number of agents you can take a look at this original paper which which proves it this is a highly readable paper and also interesting to read. So the title of this paper is on strategy proofness and single picketness so this gives a ton of results that we have proved in the previous module and in this module.