A New Approach to the Social Choice Function

by

John Clifton Lawrence
General Algorithm
POB 230351
Encinitas, CA 92023
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E-mail: jlawrence@genalg.com

 1998 by John Clifton Lawrence

January 15, 1998

Abstract
In “Social Choice and Individual Values” Arrow (1951) discusses two kinds of functions: the Social Choice Function, SCF, and the Social Welfare Function, SWF. The SCF is a function whose range is an alternative or set of alternatives. The SWF is a function whose range is an ordering of alternatives or sets of alternatives.  Arrow’s famous General Possibility Theorem, GPT,  states that no SWF exists which meets certain criteria. However, the SCF may well exist even in cases in which the SWF doesn’t because the production of an alternative or set of alternatives is less restrictive than the production of a set of orderings over the entire alternative set. 
The only role that the SCF plays in the GPT is in the specification of Condition 3, the Independence of Irrelevant Alternatives. In that condition Arrow specifies that the SCFs over a set S  T (where T contains the total number of alternatives) produce the same social result in two different cases when the individual preference orderings within the set S are identical in both cases. The set T may have different orderings in the two cases, but the alternative or set of alternatives produced by the SCF must be the same. This is the weakest possible requirement for social orderings within the set S since it only requires the top alternative or set of alternatives to be the same. Stronger requirements would require that the top two or more alternatives within the set S be ordered in the same way, and the strongest requirement would be that the entire social orderings within the set S in the two cases be the same. The SWF produces orderings over T; a generalized SCF would produce orderings over S.
Arrow provides for tie solutions via the SCF but only for ties among alternatives. Since the SWF requires orderings and not alternatives, it is natural to examine ties among orderings and not just ties among alternatives. When ties among orderings are allowed (as demanded by the completeness axiom), Condition 3 can be strengthened to the maximum and solutions which violate the GPT can still be found. We demonstrate this for m, the number of alternatives, = 3 and 4.
Introduction
We assume a set, S, of m alternatives a,b,c ... and a relationship R which means “preferred or indifferent.” Therefore, xRy means x is “preferred or indifferent” to y. For the present we will just consider binary orderings of this nature. Arrow (1951) sets down two axioms. 	

	Axiom 1: “For all x and y, either xRy or yRx.” 
	Axiom 2: “For all x, y, and z, xRy and yRz imply xRz.”

	Axiom 1 is the axiom of connection. Since xRx, R is reflexive, and since xRy or yRx for all alternatives in the set S, R is complete. Arrow also states (p. 13): “Note also that the word ‘or’ in the statement of Axiom 1 does not exclude the possibility of both xRy and yRx. That word merely asserts that at least one of the two events must occur; both may.” Therefore we can restate Axiom 1 as follows: 

For all x and y, one of the following must be true: 1) xRy; 2) yRx; 3) both xRy and yRx. We will use the nomenclature {xRy, yRx} for “both xRy and yRx.”

Arrow (p. 14) then goes on to make two definitions:

Definition 1: “xPy is defined to mean not yRx.”

Definition 2: “xIy means xRy and yRx.”

where xPy is read “x is preferred to y” and xIy is read “x is indifferent to y.”

A problem arises here in Arrow’s logic. According to Axiom 1, one of the following must be true: 1) xRy; 2) yRx; 3) both xRy and yRx. If yRx is not true, then either (1) xRy or (3) both xRy and yRx must be true. Thus, Definition 1 doesn’t make sense because xPy and (3) above cannot be true at the same time. Definition 1 conflicts with Axiom 1. Definition 2 is logically consistent but unnecessary. It also defines a tie as an indifference which limits the generality of the analysis. In addition, it stipulates a logical relationship between xRy and yRx: xIy  xRy AND yRx, where AND is the logical and, whereas Axiom 1 only requires that “both [events] may [occur]” which is heuristically a tie.

Arrow states: “Axioms 1 and 2 do not exclude the possibility that for some distinct x and y, both xRy and yRx. A strong ordering, on the other hand, is a ranking in which no ties are possible.” Clearly, Arrow intends to define a tie as an indifference or he wouldn’t imply that his analysis includes the possibility of a tie. The statement that “a strong ordering ... is a ranking in which no ties are possible” is also not true. Consider the strong ranking operator P. In an election between two candidates, x and y, in which half the voters prefer x to y and half prefer y to x, the result is clearly a tie between x and y. One can also consider the result to be a tie between the orderings xPy and yPx or using my notation: {xPy, yPx}.

Arrow further compounds the error of Definition 1 with 

Lemma 1 (e):  “For all x and y, either xRy or yPx.” 

It can not be too strongly emphasized that not xRy is yRx or {xRy, yRx} xIy and not yPx. This makes sense heuristically since if y is not “preferred or indifferent to” x, then any of the following may be true: 1) x may be “preferred or indifferent” to y or 2) x may be indifferent to y or 3) x may be preferred to y. There is no way to know that xPy when the specification is xRy or yRx or xIy. If we could know that y is not preferred to x AND y is not indifferent to x, then we would know that xPy. But we only know that y is not “preferred or indifferent” to x. This is one of the problems of Arrow’s analysis that leads to the erroneous conclusion that social choice is impossible.

