 Let us now start proving the Gibbard-Sutherbrook theorem. As we have mentioned in our previous module, it says that if you have three or more alternatives, then every social choice function that is onto and strategy proof implies and is implied by the fact that this social choice function is dictatorial. And we have also seen a corollary in the previous module it says that under strategy proofness, all these notions of parity efficiency, unanimity and ontoness are the same. So this is this corollary here. And therefore, you can restate the Gibbard-Sutherbrook theorem replacing this ontoness with parity efficiency or unanimity. So in some texts you will find similar definitions and they are all the same. Now before moving on to the proof, let us look at some of the points that we can note from this statement which is implied and which are not implied by Gibbard-Sutherbrook theorem. So the first observation that we can make is that this number of alternatives needs to be at least three or more. If you had two alternatives, then we know that there exists some certain kind of onto and strategy proof mechanisms, strategy proof social choice functions for instance, plurality or any kind of scoring rule-based mechanisms that we have discussed before. They are all non-dictatorial social choice functions yet they are onto and strategy proof. So that is one interesting point to note. The Gibbard-Sutherbrook result does not hold when you have two alternatives. The second observation that we are going to make is that this domain of this social choice function is the entire set of all possible strict preferences. So script P essentially denotes that it is the permutation of all possible alternatives. So if you have M-ordinates script P lists all possible M-factorial permutations of these alternatives which can potentially be the preferences of these agents. So this actually gives so one way to visualize or build an intuition about this result is that the domain of the social choice function is too rich and the voter has many options to misreport and that essentially gives rise to such kind of an impossibility result. Later on we will see that if your domain was limited that is not all possible strict preferences over these alternatives were feasible in your social choice function then GSTRM may not hold and we will see examples of that. The third observation is about indifference. We have so far discussed about strict preferences. You can actually generalize or include more things about indifference in preferences. So you can either expand P because indifference could actually lead to a superset of script P that we have discussed. If that happens then you already know the type of constructions that we are going to do already exist there. So in that case GSTRM will continue to hold but if you have removed some of these preferences, some of the strict preferences that we are going to consider then possibly GSTRM won't hold. So it all depends on the domain. So just remember that our social choice function was a mapping script P to the power n that means all the players have these the preferences which lives in script P and we are picking one alternative. So this is the social choice setup that we are picking one alternative out of from this preference profile. If you have this domain to be very much restricted which we will discuss very soon then GSTRM can be bypassed and we will see such examples. And the cardinalization, the last point is about cardinalization. Here we are only talking about the ordinal preferences of these agents but if there are cardinal preferences which also includes note that cardinalization is just another way of representing or maybe a richer way of representing the preferences. If the numbers are equal, the cardinal numbers are equal then you can assume that they are indifferent. If they are unequal then you can have a complete ordering between those alternatives. If by cardinalization you are not ruling out all the possible P's all the strict preferences that we are going to use in the proof then GSTRM will continue to hold. So in some sense we can say that cardinalization that admits all those strict preferences for all these agents then in that domain also we can say that an equivalent GSTRM will hold. So it does not really matter on cardinalization or ordinalization, it just matters that how these alternatives are positioned, what are the possible ways you can order them. Okay so with that let us start the proof and for this proof we are going to follow a very direct approach and this is due to SEN 2001. So at the end I will give the complete reference for this for this paper and what we are going to do in this module because the proof is essentially a little long and quite laborious to follow. So therefore we will just look at the simpler setting of two players, two voters and then apply induction. So this proof essentially uses proofs it for two agents and then applies induction. We will not discuss this induction step because that will be too involved but the two agent case is quite easy to follow and that is what we are going to do in this module to build the intuition. So in order to prove it let us go in steps. So the first sub result the lemma is that if we have three or three or more alternatives and you have two agents one and two and the fact that F is on point strategy proof then for every preference profile what we can conclude is that this outcome will either be the top most alternative of player one or the top most alternative of player two. So it cannot be any other alternative apart from these two things. So this is what it means. If it happens that these two alternatives are the same then we already know and this is something that we are going to use over and over again even though we have stated on-to-ness and strategy proofness in the statement of the of this theorem we all also know from the previous corollary that it is equivalent to say this is Pareto efficient or unanimous. That is because under strategy proofness all these results are the same and we are going to in our proof we are going to use all of these results interchangeably. So we have already made use of one of those results. So here we are assuming that F is unanimous as well. So if that is unanimous then and this two alternatives are one and the same then the theorem holds the lemma holds because then unanimity will actually imply that it is it should be the same outcome the outcome of the social choice function should be the should be the same because the both the players are agreeing on the on their top alternatives. So therefore let us assume for the general case these two alternatives are not the same. The top most alternative of one is A and the top most alternative of B of two is B. Now let us assume for contradiction that the outcome the final outcome is neither A nor B. So we are actually negating this statement. So it is not neither A nor B and notice that this requires