"Maximize Affirmed Majorities" (MAM) is Independent of Clone Alternatives.

Revised:  December 19, 2003

The criterion independence of clone alternatives was first promoted by TN Tideman.  
For more information, including the rationale that justifies the criterion, see the document 
"Independence from Similar Alternatives."

Call a subset of alternatives B Í A a set of exact clones within A given 
a collection of votes R if and only if, for all pairs of alternatives x,y Î B, 
every vote in R expresses indifference between x and y.

Call a subset of alternatives B Í A a set of clones within A given a 
collection of votes R if and only if both of the following conditions hold: 
    (1)  For all x,y Î B and all z Î A\B, every vote in R that ranks z over x also 
          ranks z over y, and every vote in R that ranks x over z also ranks y over z.
    (2)  For all z Î A\B, B È {z} is not a set of exact clones within A given R. 

(Condition 2 is an innocuous technical condition which allows us to relax condition 1, 
thereby allowing some voters, but not all, to be indifferent between clones and some 
non-clones.  Thus our definition of clone alternatives is slightly broader than Tideman's.  
This makes our independence of clones criterion slightly stronger than his; however 
the proof of its satisfaction becomes slightly more complicated.)

For all non-empty B Í A and all collections of votes R, let R|_B denote 
the restriction of R to the alternatives in B.  That is, R|_B is the same as R  
except all preferences regarding alternatives not in B are deleted.

Call a function f a decision scheme if and only if, for all finite non-empty B Í A 
and all collections of votes R, the following three conditions hold: 
    (1)  For all x Ï B, f(x,B,R) = 0. 
    (2)  For all x Î A, 0 £ f(x,B,R) £ 1. 
    (3)  S_{(x Î B)}f(x,B,R) = 1. 
    (4)  For all x Î A and all C Í A, if B Í C then f(x,B,R) = f(x,B,R|_C). 

The interpretation of f(x,B,R) is the probability that alternative x will be elected from 
a subset of "nominees" B given votes R.  Condition 1 implies only an alternative in B, 
a nominee, will be elected.  Condition 2 reflects the fact that any probability is always a 
real number between 0 and 1.  Condition 3 reflects the fact that the sum of probabilities 
over an exhaustive set of mutually exclusive events is always 1, so a decision scheme 
often called a "lottery") corresponds is a voting procedure that always elects exactly one 
alternative. (In other words, a resolute procedure.)  The procedure is not necessarily 
always deterministic.  Note that this framework encompasses both deterministic and 
non-deterministic voting procedures (including those that are "usually" deterministic 
such as MAM).  Condition 4 implies the outcome will depend only on the restriction 
of the votes to the nominees, a mild condition which simplifies the notation in the 
current context where we are exploring the effect of changing the set of nominees.)

Call a decision scheme f independent of clone alternatives if and only if both of 
the following conditions hold for all finite A, all R, all finite non-empty C Í A 
that is a set of clones within A given R, and all D which is a strict subset of C: 
    (1)  For all x Î A\C, f(x,A,R) = f(x,A\D,R|_A\D).
    (2)  S_{(x Î C)}f(x,A,R) = S_{(x Î C)}f(x,A\D,R|_A\D).  

(Condition 1 means the probability any particular "non-clone" will be elected must be 
unaffected by deletion of a strict subset of the clones.  Condition 2 means the probability 
that the elected alternative will be within the set of clones must be unaffected by deletion 
of a strict subset of the clones.  Note that condition 2 automatically holds if condition 1 
holds since, by condition 3 of the definition of a decision scheme, the sum of the clones' 
probabilities of being elected is 1 minus the sum of the non-clones' probabilities of being 
elected, which is unaffected by deletion of a strict subset of clones if condition 1 holds.)

Satisfaction of independence of clone alternatives (ICA) implies manipulation-minded 
people cannot expect to affect the outcome by nominating alternatives which are nearly 
identical to those already on the ballot.  ICA is a stronger criterion than independence 
of exact clone alternatives (IECA, which would be defined in an analogous manner).  

For more information about these criteria and some examples showing how some voting 
procedures fail to satisfy them, see the draft "Independence from Similar Alternatives." 

Example 1:  Independence of clone alternatives.