The Social Choice Function
The choice function, C(S), is defined by Arrow (p. 15) as follows: 	

Definition 3: “C(S) is the set of all alternatives in S such that, for every y in S, sRy.”
 
As such it can be used to specify ties among alternatives if the set, C(S), contains more than one element. Sen (1970, p. 48) says, "Arrow's impossibility theorem is precisely a result of demanding social orderings as opposed to choice functions." In other words, if the solutions required were simply alternatives, Arrow’s Impossibility Theorem would not apply. Since Arrow only uses it in his specification of the Condition of the Independence of Irrelevant Alternatives, it would have been more natural (and certainly stronger) to define C(S) as an ordering over a subset instead of the highest ranking alternative or set of alternatives in a subset. Why, if orderings are required in the solution, should only top-ranking alternatives be required in one of the conditions? And why, if orderings are required, should ties only be allowed among alternatives and not among orderings? We define an ordering function herein which strengthens the Condition of Independence of Irrelevant Alternatives and allows for ties among orderings as well as ties among alternatives.

Arrow (p.15) makes the following incorrect statement: “Each element of C(S) is to be preferred to all elements of S not in C(S) and indifferent to all elements of C(S); and, therefore, if x belongs to C(S), xRy for all y in S.” Assuming we have specified xRy, yRx or {xRy, yRx} for all x and y in S, we don’t know whether one element is preferred to another element—only that one element is “preferred or indifferent” to another element. Therefore, we can’t know that the elements of C(S) are “preferred to all elements of S not in C(S),” only that the elements of C(S) are “preferred or indifferent” to the elements not in C(S). For example, let’s assume the following set S and the following relationships:
			S = {x,y,z}
			xRy, yRz, xRz
Then, clearly C(S) = x. y and z are not in C(S), but x is not preferred to y and z but it is  “preferred or indifferent” to y and z and for every element, w, of S xRw which is all Definition 3 requires.

For x and y in C(S), xRy and yRx or {xRy, yRx}. For x in C(S) and z not in C(S), xRz and not zRx, not {xRz, zRx}. Arrow’s definition of C(S) makes sense. However, it would be clearer if the following statement would be appended: For y in C(S), xRy and yRx. For y not in C(S), xRy.

Clearly, for x and y in C(S), xIy according to Definition 2 and this is the basis for Arrow’s claim that he provides for ties. However, note that he has totally identified a tie as an indifference.

If the relationships among the alternatives in S are transitive according to Axiom 2 and either xRy or yRx but not both, then C(S) can contain only one element since the elements can be ordered, for example, as follows: xRyRz. If ties are possible, then, for example, we might have the following relationship:

			{xRy, yRx}, yRz, xRz

Here, according to Arrow’s Definition 2 we could write xIy and 
C(S) = {x,y}. Then the elements of C(S) are indifferent to each other but are “preferred or indifferent” (not “preferred” as Arrow states) to the element, z, not in C(S).

If the relationships are intransitive, then C(S) might be empty. For example, if xRy, yRz and zRx, then C(S) is empty since there is no element that is “preferred or indifferent” to every other element.

Consider the example: {xRy, yRx}, {yRz, zRy}, xRz or xIy, yRz, xRz according to Definition 2. What is C(S)? Clearly, C(S) is {x,y} since xRy, xRz and yRx, yRz. This does not violate Arrow’s Axiom 2 since xRy and yRz imply xRz. However, this does violate Arrow’s Lemma 1(d): If xIy and yIz, then xIz, since xIy and yIz but xRz. Clearly, Arrow’s definition of transitivity is incomplete because it doesn’t provide for the tie case.

Arrow goes on to compound the confusion over xPy and not yRx. He states: “Conversely, suppose C[x,y] contains the single element x. Since y does not belong to C[x,y]), not yRx; by Definition 1, xPy.” But not yRx is xRy or {xRy, yRx} by Axiom 1, not xPy. The root of the problem is that xPy implies xRy but not vice versa. Not yRx does not imply xPy, but it implies either xRy or {xRy, yRx}. Similarly, xIy implies xRy, but xRy does not imply xIy.

Consider xRyRz. Arrow would have us believe that since not yRx, xPy and since not zRy, yPz and, therefore, xPz. However, if x is preferred or indifferent to y and y is preferred or indifferent to z, then there is the possibility that x is indifferent to y and y is indifferent to z and, therefore, x is indifferent to z.  

Arrow’s Lemma 2 is also wrong.

LEMMA 2: A necessary and sufficient condition that xPy is that x be the sole element of C([x,y]).

However, if x is the sole element of C([x,y]), then xRy by Definition 3 and if xRy, C([x,y]) is x by definition. Therefore, xPy in Lemma 2 should be replaced by xRy.