you to use three alternatives at least three alternatives. And now we are going to look at a as before we are going to look at certain transitions. So we are going to construct certain preference profiles. We are given this particular preference profile P1 and P2 and we know that the outcome here is C by our assumption which is neither the top alternative of one neither the top alternative of two. So now what we are going to do so we are looking at this first alternative here so the top most alternative of player one and we are putting that A in this in this way. Similarly we are also constructing another preference profile where the the second so B that is the top alternative of player two has been pushed to the top to the second top and the other alternative remains the other preference profile remains the same. Similarly here it was the first place preference profile was kept the same and for the second place preference so A was maybe somewhere down below here it has been pushed to the second top. Now what can we say so we can conclude for this particular preference profile here that the outcome should be either A or B and why is that that is because of the Pareto efficiency and here we are actually using using this fact that F is also Pareto efficient so that means it will not pick something which is Pareto dominated and what is Pareto dominated so anything except B and A is Pareto dominated by both these players with respect to A so A Pareto dominates all other alternatives here so therefore none of those can be an outcome in this preference profile if because F is Pareto efficient so it has the only option of of either A or B that that should be the outcome. Now let us notice that if this outcome was B for instance then what player two gets if you look at it so here the here the outcome is B and here the outcome is C which is strictly worse than B for player two so and because player one's preference is not changing then player two will misreport from P2 to P2 prime and that will be a beneficial deviation so since A phase strategy proof as well we cannot have this alternative so B is ruled out the only option therefore is under this alternative is A so it should have A as its outcome so similar arguments you can apply here as well and look at the transition from here to here and apply the same argument for player one and you can see that here the outcome is definitely going to be B it cannot have A so B will be the outcome in this case it's a very symmetric argument now we are in a nice position to make a transition so you notice that this is P2 prime and this preference profile is nothing but collecting together this P1 prime and this P2 prime for these two pairs and what we can observe is if you trans if you transition from from this preference profile to here what is happening is that for I mean the position of B is remaining the same in this case so if you apply the equivalent definition of strategy proofness which is monotonicity then you can conclude that here the outcome would have been B but if you apply the same argument from here to here of course so what is happening is that here the the relative position of B was same here the relative position of A remains same right so here also A's relative position does not change you can again apply monotonicity and you can conclude that this outcome will be A and since A is not equal to B this is a contradiction I mean you arrive at a contradiction from one transition you are getting an alternative which is A the other transition is telling the alternative will be the outcome will be B and that cannot happen so what we have assumed that here the outcome is C which is neither A nor B then that is that is not true so we have actually proved this this lemma all right so with that lemma in our hand and now we can actually go and prove the give us other word result for two alternatives so what do we have here so we have for two alternatives we have at least three so two agents we have at least three alternatives and we know that this is an ongoing strategy proof so strategy proof means that this is equivalent to monotonicity we have used it already and on to this means that it can also be better efficient or unanimous so just to keep in mind that this this is what it means and let us assume that the top most alternative of player one and player two are not the same so this is A and this is B and let us look at a different preference profile of both these players where the top most alternative is C for player one and the top most alternative is D for player two and there are two conclusions that if the outcome in the first preference profile is the top alternative of player one that is it is A then it should continue to be the top alternative of player one so that will be C and similarly if the top alternative is of B of player two which is B then in the new preference profile the top alternative will be the same the top alternative of the same player which is D and we will we will just prove this one part because the the proof of the second part will be quite symmetric so without loss of generality we will just prove this so just to mention that this actually proves the dictatorship for two players and and then later on this this direct proof that we have mentioned actually uses an induction for more than two players now one thing that we have already observed and we have also mentioned earlier that if C and D are same so in this case we are not putting any constraints about C and D so if C and D are same then unanimity implies that this this conclusion will be true so the lemma is trivial in that case so for all all the non-trivial cases we will assume that C is not equal to D so that is what we are going to assume now let us list down all possible exhaustive cases and there are six cases where C is not equal to D and and since we are considering only only this part so the first implication of of this lemma we are going to assume so even even when we say case one you can also create an equivalent statement which involves the the second agent's case so it is it can be the same argument can be interchangeably used for player one and player two to show that if you flip the names so here this is player one this is player two but if you flip their identities make player one player two and player two player one then the similar argument will follow and we can get the similar conclusion so we are not going to prove that part in the in this proof but whenever we are using some of these cases some of these later cases here and we are referring to case one we are going to assume that that that this other counterpart is also true because sometimes we will need those cases to be used okay so so let us first list down all the cases so here the so the first case is where the C and B is equal to A and B then you keep the second alternative is fixed and the first alternative is neither in or B then these the other cases where it is the first