Suppose there are 3 alternatives x, y and z.  
Suppose the voters vote the following rankings:

Clearly {y,z} is a set of clones since every voter either ranks x over both y and z  
or both y and z over x.  ICA requires that the probability that x is elected must not 
change if z is deleted.  Note that every voting procedure that reduces to majority rule 
when there are only two nominees (that is, nearly every voting procedure) will elect x  
if {x,y} is the set of nominees.  Thus any voting procedure that reduces to majority 
rule when there are only two nominees and is independent of clones must elect x  
if {x,y,z} is the set of nominees.  Otherwise, would-be manipulators will nominate 
alternatives like z that are obviously inferior (to y).  The Borda procedure, for instance,  
elects x from {x,y} but elects y from {x,y,z}.  Thus the faction that nominates the 
most alternatives has an unfair advantage under Borda, creating an incentive to 
nominate many inferior alternatives, thereby making farces of elections. (It can be 
seen from this example that ICA is closely related to the criterion independence 
from Pareto-dominated alternatives.)  

Example 2:  Complete independence of clone alternatives in a "tied" scenario.

Suppose there are 3 alternatives x, y and z.  
Suppose the voters vote the following rankings:

It can be seen that {y,z} is a set of clones since every voter either ranked x  
over both y and z or both y and z over x.  ICA requires that the probability 
that x is elected must not change if z is deleted from the set of alternatives 
and from the votes.  Clearly, any voting procedure which is anonymous  
and neutral will give x a ½ chance of being elected if {x,y} is the set of 
nominated alternatives.  Thus any voting procedure which is anonymous, 
neutral and independent of clones must give x a ½ chance of being elected 
if {x,y,z} is the set of nominees. (By similar reasoning, since {x,y} is also 
a set of clones, any voting procedure which is anonymous, neutral and 
independent of clones must give z a ½ chance of being elected.  Thus, 
since the sum of all alternatives' probabilities of being elected must be 1, 
such a procedure must give y zero chance of being elected.)

Theorem "MAM is Independent of Clones."  

Proof:  Refer to the definitions in the document "A formal definition of MAM".  
Pick any set of alternatives A and any collection of votes R.  Assume C Í A is a 
finite non-empty set of clones within A given R and assume D is a strict subset of C.  
Abbreviate A' = A\D and R' = R|_A' ( the restriction of R to A').  Pick any x Î A\C.  
Let MAM_x denote the probability that MAM elects x from A given R.  Let MAM'_x 
denote the probability that MAM elects x from A' given R'.  To show that MAM 
satisfies ICA, we must show MAM_x = MAM'_x. (In other words, that MAM satisfies 
condition 1 of the definition of ICA.  As noted above, satisfaction of condition 2 
follows immediately if condition 1 is satisfied.)  Let W_x = {s Î L(R,A) such that 
MAM(A,R,s) = x}.  Thus MAM_x = #W_x/#L(R,A).  Let W'_x = {s' Î L(R',A') 
such that MAM(A',R',s') = x}.  Thus MAM'_x = #W'_x/#L(R',A').  We must 
establish the following proposition: 

        Proposition (1)  #W_x/#L(R,A) = #W'_x/#L(R',A').

Proof of 1:  By inspection of the definition of L(.,.), #L(R,A) = (#R)! ´ (#A)! 
and #L(R',A') = (#R')! ´ (#A')!.  Since #R' = #R, the following statement holds:

(1.2)  #L(R,A)/#L(R',A') = ((#A)!)/((#A')!) 

Thus proposition 1 will follow immediately if we establish the following proposition:

Proposition (1.3):  #W_x = #W'_x ´ ((#A)!)/((#A')!). 

For all s Î L(R,A), let s|_A' denote the restriction of s to A'.  In other words, 
s|_A' = (s_R,s_A)|_A' = (s_R|_A',s_A|_A') where s_R|_A' is the same as s_R except that 
all alternatives in D are deleted, and s_A|_A' is the same as s_A except all alternatives 
in D are deleted. (Thus s|_A' Î L(R',A').)  By 1.2, the following statement holds: 

(1.4)  For all s' Î L(R',A'), #{s Î L(R,A) such that s|_A' = s'} = ((#A)!)/((#A')!) 

Pick any s Î L(R,A).  By 1.4, proposition 1.3 will follow immediately if we 
establish the following proposition:

Proposition (1.5)  s Î W_x if and only if s|_A' Î W'_x.