Transitivity
So far we have been considering the relationship between just two alternatives: xRy. When there are three or more alternatives, then Arrow’s Axiom 2, the axiom of transitivity comes into effect. Transitivity limits the total number of logical possibilities to just those that heuristically and intuitively make sense. If x is “preferred or indifferent” to y and y is “preferred or indifferent” to z, then it makes sense that x should be “preferred or indifferent” to z. To say that z is “preferred or indifferent” to x in the last sentence would not make sense although xRy, yRz and zRx is a logical possibility. Thus transitivity imposes a rational limitation on the total number of logical possibilities. Another logical possibility, if xRy and zRy, then xRz or zRx, need not be stated since the implication does not imply a specific logical possibility. Either possibility is intuitively appropriate.

From Axiom 1 we know there are three possibilities: xRy, yRx or {xRy, yRx}. Arrow’s Axiom 2, while correct insofar as it goes, does not take into account the tie case: {xRy, yRx}. Heuristically, Axiom 2 makes sense as long as there are no ties involved. Therefore, it is fair to ask how the tie case affects the axiom of transitivity. There are 33 logical combinations of xRy, yRx and {xRy, yRx}. There are only the following cases involving ties. The relationships which are marked transitive are those which make intuitive sense and are non-trivial in that only one of the two possible logical implications makes sense. The logic is Relationship 1 and Relationship 2 imply Relationship 3.

Case	Relationship 1	Relationship 2	Relationship 3	Transitive?
	1	{xRy, yRx}	yRz	xRz	Yes
	2	{xRy, yRx}	yRz	zRx	No
	3	{xRy, yRx}	zRy	xRz	No
    4	{xRy, yRx}	zRy	zRx	Yes
	5		xRy		{yRz, zRy}	xRz	Yes
	6		xRy		{yRz, zRy}	zRx	No
	7		yRx		{yRz, zRy}	xRz	No
	8		yRx		{yRz, zRy}	zRx	Yes
	9		xRy		yRz	{xRz, zRx}	No
	10		xRy		zRy	{xRz, zRx}	Trivial
	11		yRx		yRz	{xRz, zRx}	Trivial
	12		yRx	zRy	{xRz, zRx}	No
	13	{xRy, yRx}	{yRz, zRy}	xRz	No
	14	{xRy, yRx}	{yRz, zRy}	zRx	No
	15	{xRy, yRx}	yRz	{xRz, zRx}	No
	16	{xRy, yRx}	zRy	{xRz, zRx}	No
	17		xRy	{yRz, zRy}	{xRz, zRx}	No
	18		yRx	{yRz, zRy}	{xRz, zRx}	No
	19	{xRy, yRx}	{yRz, zRy}	{xRz, zRx}	Yes

There are 5 non-trivial relationships from the above table:

1) If {xRy, yRx} and yRz, then xRz.
2) If {xRy, yRx} and zRy, then zRx.
3) If xRy and {yRz, zRy}, then xRz.
4) If yRx and {yRz, zRy}, then zRx.
5) If {xRy, yRx} and {yRz, zRy}, then {xRz, zRx}.

These 5 relationships along with the relationship from Axiom 2 define transitivity when ties are taken into consideration. Since 
xIy  {xRy, yRx} by Arrow, we can write the following transitive relationships.

1) xRyRz
2) xIyRz
3) zRxIy
4) xRyIz
5) yRxIz
6) xIyIz

Any permutation of x,y and z is allowed in the above relationships. These taken together with the trivial relationships are the only relationships allowed by a generalized version of Axiom 2. Instead of Arrow’s definition of indifference given above we can define a tie operator, T, such that xTy  {xRy, yRx}. This would be a more general relationship way of defining {xRy, yRx} than as an indifference.

A generalized version of Axiom 2, then, limits the relationships among three alternatives to the ones given above which can be expressed as alternative operator alternative operator alternative operator. Any relationship among three variables that cannot be expressed in this way is intransitive.

For three alternatives, Axiom 1 needs to be generalized also. There are six possible outcomes not involving ties: xRyRz, xRzRy, yRxRz, yRzRx, zRxRy, zRyRx. These correspond to xRy and yRx in the binary case. In order to be complete, we need to include all combinations of these outcomes as possible outcomes similar to the binary case.

Axiom 1 (generalized): For all x, y, z, either xRyRz or xRzRy or yRxRz or yRzRx or zRxRy or zRyRx or combinations of the type {xRyRz, xRzRy} or combinations of the type {xRyRz, xRzRy, yRxRz} or   combinations of the type {xRyRz, xRzRy, yRxRz, yRzRx} or  combinations of the type {xRyRz, xRzRy, yRxRz, yRzRx, zRxRy} or {xRyRz, xRzRy, yRxRz, yRzRx, zRxRy, zRyRx} where  is the binomial coefficient 

The Social Welfare Function (SWF)
Arrow defines a SWF as follows: 

Definition 4: By a social welfare function will be meant a process or rule which, for each set of individual orderings R1, ..., Rn for alternative social states (one ordering for each individual), states a corresponding social ordering of alternative social states, R.

