tag:blogger.com,1999:blog-4978426466915379555.post8766908686691717864..comments2015-12-24T10:17:39.421-05:00Comments on Formal Methods in Political Philosophy: COLLECTIVE CHOICE FOURTH INSTALLMENTRobert Paul Wolffhttp://www.blogger.com/profile/11970360952872431856noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-4978426466915379555.post-1746388350489254042010-08-09T07:10:47.458-04:002010-08-09T07:10:47.458-04:00I was stuck at a similar point. Can I check I unde...I was stuck at a similar point. Can I check I understand? <br /><br />At first glance, it looks like the proof is invalid. The condition for i to dictate over (x,z) is that in any setting where i strongly prefers x to z, society strongly prefers x to z. But, looking at steps (i)-(v), it seems like all we have is that when i strongly prefers x to z *and a bunch of further assumptions are in place* then society prefers x to z. <br /><br />But I'm thinking the trick concerns Condition 3 of the assumptions. By that assumption, how society ranks x against z is determined by individuals' ranking of x against z. So let's look just at {x,z}, assume that i strongly prefers x over z, and assume nothing about what anyone else thinks. Now, because of the independence assumption (condition 3) we're allowed to fill in the details any way we choose as to preferences over options not in {x,z} (where individuals rank y, for example)---and this can't change anything about the societal ranking of x and z. <br /><br />What the proof does is fix on a particular way of adding in details about where y fits in to each person's preferences, and showing that in that one case, the assumptions force society to rank x over z. But, as we said, the one case is representative for all cases, since the societal ordering can't depend on these sort of extraneous details. So society prefers x to z in all rankings. <br /><br />Is that how the argument goes? Sorry if this is nonsense...Robbie Williamshttp://www.blogger.com/profile/02081389310232077607noreply@blogger.comtag:blogger.com,1999:blog-4978426466915379555.post-58475085071338543462010-07-23T10:14:33.536-04:002010-07-23T10:14:33.536-04:00Simo, This is the problem with trying to do this ...Simo, This is the problem with trying to do this online. For me, at any rate, it would be much easier if we could talk about it face to face. About the first point: The trick in the proof is that if, assuming some particular set of preference rankings, Arrow can prove that i dictates over some ordered pair, then i is a dictator over that ordered pair regardless of how everybody else's rankings change. i is not just a dictator over the ordered pair for some special set of rankings. This is another way of saying how powerful Lemma 3 is, and therefore how powerful an assumption it is to say that i is decisive over a single ordered pair.<br /><br />As for the last part of the comment,I think [I emphasize think] the point is that in P, for example, the "if then" is an entailment, not a truth functional connective. Does that help at all?Robert Paul Wolffhttp://www.blogger.com/profile/11970360952872431856noreply@blogger.comtag:blogger.com,1999:blog-4978426466915379555.post-48562724123997974702010-07-23T09:46:37.623-04:002010-07-23T09:46:37.623-04:00I have problems understanding, why this proof work...I have problems understanding, why this proof works: steps i-v seem to prove the following: if xDy for i and individual preference orderings are thus-and-thus, then xḎz. So is step viii ok? In step vi you are assuming a different set of individual orderings and the following doesn't seem to be a correct inference <br /><br />1.if xDy for i and individual preference orderings are thus-and-thus, then xḎz. (step v)<br />2. xDy for i (assumed t the beginning) and now some different individual preference orderings (step vi) <br />3. therefore, xḎz (step viii) <br /><br />Besides, I dont see why this proof would work for all sets of individual prefence orderings: in all your constructions of different preference individual orderings, you either assume that aPib for all i, or that aPjb for all j who aren't i, and bPiA, so that you can always infer aPb, either with Condition P or the assumption that aDb for i. Yet since condition P and the assumtption aDb are both if-then statements, they are true when their antecedents are false. So the assumption aDb for i is true, even when (some of) the others agree with i. But in that case, you cant infer via modus ponens that aPb (since the antecedent is false). But how do you prove in these cases, that aḎb for i for all a and b?Simohttp://www.blogger.com/profile/17192553001632092435noreply@blogger.com