Let XPairs denote Pairs(A)\Pairs(C). (In other words, XPairs is the ordered pairs of 
alternatives involving at least one non-clone.)  Let XPairs' denote Pairs(A')\Pairs(C). 
(The ordered pairs of alternatives in A' involving at least one non-clone.)  Make the 
following abbreviations:  

Maj = Majorities(A,R). 
RVH = RVH(A,R,s). 
Precede(a,b) = Precede(a,b,A,R,s) for all a,b Î A.
Aff = Affirmed(A,R,s). 
Top = Top(A,R,s). 
MAM = MAM(A,R,s). 

s' = s|_A'.  
Maj' = Majorities(A',R'). 
RVH' = RVH(A',R',s'). 
Precede'(a,b) = Precede(a,b,A',R',s') for all a,b Î A'.
Aff' = Affirmed(A',R',s'). 
Top' = Top(A',R',s'). 
MAM' = MAM(A',R',s'). 

The following propositions, 1.6 through 1.18, will facilitate the proof of 1.5:

Proposition (1.6): "Random Voter Hierarchy ranks clones together."  
That is, for all y Î A\C, RVH either ranks y over every alternative in C 
or ranks every alternative in C over y. 

Proof of 1.6:  Pick any y Î A\C and any c,c' Î C.  By condition 2 of 
the definition of clones, at least one vote ranks y over c or c over y.  
This means R^(yc) is not empty.  Thus, by condition 1 of the definition 
of clones, R^(yc') is not empty.  Furthermore, condition 1 of the definition 
of clones implies R^syc = R^syc'. (That is, the vote in R^(yc) ranked highest 
by s_R is the same as the vote in R^(yc') ranked highest by s_R.)  Furthermore, 
condition 1 of the definition of clones implies R^syc ranks y relative to c 
the same as R^syc' ranks y relative to c'.  It follows by the definition of 
Random Voter Hierarchy that RVH either ranks y over both c and c' 
or ranks both c and c'  over y.  Since y, c and c' were picked arbitrarily, 
1.6 is established.         QED 

Proposition (1.7):  For all a,b Î A', R(a,b)|_A' = R'(a,b). (That is, 
the restriction to A' of the votes in R that rank a over b is the same 
as the votes in R' that rank a over b.) 

Proof of 1.7:  This follows by inspection of the definitions of A' and R'.   QED 

Proposition (1.8):  Maj Ç Pairs(A') = Maj'. 

Proof of 1.8:  This follows by 1.7 and the definition of Majorities().      QED 

Proposition (1.9):  For all a,b Î A', RVH ranks a over b 
if and only if RVH' ranks a over b. 

Proof of 1.9:  This follows immediately from theorem "Random Voter 
Hierarchy is Independent of Irrelevant Alternatives (1)".           QED 

Proposition (1.10):  For all a,b Î A', Precede(a,b) Ç Pairs(A') = Precede'(a,b). 

Proof of 1.10:  This follows immediately from theorem "Precedence is 
Independent of Irrelevant Alternatives."    QED 

Proposition (1.11):  For all y Î A\C and all c,c' Î C, 
the following four statements hold: 
       (1.11.1)  R(y,c) = R(y,c'). 
       (1.11.2)  R(c,y) = R(c',y).  
       (1.11.3)  (y,c) Î Maj if and only if (y,c') Î Maj. 
       (1.11.4)  (c,y) Î Maj if and only if (c',y) Î Maj. 

Proof of 1.11:  Statements 1.11.1 and 1.11.2 are implied by condition 1 of 
the definition of clones.  Statements 1.11.3 and 1.11.4 follow from 1.11.1 
and 1.11.2 and inspection of the definition of Majorities().     QED 

Proposition (1.12):  For all a Î A, all y,z Î A\C and all c,c' Î C, 
all four of the following statements hold: 
       (1.12.1)  (c,y) Î Precede(z,a) if and only if (c',y) Î Precede(z,a).  
       (1.12.2)  (y,c) Î Precede(a,z) if and only if (y,c') Î Precede(a,z).  
       (1.12.3)  (y,a) Î Precede(c,z) if and only if (y,a) Î Precede(c',z).  
       (1.12.4)  (a,y) Î Precede(z,c) if and only if (a,y) Î Precede(z,c'). 

Proof of 1.12:  By symmetry, we need only establish the "only if" clauses.  
First we establish 1.12.1:  Pick any a Î A, any y,z Î A\C and any c,c' Î C.  
Assume (c,y) Î Precede(z,a).  We must show (c',y) Î Precede(z,a).  
Since (c,y) Î Precede(z,a), one of the following four conditions must hold:

(1.12.1.1)  #R(c,y) > #R(z,a). 
(1.12.1.2)  #R(c,y) = #R(z,a) and #R(y,c) < #R(a,z). 
(1.12.1.3)  #R(c,y) = #R(z,a) and #R(y,c) = #R(a,z) 
                 and RVH ranks a over y. 
(1.12.1.4)  #R(c,y) = #R(z,a) and #R(y,c) = #R(a,z) 
                 and a = y and RVH ranks c over z.