Therefore, the scenario is that each individual voter/consumer specifies either xRiy or yRix or possibly, for the sake of completeness, {xRiy, yRix}. There is absolutely no way to determine whether or not a voter prefers x to y or y to x because his specification is that he “prefers or is indifferent” between x and y. We cannot ever conclude that not yRix is xPiy because we never know particular preference and indifference information—only “preference or indifference” information. We know from Axiom 1 that either xRiy or yRix or both which we write {xRiy, yRix}. Either we have xRiy or yRix or a tie. Arrow goes on to define a tie as an indifference. In any case, if yRix is not true, then xRiy or {xRiy, yRix} must be true. We can never get preference information out of the individual orderings R1, ..., Rn since it’s not called for in Definition 4.

Even with a formal logical consideration, preference information is not available. Let’s define xRy formally as xPy OR xIy where OR is logical or. Then if each individual specifies according to Axiom 1 either 1) xRiy, 2) yRix or 3) {xRiy, yRix} xIiy, one of the above three must be true and the other two false. Therefore, if NOT yRix is true where NOT is the logical not, then either xRiy or xIiy must be true. If individual i specifies xRiy, one cannot conclude xPiy; one can only conclude that i prefers x to y or is indifferent between x and y. We can only conclude that xPiy if we know that xRiy AND {NOT xIiy} where AND is the logical and. But by Definition 4 and Axiom 1, we do not elicit that information from individual i. We only elicit 1), 2) or 3) above and not {(2) AND NOT (3)}. 

Even without ties, NOT yRix  xPiy is not true. If we assume that each individual can only specify xRiy or yRix and cannot specify  {xRiy, yRix} xIiy, we still cannot extract preference information. If y is not preferred or indifferent to x, y can still be indifferent to x. Therefore, NOT yRix does not imply xPiy. The truth table is the following:

	yRix	NOT yRix 	yPix 	xPiy	xIiy	
	0		1	0	0	1
	0	1		0	1	0
	
Considering the SCF and Definition 4, if the SWF were xRyRz, the SCF would be x. If the SWF were {xRy, yRx}Rz or in Arrow’s notation xIyRz, the SCF would be [x,y]. The SCF truncates the SWF and yields the top alternative or set of alternatives in case the top alternatives are tied. With three alternatives it is possible in accordance with generalized Axiom 1 to have solutions of the form {xRzRy, zRyRx} or {zRyRx, yRxRz, yRzRx}. This generalizes directly from the binary case, {xRy, yRx}. The most general tie with three alternatives would be {xRyRz, xRzRy, yRxRz, yRzRx, zRxRy, zRyRx}.

With generalized Axiom 1, a tie refers to orderings and not to alternatives. The choice function C(S) would specify a tie between the alternatives x and y if xIyRz were the social ordering, for example. We are considering here ties among the orderings themselves and not just among the "top slot" of those orderings.

Let us assume there are six voter/consumers and that each specifies a different ordering. There is then one specification for each possible individual ordering. The common sense, heuristic solution is a tie among all the possible orderings. Similarly, there are 24  possible non-tie orderings for 4 alternatives, and, for the case of 24 voter/consumers, each specifying a different ordering, common sense would dictate a tie among all the possible social orderings. A similar case can be made for m=5, 6, ... . These are the broadest conceivable tie sets, and will be called maximal tie sets. Tie sets involving less than the total number of orderings are also possible and demanded by the completeness requirement.

Independence of Irrelevant Alternatives
Arrow postulates five conditions with which, in addition to Axioms 1 and 2, a SWF should be compliant. The only one of these that involves the SCF is Condition 3, the Independence of Irrelevant Alternatives.

Condition 3: “Let R1, ..., Rn and R'1, ..., R'n be two sets of individual orderings and let C(S) and C'(S) be the corresponding social choice functions. If, for all individuals i and x and y in a given environment S, xRiy if and only if xR'iy, then C(S) and C'(S) are the same.”

Arrow postulates that only the top slot of the orderings in environment S be the same since C(S) only selects the top slot. It might be asked why it shouldn’t be required that the top two slots of the orderings be the same or the top three slots etc. Therefore, for a general SWF solution such as 
		
			x1Rx2R ... xjRxj+1R ... xm-1Rxm

we can generalize the SCF as follows:

		C1 = x1
		C2 = x1Rx2
	.
	.
	.
	Cj = x1Rx2R ... Rxj

Cj is the ordering of the top j alternatives. In particular, if there are s alternatives in the environment S, then Cs would be the ordering of all those alternatives. C1 = C(S), Arrow’s social choice function.