alternative of this player is remaining the same and the second or the second player's top alternative is neither B not equal to B and similarly you can you can list out all possible cases so just check that this actually covers all the possible cases that can potentially happen when C is not equal to D and we have listed it in this way because that this will give us some advantage in proving we will use some of this the proofs of some of these cases in the later cases as well okay so let us go over these cases one by one so the first case is where C is equal to A and D is equal to B so when C is equal to A and D is equal to B so this is the the modified profile so this is the original profile and we are going to assume that the outcome here was A so the outcome here is A now we are we are we are supposed to say what is the out what is going to be the outcome here and C that is this the top most alternative of player one in this in this new profile is A and that for player two is B now we are going to construct a different preference profile and that is very carefully chosen what we have done is we have pushed B onto the second top position and similarly for player two A to the second top position now we already know by the previous lemma the lemma that we proved in the beginning of this module that the in this preference profile the outcome will be either A or B that is all that is something that we already know now our conclusion is that it should be the outcome will be A and now let us for contradiction contradiction of the current lemma let us assume that this outcome is B so that is even though the in the first preference profile the top most alternative of player one was chosen say in the second preference profile the top most alternative for player two was chosen so let us do that then let us see what is the contradiction then what you can say here is from if you go from this preference profile so from this preference profile to this preference profile that we have constructed we see that the A's relative position has weakly increased so A was somehow down below in this case and it has moved to the second top position because B was on the top in this previous one A's relative position has weakly improved so we can apply monotonicity onto it and because A was the outcome here A will continue to be the outcome according to monotonicity but now we can apply the same thing between these two preference profiles because here B was somewhere down below and B has been pushed to the second top position for player one using the same argument and same monotonicity we can conclude that the outcome will be B this case so that is essentially a contradiction so in one of these transitions we conclude that this is B and the other transition we conclude that is A which is not which is not possible so we cannot have this this condition B to be true so this has to be A so this this question mark is essentially A so we have approved case one in this case so let us look at case two where C is neither A nor B and the second alternative is B now we know that this C this is a different alternative neither A nor B and we also know that this top most alternative in this this two top most alternatives one of them will be the outcome and again as before we are going to assume for contradiction this is the outcome then again we we can construct a P1 hat and notice that P2 we are going to keep the same so this is the same as this P2 now what what did we do in this P1 hat we have actually we kept A somewhere here and we have pushed A onto the top now again if you apply the condition from P1 P2 right so P1 P2 to this condition that we already know that using this case one so what was case one so the top most alternatives for both these players are remaining identical maybe the position of the other alternatives are changing but their top most positions are remaining the same therefore using case one we can conclude because and this is we are using the same case one for player player two there we were we were proving it for player one but player two we have we have said that a similar argument will hold even for player two so we are using that so here the top most alternative of player B player two that is B was the outcome so therefore that will continue to be the outcome here that was that was our case one so because this using case one we have an outcome so here the outcome is going to be B now what one can observe is that here the outcome was A and for player one B was leaving somewhere below so because A was the so we know that C is neither A nor B A is in the second position so B will leave somewhere below it's lower in preference than both C and A for player one and the outcome is going to be B so player one might as well be tempted to misreport its preference when it is when its true preference profile is B one and B two in that world player one will misreport to B one because the outcome is going to be A which it strictly prefers over B which is the current outcome so we have actually constructed an example that this is actually manipulable this social choice function is actually manipulable which is a contradiction so therefore we cannot have this this condition to be true the outcome will certainly be C in this case that proves our case two as well so for case three we have to use this case two in a repeated manner so we know what was what was case three we have C which is neither A nor B and D which is not equal to B so we know that D this is not equal to B and C as before neither A nor B right so now we can construct this preference profile as as we have done in in in the previous cases so P1 hat P2 hat and their top most alternate is the C and B right so we are keeping the second player's top most alternative to be the same and the first player's top most alternative has moved to C which is neither A nor B now as before let us assume for contradiction that this outcome here even though the outcome here is A the outcome here is let's say D then if we look at this transition from P prime from this preference profile to this one then what we can see using case two so remember what was case two so case two was the the situation where the so player wants so one of the players top alternative does not change which is which is happening in the case of player one so its top most alternative is not changing the other agents top most alternative is changing and that player's top most alternative was the outcome here so if that holds so using case two we have seen that if you change so whichever players outcome was here and that the top most alternative was chosen as the outcome that players top most alternative will also be chosen here so therefore we can conclude from this transition that the outcome here will be B but similarly we can use the same case two between these two transition between the transition between these two alternatives preference profiles so here because the the top outcome was A which