By 1.11.1, #R(c,y) = #R(c',y).  By 1.11.2, #R(y,c) = #R(y,c').  
By 1.6, RVH ranks c' over z if and only if RVH ranks c over z.  
Therefore one of the following four conditions must hold:

(1.12.1.5)  #R(c',y) > #R(z,a). 
(1.12.1.6)  #R(c',y) = #R(z,a) and #R(y,c') < #R(a,z). 
(1.12.1.7)  #R(c',y) = #R(z,a) and #R(y,c') = #R(a,z) 
                 and RVH ranks a over y. 
(1.12.1.8)  #R(c',y) = #R(z,a) and #R(y,c') = #R(a,z) 
                 and a = y and RVH ranks c' over z.

By the definition of precedence, (c',y) Î Precede(z,a), establishing the 
"only if" clause of 1.12.1.  By symmetry, the "if" clause must also hold.  
Thus 1.12.1 has been established.  By similar reasoning, 1.12.2, 
1.12.3 and 1.12.4 can be established.       QED  

Proposition (1.13):  All four of the following statements hold: 
        (1.13.1)  For all y Î A\C and all c,c' Î C, 
                       (y,c) Î Aff if and only if (y,c') Î Aff. 
        (1.13.2)  For all y Î A\C and all c,c' Î C, 
                      (c,y) Î Aff if and only if (c',y) Î Aff.  
        (1.13.3)  For all y Î A\C and all c,c' Î C\D, 
                       (y,c) Î Aff' if and only if (y,c') Î Aff'. 
        (1.13.4)  For all y Î A\C and all c,c' Î C\D, 
                       (c,y) Î Aff' if and only if (c',y) Î Aff'.  

Proof of 1.13:  First we prove 1.13.1 and 1.13.2:  For all a,b Î A, 
abbreviate Aff_ab = Aff Ç Precede(a,b).  Let Q denote the union of 
the following four sets:   
        Q₁ = {(y,c) Î Aff such that y Î A\C and c Î C  
                  and there exists c' Î C such that (y,c') Ï Aff}. 
        Q₂ = {(y,c) Î Pairs(A)\Aff such that y Î A\C and c Î C  
                  and there exists c' Î C such that (y,c') Î Aff}. 
        Q₃ = {(c,y) Î Aff such that y Î A\C and c Î C  
                  and there exists c' Î C such that (c',y) Ï Aff}. 
        Q₄ = {(c,y) Î Pairs(A)\Aff such that y Î A\C and c Î C  
                  and there exists c' Î C such that (c',y) Î Aff}.  
To establish 1.13.1 and 1.13.2, we must show Q is empty.  Suppose the contrary.  
By theorem "Precedence is a Strict Ordering" we can pick (a,b) Î Q such that 
Q Ç Precede(a,b) is empty. (That is, (a,b) is the pair in Q preceded by no other 
pair in Q.)  There are four cases to consider:  

Case 1.13.5:  (a,b) ΠQ₁.  This means a Î A\C and b Î C and (a,b) Î Aff 
and there exists c Î C such that (a,c) Ï Aff.  By the definition of Affirmed(), 
Aff Í Maj, so (a,b) Î Maj.  By 1.11.3, (a,c) Î Maj.  Since (a,c) Î Maj\Aff, 
by the definition of Affirmed() (a,c) must be inconsistent with Aff_ac.  This 
means there exist alternatives z₁,z₂,...,z_k Î A such that z₁ = c and z_k = a 
and {(z₁,z₂),(z₂,z₃),...,(z_k-1,z_k)} Í Aff_ac.  Since z₁ Î C, we can let i denote 
the largest integer in {1,2,...,k} such that z_i Î C.  Since z_k = a Î A\C, i < k.  
Since (z_k,c) Î Maj, (c,z_k) Ï Maj.  By 1.11.4, (c',z_k) Ï Maj for all c' Î C.  
Since (z_k-1,z_k) Î Maj, this means z_k-1 Ï C, so i < k-1.  Thus z_i+1 Î A\C and 
z_i+2 Î A\C.  Thus we can let P denote {(z_i+1,z_i+2),...,(z_k-1,z_k)} È {(z_k,b)}.  
Since P Í Aff, (b,z_i+1) is inconsistent with Aff.  By theorem "Consistency 
Maintained", Aff is internally consistent, so (b,z_i+1) Ï Aff.  Since (z_i,z_i+1) Î Aff 
and (b,z_i+1) Ï Aff, (z_i,z_i+1) Î Q₃.  Since (z_i,z_i+1) Î Precede(a,c), by 1.12.4 
(z_i,z_i+1) Î Precede(a,b).  Thus (z_i,z_i+1) Î Q Ç Precede(a,b).  But this is a 
contradiction since Q Ç Precede(a,b) is empty, so this case is impossible.  