As an example, let us assume that R corresponding to R1, ..., Rn is uRvRwRxRyRz and R' corresponding to R'1, ..., R'n is uRvRyRxRwRz and that S = {v, w, x, y}. Also all individuals vote exactly the same for v, w, x and y in the two sets of individual orderings. Then Arrow’s Condition 3 is met since C(S) = C'(S) = v. However, C2(S) = vRw ≠ C'2(S) = vRy. Similarly, C3(S) ≠ C'3(S) and C4(S) ≠ C'4(S). Arrow’s Condition 3 is the weakest possible requirement!

Consider another example. Let us assume that R corresponding to R1, ..., Rn is uRvRwRxRyRz and R' corresponding to R'1, ..., R'n is zRvRwRxRyRu and that S = {v, w, x, y}. Also all individuals vote exactly the same for v, w, x and y in the two sets of individual orderings. Then C(S) = C'(S) = v. C2(S) = C'2(S) = vRw. C3(S) = C'3(S) = vRwRx and C4(S) = C'4(S) = vRwRxRy.

When ties are considered, Arrow’s social choice function is inadequate. Consider the following example. R = {xRyRz, yRzRx, zRxRy}. R' = {zRxRy, yRzRx, xRzRy}. Let S = {x,y} and assume that all individuals vote exactly the same for x and y in the two cases. In S we have R = {xRy, yRx, xRy} and R' = {xRy, yRx, xRy}, and C(S) = C'(S) since R and R' are identical. However, there is no way of determining what C(S) is unless an additional rule is laid down for combining terms in the expression  {xRy, yRx, xRy}. Let us use the following rule: For m = 3 and S = {x,y}, let C2(S) = the maximum over R of xRy and yRx. In the tie expression above, {xRy, yRx, xRy}, there are 2 xRys and 1 yRx. Therefore, C2(S) = xRy.

The generalized SCF then consists of orderings and a rule for selecting these orderings when the SWF is singular or a tie.

The Ternary Case — n odd
We now proceed to demonstrate solutions which are social orderings for a specific SWF for the case m = 3 which satisfy a strengthened version of Arrow’s conditions. Let us assume alternatives x, y and z and n (odd) voter/consumers. As a consequence of Arrow’s Condition 3, the independence of irrelevant alternatives, we know that “knowing the social choices made in pairwise comparisons determines the entire social ordering.” Accordingly, we consider the social choices of the alternatives two by two. Our SWF is as follows. If N(x,y) > N(y,x), then xRy. If N(y,x) > N(x,y), then yRx. At the ternary level we have 8 cases:
		Case 1:	xRy, xRz, yRz
		Case 2:	xRy, xRz, zRy
		Case 3:	xRy, zRx, yRz
		Case 4:	xRy, zRx, zRy
		Case 5:	yRx, xRz, yRz
		Case 6:	yRx, xRz, zRy
		Case 7:	yRx, zRx, yRz
		Case 8:	yRx, zRx, zRy

According to the Condorcet (1785) method for determining the outcome of an election, we consider each of the alternatives in pairs, determine the winner for each pair and then determine the final social ordering by combining these results. We use the Condorcet method in our SWF for the above cases in which it actually produces a result. Therefore, we have the following:

		Case			Social Ordering
		1				xRyRz
		2				xRzRy
		4				zRxRy
		5				yRxRz
		7				yRzRx
		8				zRyRx

This leaves only cases 3 and 6. Consider the solution {xRyRz, yRzRx, zRxRy} for Case 3. We call a reduced ordering or reduced solution an ordering with one or more alternatives removed.  If we consider  {xRyRz, yRzRx, zRxRy} and remove z, we get {xRy, yRx, xRy}. Combining terms we have {2xRy, yRx}. If we choose the most numerous of xRy and yRx as the solution, we get xRy by 2 to 1 which is in accordance with the binary result.

Likewise, if we reduce {xRyRz, yRzRx, zRxRy} by y, we get {xRz, zRx, zRx} or {xRz, 2zRx}. 2zRx > xRz and we take zRx as the reduced solution which agrees with the known binary solution. Similarly, if we remove x from the social solution, we have {yRz, yRz, zRy} which yields yRz. Accordingly, our SWF algorithm is as follows:

	1) Choose the Condorcet solution if it exists.
	2) If the Condorcet solution doesn’t exist, construct a solution 	
	such that, when the solution is reduced by any single 	alternative, 	the most numerous of the remaining binary
	relationships is the same as the known binary solution.

Notice that our algorithm will always produce consistent results if the ternary solution is generated from the binary solution in such a way that there is a 2 to 1 ratio between the correct binary solution and the incorrect binary solution and then we take the larger of the two as our reduced solution. We construct our solutions in this manner in order to be compliant with Arrow’s Condition 3. Satisfying the other Conditions is then trivial and is easily shown. Whether or not such a solution always exists for m > 3 will be answered affirmatively elsewhere (Lawrence, 1998). Here all we need to show is the existence of a solution for Case 6. Consider the solution {yRxRz, xRzRy, zRyRx}. Reduction by z yields yRx; by y, xRz; by x, zRy. These all agree with the known binary cases and are consistent with the above definition.

