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The Republicans held a filibuster the other night to block the Reid-Levin Amendment from a vote. The aim of this article is to axiomatically show, using Proof Theory, how Majority Rule works and, that, the mathematics behind Majority Rule is in fact the easy part; the really, really difficult part are the social and game theory-like games that we play with each other, as the recent filibuster demonstrates.
Of the filibuster, Senator Harry Reid said:
It would be one thing for Republicans to vote against this bill. If they honestly believe that “stay the course” is the right strategy — they have the right to vote “no.” But now, Republicans are using a filibuster to block us from even voting on an amendment that could bring the war to a responsible end. They are protecting the President rather than protecting our troops. They are denying us an up or down — yes or no — vote on the most important issue our country faces. I would like to inform the Republican leadership and all my colleagues that we have no intention of backing down. If Republicans do not allow a vote on Levin/Reed today or tomorrow, we will work straight through the night on Tuesday. The American people deserve an open and honest debate on this war, and they deserve an up or down vote on this amendment to end it.
Well, the filibuster worked. Senator Harry Reid pulled the entire Defense Authorization Bill from off the floor today:
Because Republicans continue to block votes on important amendments to the Defense Authorization bill, we can make no further progress on Iraq and this bill at this time.
For these reasons, I have temporarily laid aside the Defense Authorization bill and have entered a motion to reconsider.
But let me be clear to my Republican colleagues — I emphasize the word “temporarily”. We will do everything in our power to change course in Iraq. We will do everything in our power to complete consideration of a Defense Authorization bill. We must do both.
And just to remind my Republican colleagues — even if this bill had passed yesterday, its provisions would not take effect until October.
So we will come back to this bill as soon as it is clear we can make real progress. To that end, I have asked the Democratic Whip and Democratic Manager of the bill to sit down with their counterparts to work on a process to address all outstanding issues related to this bill so the Senate can return to it as soon as possible.
I don’t have an opinion on the Bill or amendment that I’m willing to share or defend at this time, but this political posturing by the Republicans and the response from the Democrats is an amazing demonstration of Game Theory at work. But, I’m not here to speak of Game Theory; instead, I’d like to revisit Majority Rule.
Majority Rule works. Many, many years ago, I wrote a paper while I was an undergraduate that I submitted for publication to the Notre Dame Journal of Symbolic Logic. I was given a revise/resubmit, which for an undergraduate is a great feat to accomplish. I decided not to resubmit and, focused, instead on trying to graduate. Anyway, that paper was an attempt at axiomatizing Majority Rule.
In what follows, I will axiomatically show how Majority Rule works. Unlike real life, such as the filibuster and other social games that people play, the stuff below is actually the easy stuff. The really dificult stuff are the social games that people play. Math is easy; People-stuff is hard.
Why are we attracted to the majority rule system? Some argue that it is because the majority rule system is fair and effective. Proponents argue that it is fair because it satisfies two of its three main properties: option equality, voter equality, and sensitivity.
Option equality means that the options are equal, i.e., if the same amount of people vote for choice a, the results will switch. Voter equality means that no person’s vote is more influential than that of another’s vote. Furthermore, supporters of the majority rule system also argue that the system is effective because it satisfies the third property, sensitivity. Sensitivity means that if the voting is tied and one more person votes, then the option he votes for wins. This paper will show that the majority rule system does satisfy the three main properties mentioned above and, by doing so, proves to be both fair and effective. Note: this paper does not try to elaborate or address socio-political or socio-economical issues — such issues are appropriate and should be addressed in a more wholesale social scientific paper. I simply will show through the use of the proof that the Majority Rule System satisfies the properties it claims to uphold.
Before showing that majority rule satisfies option equality, voter equality, and sensitivity, I will need to set out my definitions and presuppositions. One major presupposition in the notion of majority rule is that of choice. Social scientists call the method of choosing a social choice function. A social choice function is a rule for choosing between options where some of the citizens prefer one option and others prefer the other. The majority rule system chooses one option over another when more citizens prefer the one option than the other. From this we get the definition of a social choice function (SCF):
SCF: C is a social choice function for options a and b and citizens V iff a≠b, and V is finite and non-empty, and for all disjoint subsets A and B of V, C(A, B) equals {a} or {b} or {a, b}.
We also get the definition of the majority rule social choice function:
mrSCF: C is a majority rule social choice function for options a and b and citizens V iff C is a social choice function for options a and b and citizens V, and for all disjoint subsets A and B of V, C(A, B)={a, b} if A≈B, C(A, B)={a} if A>B, and C(A, B)={b} if B>A.
In addition to the definition of majority rule social choice function, there are three properties of majority rule. The first is called option equality (OE). This may be defined thus:
OE: C satisfies option equality for options a and b and citizens V iff for all disjoint subsets A and B of V, C(A, B)={a} iff C(B, A)={b}.
The second property of majority rule is called voter equality (VE). Voter equality, defined, is:
VE: C satisfies voter equality for options a and b and citizens V iff for all disjoint subsets A and B of V, if A≈B, then C(A, B)=C(B, A).