Case 1.13.6:  (a,b) ΠQ₂.  This means a Î A\C and b Î C and (a,b) Ï Aff 
and there exists c Î C such that (a,c) Î Aff.  By the definition of Affirmed(), 
Aff Í Maj, so (a,c) Î Maj.  By 1.11.3, (a,b) Î Maj.  Since (a,b) Î Maj\Aff, 
by the definition of Affirmed() (a,b) must be inconsistent with Aff_ab.  This 
means there exist alternatives z₁,z₂,...,z_k Î A such that z₁ = b and z_k = a and 
{(z₁,z₂),(z₂,z₃),...,(z_k-1,z_k)} Í Aff_ab.  Since z₁ Î C, we can let i denote the 
largest integer in {1,2,...,k} such that z_i Î C.  Since z_k = a Î A\C, i < k.  
Since (z_k,c) Î Maj, (c,z_k) Ï Maj.  By 1.11.4, (c',z_k) Ï Maj for all c' Î C.  
Since (z_k-1,z_k) Î Maj, this means z_k-1 Ï C, so i < k-1.  Thus z_i+1 Î A\C and 
z_i+2 Î A\C.  Thus we can let P denote {(z_i+1,z_i+2),...,(z_k-1,z_k)} È {(z_k,c)}.  
Since P Í Aff, (c,z_i+1) is inconsistent with Aff.  By theorem "Consistency 
Maintained", Aff is internally consistent, so (c,z_i+1) Ï Aff.  Since (z_i,z_i+1) Î Aff 
and (c,z_i+1) Ï Aff, (z_i,z_i+1) Î Q₃.  Thus (z_i,z_i+1) Î Q Ç Precede(a,b).  
But this is a contradiction since Q Ç Precede(a,b) is empty, so this case 
is impossible.  

Case 1.13.7:  (a,b) ΠQ₃.  This means a Î C and b Î A\C and (a,b) Î Aff 
and there exists c Î C such that (c,b) Ï Aff.  By the definition of Affirmed(), 
Aff Í Maj, so (a,b) Î Maj.  By 1.11.4, (c,b) Î Maj.  Since (c,b) Î Maj\Aff, 
by the definition of Affirmed() (c,b) must be inconsistent with Aff_cb.  This 
means there exist alternatives z₁,z₂,...,z_k Î A such that z₁ = b and z_k = c and 
{(z₁,z₂),(z₂,z₃),...,(z_k-1,z_k)} Í Aff_cb.  Since z_k Î C, we can let i denote the 
smallest integer in {1,2,...,k} such that z_i Î C.  Since z₁ = b Ï C, i > 1.  
Since (c,z₁) Î Maj, (z₁,c) Ï Maj.  By 1.11.3, (z₁,c') Ï Maj for all c' Î C.  
Since (z₁,z₂) Î Maj, this means z₂ Ï C, so i > 2.  Thus z_i-1 Î A\C and 
z_i-2 Î A\C.  Thus we can let P denote {(a,z₁)} È {(z₁,z₂),...,(z_i-2,z_i-1)}.  
Since P Í Aff, (z_i-1,a) is inconsistent with Aff.  By theorem "Consistency 
Maintained", Aff is internally consistent, so (z_i-1,a) Ï Aff.  Since (z_i-1,z_i) Î Aff 
and (z_i-1,a) Ï Aff, (z_i-1,z_i) Î Q₁.  Since (z_i-1,z_i) Î Precede(c,b), by 1.12.3 
(z_i-1,z_i) Î Precede(a,b).  Thus (z_i-1,z_i) Î Q Ç Precede(a,b).  But this is a 
contradiction since Q Ç Precede(a,b) is empty, so this case is impossible.  