Therefore, we have demonstrated a consistent algorithm for the SWF which yields the same social orderings when reduced from the ternary case to the binary case as those produced at the binary level directly from the domain. There is complete consistency of social orderings and not just of alternatives produced by the choice function. The choice function only produces the top position in an ordering.  We demand consistency over all orderings which can be produced by reducing a social ordering and this strengthens Arrow's Condition 3.

Arrow (1951, p. 26) states that “...suppose that an election system has been devised whereby each individual lists all the candidates in order of his preference and then, by a preassigned procedure, the winning candidate is derived from these lists. ...Suppose an election is held, with a certain number of candidates in the field, each individual filing his list of preferences, and then one of the candidates dies. Surely the social choice should be made by taking each of the individual's preference lists, blotting out completely the dead candidate's name, and considering only the orderings of the remaining names in going through the procedure of determining a winner.” This is precisely what we have done in choosing our SWF. Notice that it is completely consistent with the solutions for those cases determined by the Condorcet method and yields consistent results when the “dead” candidate is “blotted out” of the social ordering as well as when the “dead” candidate is blotted out of the individual orderings.

Let us examine the result produced by the SCF function for the above cases. Let S ={x,y}

Case	Social Ordering			C1		C2
1		xRyRz			x		xRy
2		xRzRy			x		xRy
3	{xRyRz, yRzRx, zRxRy}		x		xRy
4		zRxRy			z		xRy
5		yRxRz			y		yRx
6	{yRxRz, xRzRy, zRyRx}		y		yRx
7		yRzRx			y		yRx
8		zRyRx			y		yRx

Now let’s consider Arrow’s Condition 3 and let Ri  be such as to produce xRy, yRz, xRz (Case 1) and R'i be such as to produce xRy, yRz zRx (Case 3). S = {x,y}. Then C1(S) should equal C'1(S).  It does and also C2(S) = C'2(S) which is more than Arrow requires. Similarly, all the above cases satisfy not only Arrow’s Condition 3 which requires C1 but a stronger condition requiring C2. Examining the above relationships we notice that, regardless of the relationship between x and z and y and z, as long as xRy, C1 = x and C2 = xRy.

The Ternary Case — n even
When n is even we have a total of 27 cases. We have already considered the first 8 cases above. For convenience we define {xRy, yRx} as xTy instead of xIy since it is more general. In addition there is one more tie possibility, a three way tie: N(x,y) = N(y,x) = N(y,z) = N(z,y) = N(x,z) = N(z,x). We write this as {xRy, yRx, yRz, zRy, xRz, zRx} and define this as xTyTz. Solutions for the remaining cases are shown below. We also note the following transitivity requirements:

		1) If xRy and yTz, then xRz;
		2) If xTy and yRz, then xRz;
		3) If xTy and yTz, then xTz.

Case	Binary Solutions	Ternary Solution
 9:	xRy, xRz, yTz			xRyTz
 10:	xRy, zRx, yTz			{zRxRy, xRyTz, yTzRx}
 11:	yRx, xRz, yTz			{yRxRz, xRyTz, yTzRx}
 12:	yRx, zRx, yTz			yTzRx
 13:	xRy, xTz, yRz			{xRyRz, yRxTz, xTzRy}
 14:	xRy, xTz, zRy			xTzRy
 15:	yRx, xTz, yRz			yRxTz
 16:	yRx, xTz, zRy			{zRyRx, yRxTz, xTzRy}
 17:	xTy, xRz, yRz			xTyRz
 18:	xTy, xRz, zRy			{xRzRy, xTyRz, zRxTy}
 19:	xTy, zRx, yRz			{yRzRx, xTyRz, zRxTy}
 20:	xTy, zRx, zRy			zRxTy
 21:	xRy, xTz, yTz			{xRyTz, xTzRy, xTyTz}
 22:	yRx, xTz, yTz			{yRxTz, yTzRx, xTyTz}
 23:	xTy, xRz, yTz			{xRyTz, xTyRz, xTyTz}
 24:	xTy, zRx, yTz			{zRxTy, yTzRx, xTyTz}
 25:	xTy, xTz, yRz			{yRxTz, xTyRz, xTyTz}
 26:	xTy, xTz, zRy			{zRxTy, xTzRy, xTyTz}
 27:	xTy, xTz, yTz			xTyTz

Now, if xRy and S = {x,y}, we require C2 = C'2 regardless of the relationship between x and z and y and z. An examination of the above cases shows this to be true.

m = 4 — n odd
For m = 4 and no ties considered (n odd), we have the following solutions. (Note that abcd  aRbRcRd.) The generalized SCF amounts to starting with the result produced by the SWF and reducing the solution down to the set S where ST according to the above algorithm. This is completely in accordance with starting with the set S and generating a result according to the above SWF. In other words we can reduce down to S or expand up to S with the same results using the algorithm presented here. The results produced are identical and Cj(S) = C'j(S) in every case for 1 ≤ j≤ 3. 