The last property of majority rule is called sensitivity (S). This is defined as the following:
S: C satisfies sensitivity for options a and b and citizens V iff for all disjoint subsets A and B of V, if C(A, B)={a, b}, then for every non-empty subset X of V-(A∪B), C(A∪X, B)={a}.
As I proceed, please keep in mind the aim of this paper: to show that the majority rule social choice function satisfies the three properties of majority rule, namely, option equality, voter equality, and sensitivity. In other words,
Theorem: C is the Majority Rule social choice function for options a and b and citizens V iff C is a social choice function that satisfies Option Equality, Voter Equality, and Sensitivity for options a and b and citizens V.
To accomplish this aim, I must proceed in two phases. The first phase will contain the three properties of majority rule. The second phase will contain the three conditions of majority rule social choice function.
Phase I
1.1 (OE) C(A, B)={a} iff C(B, A)={b}
1.2 (VE) If A≈B, then C(A, B)=C(B, A)
1.3 (S) If C(A, B)={a, b}, then for every non-empty subset X of V-(A∪B), C(A∪X, B)={a}
Phase II
2.1 C(A, B)={a, b} if A≈B
2.2 C(A, B)={a} if A>B
2.3 C(A, B)={b} if B>A
What I have done by setting up both Phase I and II is to put the aim of this paper in a slightly different way: I now need to show how the elements in Phase I satisfy the elements in Phase II. To do this, I will assume the elements in Phase II and prove the elements in Phase I. Also, I need to show how the elements in Phase II satisfy those in Phase I. To do this, I will assume the elements in Phase I and prove the elements in Phase II. I can prove the theorem this way because it is a bi-conditional statement. That is, if I assume the left-hand side, the right-hand side should follow. Similarly, if I assume the right-hand side, the left-hand side should also follow. By showing how Phase I and II prove each other, I will ultimately show that the majority rule social choice function does satisfy option equality, voter equality, and sensitivity and hence demonstrate fairness and effectiveness.
Phase I
1.1: I need to show that C(A, B)={a} iff C(B, A)={b}.
Proof: Assume that C(A, B)={a}; show C(B, A)={b}. If A≈B, then C(A, B)={a, b}. If B>A, then C(A, B)={b}. By process of elimination and cases, since C(A, B)={a}, A>B. So, C(B, A)={b}. Now, assume the other side of the bi-conditional, i.e., C(B, A)={b}; show C(A, B)={a}. If A≈B, then C(A, B)={a, b}. If A>B, then C(A, B)={a}. By cases and process of elimination, B>A. So, C(A, B)={a}. Q.E.D.
1.2: If A≈B, then C(A, B)=C(B, A).
Proof: Assume A≈B; show C(A, B)=C(B, A). So, C(A, B)={a, b}. Also, C(B, A)={a, b}. Hence, C(A, B)=C(B, A). Q.E.D.
1.3: If C(A, B)={a, b}, then for every non-empty subset x of V-(A∪B), C(A∪X, B)={a}.
Proof: Assume C(A, B)={a, b}; show that for every non-empty subset X of V-(A∪B), C(A∪X, B)={a}. Since X≠∅ and x⊆(V-(A∪B)), A∪X>A. If A>B, then C(A, B)={a}.
If B>A, then C(A, B)={b}. So, by the process of elimination and cases, since C(A, B)={a, b}, A≈B. Hence, A∪X>B. Consequently, C(A∪X, B)={a}. Q.E.D.
Thus, by proving Option Equality, Voter Equality, and Sensitivity, I have shown that they are derived from the notion of majority social choice function. But I am only halfway finished.
I now need to prove the elements in Phase II. Remember, that in doing so, I will be assuming the elements in Phase I.
Phase II
2.1 If A≈B, then C(A, B)={a, b}.
Proof: Assume A≈B; show C(A, B)={a, b}. Since A≈B, by voter equality I get C(A, B)=C(B, A). Also, bear in mind the definition of a social choice function, namely, that C(A, B) must equal {a} or {b} or {a, b}. By option equality, C(A, B)={a} iff C(B, A)={b}. Yet, {a}≠{b}. So, C(A, b)={a, b}. Q.E.D.
2.2 If A>B, then C(A, B)={a}.
Proof: Assume A>B; show C(A, B)={a}. Sensitivity claims that if C(A, B)={a, b}, then for every non-empty subset X of V-(A∪B), C(A∪X, B)={a}. Let A’ be a subset of A such that A’≈B. Accordingly, by Phase 2.1, I get C(A’, B)={a, b}. So, C(A, B)={a}. Q.E.D.
2.3 If B>A, then C(A, B)={b}.
Proof: Assume B>A; show C(A, B)={b}. Sensitivity dictates that if C(A, B)={a, b}, then for every non-empty subset X of V-(A∪B), C(A∪X, B)={a}. Let B’ be a subset of B such that A≈B’. By Phase 2.1, I get C(B’, A)={a, b}. But this means that C(B, A)={a}. So, by option equality, C(A, B)={b}. Q.E.D.
Now, both Phases have been proven. Since this is true, I have shown that the majority rule social choice function does satisfy option equality, voter equality, and sensitivity. By satisfying these three properties, the majority rule proves to be both fair and effective.
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