Case 1.13.8:  (a,b) ΠQ₄.  This means a Î C and b Î A\C and (a,b) Ï Aff 
and there exists c Î C such that (c,b) Î Aff.  By the definition of Affirmed(), 
Aff Í Maj, so (c,b) Î Maj.  By 1.11.4, (a,b) Î Maj.  Since (a,b) Î Maj\Aff, 
by the definition of Affirmed() (a,b) must be inconsistent with Aff_ab.  This 
means there exist alternatives z₁,z₂,...,z_k Î A such that z₁ = b and z_k = a and 
{(z₁,z₂),(z₂,z₃),...,(z_k-1,z_k)} Í Aff_ab.  Since z_k Î C, we can let i denote the 
smallest integer in {1,2,...,k} such that z_i Î C.  Since z₁ = b Ï C, i > 1.  
Since (c,z₁) Î Maj, (z₁,c) Ï Maj.  By 1.11.3, (z₁,c') Ï Maj for all c' Î C.  
Since (z₁,z₂) Î Maj, this means z₂ Ï C, so i > 2.  Thus z_i-1 Î A\C and 
z_i-2 Î A\C.  Thus we can let P denote {(c,z₁)} È {(z₁,z₂),...,(z_i-2,z_i-1)}.  
Since P Í Aff, (z_i-1,c) is inconsistent with Aff.  By theorem "Consistency 
Maintained", Aff is internally consistent, so (z_i-1,c) Ï Aff.  Since (z_i-1,z_i) Î Aff 
and (z_i-1,c) Ï Aff, (z_i-1,z_i) Î Q₁.  Thus (z_i-1,z_i) Î Q Ç Precede(a,b).  But this 
is a contradiction since Q Ç Precede(a,b) is empty, so this case is impossible.  

Since all four cases are impossible, the contrary assumption that Q is not empty 
cannot hold.  Thus 1.13.1 and 1.13.2 are established.

Next we prove 1.13.3 and 1.13.4:  Note that C\D is a set of clones within A'  
given R'.  Since 1.13.1 and 1.13.2 hold for any A, any R, and any C which is 
a set of clones within A given R, 1.13.3 and 1.13.4 can be established by 
substituting A' for A, R' for R, and C\D for C within 1.13.1 and 1.13.2.     QED

Proposition (1.14):  For all (a,b) Î XPairs and all P Í Aff Ç Precede(a,b), 
if (a,b) is inconsistent with P and (a,b) is consistent with all subsets of 
Aff Ç Precede(a,b) smaller than P, then at most one alternative in C is 
involved in any pairs in P.  

Proof of 1.14:  For all a,b Î A, abbreviate Aff_ab = Aff Ç Precede(a,b).  
Pick any (a,b) Î XPairs and any P Í Aff_ab.  Assume (a,b) is inconsistent 
with P and consistent with all  subsets of Aff_ab smaller than P.  This means 
there exist alternatives z₁,z₂,...,z_k Î A such that z₁ = b and z_k = a and 
P = {(z₁,z₂),(z₂,z₃),...,(z_k-1,z_k)}.  We must show that at most one alternative 
in C is in {z₁,z₂,...,z_k}.  Suppose the contrary.  This implies means we can 
let s denote the smallest integer in {1,2,...,k} such that z_i Î C and we can 
let l denote the largest integer in {1,2,...,k} such that z_i Î C.  Clearly s < l.  
There are three cases to consider:  

Case 1.14.1:  a Î C.  This means l = k.  Also, since (a,b) Î XPairs, b Ï C.  
Thus s > 1.  Since (z_s-1,z_s) Î Aff_ab, it follows by 1.12.2 and  1.13.1 that 
(z_s-1,a) Î Aff_ab.  But since {(z_i,z_i+1) Î P such that 1 £ i < s - 1} È {(z_s-1,a)} 
is a subset of Aff_ab smaller than P and (a,b) is inconsistent with it, this is 
a contradiction, which means this case is impossible.   

Case 1.14.2:  b Î C.  This means s = 1.  Also, since (a,b) Î XPairs, a Ï C.  
Thus l < k.  Since (z_l,z_l+1) Î Aff_ab, it follows by 1.12.1 and 1.13.2 that 
(b,z_l+1) Î Aff_ab.  But since {(b,z_l+1)} È {(z_i,z_i+1) Î P such that l + 1 £ i < k} 
is a subset of Aff_ab smaller than P and (a,b) is inconsistent with it, this is 
a contradiction, which means this case is impossible.   

Case 1.14.3:  a Ï C and b Ï C.  Thus 1 < s < l < k.  Since (z_s-1,z_s) Î Aff_ab, 
it follows by 1.12.2 and  1.13.1 that (z_s-1,z_l) Î Aff_ab.  But since {(z_i,z_i+1) Î P 
such that 1 £ i < s - 1} È {(z_s-1,z_l)}  È {(z_i,z_i+1) Î P such that l + 1 £ i < k} 
is a subset of Aff_ab smaller than P and (a,b) is inconsistent with it, this is 
a contradiction, which means this case is impossible.   