Case 1:	aRb, aRc, aRd, bRc, bRd , cRd		Solution:	abcd
Case 2:	aRb, aRc, aRd, bRc, bRd, dRc		Solution:	abdc
Case 3:	aRb, aRc, aRd, bRc, dRb, cRd		Solution:	abcd, acdb, adbc
Case 4:	aRb, aRc, aRd, bRc, dRb, dRc		Solution:	adbc
Case 5:	aRb, aRc, aRd, cRb, bRd , cRd		Solution:	acbd
Case 6:	aRb, aRc, aRd, cRb, bRd , cRd		Solution:	abdc, acbd, adcb
Case 7:	aRb, aRc, aRd, cRb, dRb , cRd		Solution:	acdb
Case 8:	aRb, aRc, aRd, cRb, dRb , dRc		Solution:	adcb
Case 9:	aRb, aRc, dRa, bRc, bRd , cRd		Solution:	abcd, bcda, dabc
Case 10:	 aRb, aRc, dRa, bRc, bRd , dRc		Solution:	abdc, bdac, dabc
Case 11:	 aRb, aRc, dRa, bRc, dRb, cRd		Solution:	abcd, dabc, cdab
Case 12:	 aRb, aRc, dRa, bRc, dRb, dRc		Solution:	dabc
Case 13:	 aRb, aRc, dRa, cRb, bRd , cRd		Solution:	acbd, cbda, dacb
Case 14:	 aRb, aRc, dRa, cRb, bRd , dRc		Solution:	acbd, bdac, dacb
Case 15:	 aRb, aRc, dRa, cRb, dRb , cRd		Solution:	acdb, cdab, dacb
Case 16:	 aRb, aRc, dRa, cRb, dRb , dRc		Solution:	dacb
Case 17:	 aRb, cRa, aRd, bRc, bRd , cRd		Solution:	abcd, bcad, cabd
Case 18:	 aRb, cRa, aRd, bRc, bRd , dRc		Solution:	abdc, bdca, cabd
Case 19:	 aRb, cRa, aRd, bRc, dRb , cRd		Solution:	adbc, bcad, cadb
Case 20:	 aRb, cRa, aRd, bRc, dRb , dRc		Solution:	adbc, dbca, cadb
Case 21:	 aRb, cRa, aRd, cRb, bRd , cRd		Solution:	cabd
Case 22:	 aRb, cRa, aRd, cRb, bRd , dRc		Solution:	cabd, abdc, dcab
Case 23:	 aRb, cRa, aRd, cRb, dRb , cRd		Solution:	cadb
Case 24:	 aRb, cRa, aRd, cRb, dRb , dRc		Solution:	cadb, dcab, adcb
Case 25:	 aRb, cRa, dRa, bRc, bRd , cRd		Solution:	abcd, bcda, cdab
Case 26:	 aRb, cRa, dRa, bRc, bRd , dRc		Solution:	abdc, bdca, dcab
Case 27:	 aRb, cRa, dRa, bRc, dRb , cRd		Solution:	cdab, bcda, dabc
Case 28:	 aRb, cRa, dRa, bRc, dRb , dRc		Solution:	dcab, dbca, dabc
Case 29:	 aRb, cRa, dRa, cRb, bRd , cRd		Solution:	cabd, cbda, cdab
Case 30:	 aRb, cRa, dRa, cRb, bRd , dRc		Solution:	cabd, bdca, dcab
Case 31:	 aRb, cRa, dRa, cRb, dRb , cRd		Solution:	cdab
Case 32:	 aRb, cRa, dRa, cRb, dRb , dRc		Solution:	dcab
Case 33:	 bRa, aRc, aRd, bRc, bRd , cRd		Solution:	bacd
Case 34:	 bRa, aRc, aRd, bRc, bRd , dRc		Solution:	badc
Case 35:	 bRa, aRc, aRd, bRc, dRb , cRd		Solution:	acdb, bacd, dbac
Case 36:	 bRa, aRc, aRd, bRc, dRb , dRc		Solution:	adbc, badc, dbac
Case 37:	 bRa, aRc, aRd, cRb, bRd , cRd		Solution:	acbd, bacd, cbad
Case 38:	 bRa, aRc, aRd, cRb, bRd , dRc		Solution:	adcb, badc, cbad
Case 39:	 bRa, aRc, aRd, cRb, dRb , cRd		Solution:	acdb, bacd, cdba
Case 40:	 bRa, aRc, aRd, cRb, dRb , dRc		Solution:	adcb, badc, dcba
Case 41:	 bRa, aRc, dRa, bRc, bRd , cRd		Solution:	bacd, bcda, bdac
Case 42:	 bRa, aRc, dRa, bRc, bRd , dRc		Solution:	bdac
Case 43:	 bRa, aRc, dRa, bRc, dRb , cRd		Solution:	bacd, cdba, dbac
Case 44:	 bRa, aRc, dRa, bRc, dRb , dRc		Solution:	dbac
Case 45:	 bRa, aRc, dRa, cRb, bRd , cRd		Solution:	acbd, bdac, cbda
Case 46:	 bRa, aRc, dRa, cRb, bRd , dRc		Solution:	dacb, bdac, cbda
Case 47:	 bRa, aRc, dRa, cRb, dRb , cRd		Solution:	acdb, dbac, cdba
Case 48:  bRa, aRc, dRa, cRb, dRb , dRc		Solution:	dacb, dbac, dcba
Case 49:	 bRa, cRa, aRd, bRc, bRd , cRd		Solution:	bcad
Case 50:	 bRa, cRa, aRd, bRc, bRd , dRc		Solution:	badc, bcad, bdca
Case 51:	 bRa, cRa, aRd, bRc, dRb , cRd		Solution:	cadb, bcad, dbca
Case 52:	 bRa, cRa, aRd, bRc, dRb , dRc		Solution:	adbc, bcad, dbca
Case 53:	 bRa, cRa, aRd, cRb, bRd , cRd		Solution:	cbad
Case 54:	 bRa, cRa, aRd, cRb, bRd , dRc		Solution:	badc, cbad, dcba
Case 55:	 bRa, cRa, aRd, cRb, dRb , cRd		Solution:	cadb, cbad, cdba
Case 56:	 bRa, cRa, aRd, cRb, dRb , dRc		Solution:	adcb, cbad, dcba
Case 57:	 bRa, cRa, dRa, bRc, bRd , cRd		Solution:	bcda
Case 58:	 bRa, cRa, dRa, bRc, bRd , dRc		Solution:	bdca
Case 59:	 bRa, cRa, dRa, bRc, dRb , cRd		Solution:	bcda, cdba, dbca
Case 60:	 bRa, cRa, dRa, bRc, dRb , dRc		Solution:	dbca
Case 61:	bRa, cRa, dRa, cRb, bRd , cRd		Solution:	cbda
Case 62:	 bRa, cRa, dRa, cRb, bRd , dRc		Solution:	bdca, dcba, cbda
Case 63:	 bRa, cRa, dRa, cRb, dRb , cRd		Solution:	cdba
Case 64:  bRa, cRa, dRa, cRb, dRb , dRc		Solution:	dcba