was the top alternative of player one then it should continue to be the top alternative of the of the same player so which is C now we know that C is not equal to B so therefore this is a contradiction and the rest of the proof is essentially trying to argue about such kind of proof by contradiction we are going to assume something which is which is contrary to the conclusion of this lemma and we'll construct certain preference profiles where we can show that that it leads to a contradiction so the the case four is when the the top alternative of player one is remaining the same both of them are A and the second alternative the top alternative of the second player is neither A nor B so it is moving to some D and now we can look at this the same different constructed preference profile the top alternative of player one and two are the same as the original ones so now we can apply from for this transition from here to here we can apply case one which will conclude that this is going to be A and from this this case to this case you can use case two because here the one of the player's top alternatives is remaining fixed the other alternative has actually changed and for contradiction we have assumed that this outcome is D therefore this will be equal to D and because D is not equal to A by assumption here then this is a contradiction so we have actually proved case one two three four so case four so case five is essentially the case where you have a kind of a flip so the top alternative of player two has become the top alternative of player one in this case and D is as before neither equal to A neither not equal to B and we can apply a very similar so we can actually construct a preference profile for for these two players so here this will be A so the top most alternative of player one is A and that for player two is D so this is the same as this D and this is the same as the original preferences preference of player one we don't really care about the rest of the alternatives so now if you look at this transition from P hat P prime to P hat then what we can observe again using case case two essentially this will be case two because one of the player's top alternative is not changing the other one's top alternative is changing and we have actually used the fact that so we are assuming that this is the this is the top alternative the outcome here so in that case in the new preference profile also the outcome would be D and if you are looking at the transition from the original preference profile to this this last preference profile and then what we can observe is that this agent so this is coming from this case four so that the outcome was A and whatever we are changing here is essentially we are looking at that agent whose top alternative is not changing so essentially this will also be so the first case was also for case four so this was also a conclusion from the case four because here the the top most the outcome was the top most alternative of that agent whose top alternative is not changing the other agent's alternative is changing and that is exactly what we have proved in the in case four and then by using this transition from P1 P2 to P P hat we can conclude that this outcome is going to be A and that is because these two are not the same we actually hit a contradiction here and the final case that is case six is the following when you have a C to be equal to B and D to be equal to A so the top alternatives of these two players have actually flipped so what what can we do in this context is that so since the outcome here is the top alternative of player one and let us assume for contradiction that the top alternative the outcome here is the top alternative of player two and let us assume another alternative which is neither N or B we can pick such an alternative because there are three or more alternatives in this in the assumption itself so now we have created some new preference profiles as you can see so you have pushed that so X was somewhere below so we have constructed a preference profile where B is on the top and that is this that is the same B as the top alternative of player two in the original preference and we have pushed X to the second highest position and in the second one we are keeping this preference so here all these preferences are the same for player two and we are pushing X to the top so in one case X is in the second top position and in the last one it is on the top most position now let us look at the transitions and we will be using all the different cases that we have already proved here so let us look at this transition from P prime to to this transition this preference profile transition we already know that here the top alternative is D and because the top most alternatives are not changing so here B and A and here also they are B and A so we can use the first case that is case one that the top alternatives are not changing and therefore if the outcome was D in this case which was which was the top alternative of player two then it will continue to be the top alternative of player two so we will have A here so that is the that is the conclusion similarly if you look at the transition from this preference profile to the last preference profile here what is happening there we can we can see that this is this is falling under the condition of case three and because A was the outcome here which was the top alternative of player one and that should be the outcome here so the outcome here will be X now if you if you look at these two alternatives here so you can already begin to see that for player one there is a profitable deviation in in these two cases so suppose it was currently looking at this preference profile so player once true preference profile was P1 at and in that case the A was both below B and X because X is not neither A nor B and that is going to be the outcome here right but if if it transitions to and misreports its preference to P1 tilde then at least it knows that the outcome will be X which is more preferred than A in in its original preference profile so player one finds it beneficial to deviate and misreport its preference and which is going to lead to the fact that this F is not a strategy proof mechanism so that cannot happen so therefore what we have assumed this is not true so it must be the case that the outcome is C which is equal to P in this case okay so that actually concludes the proof of all the exhaustive cases and we have proved the keyboard side of the board result for for two agents so this is the the main intuition how it is it is done the the actual proof with the with more than two agents essentially uses induction on this on this number of agents and I am not going to prove that if you are interested you can look at this the this paper which is titled a direct proof of GS theorem and that will give you the complete details it's just you'll have to do a lot more accounting than than what we have done in this case