Since all three cases are impossible, the contrary assumption cannot hold.  
Thus 1.14 is established.                                                                      QED

Proposition (1.15):  For all (a,b) Î XPairs' Ç Maj, (a,b) Ï Aff if and only if 
there exists P Í Aff Ç Precede(a,b) such that both of the following hold:

(1.15.1)  (a,b) is inconsistent with P. 
(1.15.2)  P Í XPairs'.

Proof of 1.15:  The "if" clause follows immediately from condition 1.15.1 
and the definition of Affirmed().  We next prove the "only if" clause:  
Assume (a,b) Î XPairs' Ç Maj and (a,b) Ï Aff.  We must show there exists 
P Í Aff Ç Precede(a,b) Ç XPairs' such that (a,b) is inconsistent with P.  
Since (a,b) Î Maj\Aff, (a,b) must be inconsistent with Aff Ç Precede(a,b).  
By induction, there must exist P₀ Í Aff Ç Precede(a,b) such that (a,b) is 
inconsistent with P₀ and consistent with all subsets of Aff Ç Precede(a,b) 
smaller than P₀.  By the definition of Affirmed(), Aff Í Maj.  Thus P₀ Í Maj.  
By 1.14, at most one alternative in C is involved in any pairs in P₀.  
Since (z,z) Ï Maj for all z Î A, this implies P₀ Í XPairs.  
There are two cases to consider: 

Case 1.15.3:  No alternative in C is involved in any pair in P₀.  This  
means P₀ Í Pairs(A\C).  Since Pairs(A\C) Í XPairs', P₀ Í XPairs'.  
Relabel P = P₀.  Clearly P Í Aff Ç Precede(a,b) Ç XPairs'  
and (a,b) is inconsistent with P, the desired conclusion.

Case 1.15.4:  Exactly one alternative in C is involved in any pairs in P₀.  
Let c denote this alternative.  There are two subcases to consider:

Case 1.15.4.1:  c Ï D.  This implies P₀ Í XPairs'.  Relabel P = P₀.  
Clearly P Í Aff Ç Precede(a,b) Ç XPairs' and (a,b) is inconsistent 
with P, the desired conclusion.

Case 1.15.4.2:  c Î D.  Represent P₀ as {(a₁,a₂),(a₂,a₃),...,(a_k-1,a_k)} 
where a₁ = b and a_k = a and there exists at least one j Î {1,2,...,k} such 
that a_j = c.  Since (a,b) Î XPairs', a₁ Ï D and a_k Ï D.  Thus we can 
let i denote the smallest integer in {2,3,...,k-1} such that a_i = c and 
we can let j denote the largest integer in {2,3,...,k-1} such that a_j = c.   
Clearly 1 < i £ j < k.  By construction, a_i-1 Î A\C and a_j+1 Î A\C.  
Pick any c' Î C\D.  Since (a_i-1,a_i) Î Aff, by 1.13.1 (a_i-1,c') Î Aff.  Since 
(a_j,a_j+1) Î Aff, by 1.13.2 (c',a_j+1) Î Aff.  Since (a_i-1,a_i) Î Precede(a,b), 
by 1.12.2 (a_i-1,c') Î Precede(a,b).  Since (a_j,a_j+1) Î Precede(a,b), 
by 1.12.1 (c',a_j+1) Î Precede(a,b).  Let P denote the following set: 
       {(a₁,a₂),(a₂,a₃),...,(a_i-1,c'),(c',a_j+1),(a_j+1,a_j+2),...,(a_k-1,a_k)} 
Clearly P Í Aff Ç Precede(a,b) Ç XPairs' and (a,b) is inconsistent 
with P, the desired conclusion.

Thus in all cases there exists P Í Aff Ç Precede(a,b) Ç XPairs' such that (a,b) is 
inconsistent with P, establishing the "only if" clause of 1.15.  Since both the "if" 
and "only if" clauses of 1.15 have been established, 1.15 is established.    QED

Proposition (1.16):  Aff Ç XPairs' = Aff' Ç XPairs'.

Proof of 1.16:  Let P denote {(a,b) Î XPairs' such that (a,b) Î Aff\Aff'  
or (a,b) Î Aff'\Aff}.  We must show P is empty.  Suppose the contrary.  
By theorem "Precedence is a Strict Ordering" we can pick (a,b) Î P  
such that the following condition holds:

(1.16.1)  P Ç Precede'(a,b) is empty.  