Let S1 = {a}, S2 = {a, b}, S3 = {a, b, c}. Then C1(S1) = C'1(S1); C2(S2) = C'2(S2); C3(S3) = C'3(S3) for every possible case above by inspection. For example, consider Cases 59 and 60 as R and R', respectively. C1(S1) = C'1(S1) = a; C2(S2) = C'2(S2) = bRa; C3(S3) = C'3(S3) = bRcRa.

Conclusion
A generalized SCF was developed in which orderings can be extracted from a SWF for a set ST where the set T contains all m alternatives under consideration. The relationship between the SCF and the SWF was examined, and it was shown that solutions produced by a particular SWF for S are the same as the solutions produced by the highest order, generalized SCF for S. In other words, using the SWF and generating solutions for ST, yields the same results as taking the solutions produced by the SWF for the set T and applying the generalized SCF.

Arrow claims to provide for tie solutions using the SCF. However, the SCF can only produce ties among alternatives. Since the purpose of the SWF is to produce orderings, it would seem natural to explore the possibility of ties among orderings. In his Axiom 1 Arrow allows for ties between orderings for two alternatives. We have generalized this to allow for ties among orderings of three or more alternatives.

We have also generalized the axioms of transitivity and completeness to take into account tie solutions. If a solution produced by a SWF is a tie, then further measures can be taken to winnow the solution set to a singular solution at least in some cases. This is the subject of another paper and was not considered here.

We have pointed out several errors in Arrow’s logic—in particular involving his assertion that NOT yRx  xPy which is untrue. The logical errors throw into question Arrow’s entire analysis which, therefore, should not be considered cast in concrete.

We have shown that, when ties are properly considered, a SWF exists for m = 3 which complies with a strengthened version of Arrow’s Condition 3, the Independence of Irrelevant Alternatives. Elsewhere (Lawrence 1998), it has been shown that this result complies with Arrow’s other conditions and generalizes for any value of m and n. Therefore, social choice or, more precisely, a SWF does exist when ties among orderings are considered correctly.
References

1. K. J. Arrow, “Social Choice and Individual Values,”  John Wiley & Sons Inc., New York, 1951.
2. J. C. Lawrence, Neutrality and the Possibility of Social Choice, 1998 unpublished.
3. J. C. Lawrence, The Possibility of Social Choice for 3 Alternatives, 1998 unpublished.
4. J. C. Lawrence, The Possibility of Social Choice, 1998 unpublished.
5. Y. Murakami, “Logic and Social Choice,”  Routledge & Kegan Paul Ltd., London, 1968.
6. A. K. Sen, “Collective Choice and Social Welfare,”  Holden-Day, San Francisco, 1970.