There are two cases to consider:

Case 1.16.2:  (a,b) Î Aff.  This implies (a,b) Ï Aff'.  By the definition 
of Affirmed(), Aff Í Maj.  Thus (a,b) Î Maj.  By 1.8, (a,b) Î Maj'.  
Since (a,b) Î Maj'\Aff', (a,b) must be inconsistent with Aff' Ç Precede'(a,b).  
By 1.16.1, Precede'(a,b) Ç P is empty.  Thus Aff' Ç Precede'(a,b) Í Aff.  
By 1.10, Precede'(a,b) Í Precede(a,b).  Combining the last two statements, 
Aff' Ç Precede'(a,b) Í Aff Ç Precede(a,b).  This implies (a,b) is inconsistent 
with Aff Ç Precede(a,b), which implies (a,b) Ï Aff.  But (a,b) Î Aff in 
case 1.16.2, a contradiction.  Thus case 1.16.2 is impossible.

Case 1.16.3:  (a,b) Ï Aff.  This implies (a,b) Î Aff'.  By the definition 
of Affirmed(), Aff' Í Maj'.  Thus (a,b) Î Maj'.  By 1.8, (a,b) Î Maj.  
By 1.15, there must exist P₂ Í Aff Ç Precede(a,b) Ç XPairs' such that (a,b)  
is inconsistent with P₂.  By 1.10, P₂ Í Precede'(a,b).  By 1.16.1, this implies 
no pair in P₂ is in P.  This implies P₂ Í Aff'.  Thus P₂ Í Aff' Ç Precede'(a,b), 
which implies (a,b) is inconsistent with Aff' Ç Precede'(a,b).  This implies 
(a,b) Ï Aff'.  But (a,b) Î Aff' in case 1.16.3,  a contradiction.  
Thus case 1.16.3 is impossible.

Since both cases are impossible, the contrary assumption that P is not empty 
cannot hold.  Thus P must be empty, and 1.16 follows immediately.        QED 

Proposition (1.17):  A\C Ç Top = A\C Ç Top'.

Proof of 1.17:  First we show A\C Ç Top Í A\C Ç Top'.  
Pick any y Î A\C Ç Top.  We must show y ΠTop'.  Suppose the contrary.  
By the definition of Top(), y is not second in any pair in Aff, and y is second 
in at least one pair in Aff'.  This means there exists z Î A'\{y} such that 
(z,y) Î Aff'.  Clearly (z,y) Î XPairs'.  By 1.16, (z,y) Î Aff, which implies 
y Ï Top.  But this is a contradiction, which implies the contrary assumption 
cannot hold, establishing A\C Ç Top Í A\C Ç Top'.  
Next we show A\C Ç Top' Í A\C Ç Top.  Pick any y Î A\C Ç Top'.  
We must show y ΠTop.  Suppose the contrary.  By the definition of Top(), 
y is not second  in any pair in Aff', and y is second  in at least one pair 
in Aff.  This means there exists z Î A\{y} such that (z,y) Î Aff.  
There are two cases to consider:

Case 1.17.1:  z Î D.  Since D is a strict subset of C, by 1.13.2 there 
must exist c Î C\D such that (c,y) Î Aff.  Clearly (c,y) Î XPairs'.  
By 1.16, (c,y) Î Aff'.  This implies y Ï Top', a contradiction.  
Thus case 1.17.1 is impossible.  

Case 1.17.2:  z Ï D.  Clearly (z,y) Î XPairs'.  By 1.16, (z,y) Î Aff'.  
This implies y Ï Top', a contradiction.  Thus case 1.17.2 is impossible.

Since both cases are impossible, the contrary assumption cannot hold, 
which means y Î Top.  Thus A\C Ç Top' Í A\C Ç Top.  
Thus A\C Ç Top = A\C Ç Top', establishing 1.17.                     QED 

Proposition (1.18):  C Ç Top is empty if and only if C Ç Top' is empty.

Proof of 1.18:   First we prove the "only if" clause:  Assume C Ç Top is empty.  
We must show C Ç Top' is empty.  By theorem "Consistency Maintained (2)," 
Aff is internally consistent.  Since Pairs(C) Ç Aff Í Aff, Pairs(C) Ç Aff must 
also be internally consistent.  By theorem "At Least One Top (1)" there must 
exist c Î C that is not second in any pair in Pairs(C) Ç Aff.  Since c Ï Top, 
c must be second in at least one pair in Aff\Pairs(C).  This means there exists 
y Î A\C such that (y,c) Î Aff.  By 1.13.1, (y,c') Î Aff for all c' Î C.  By 1.16, 
(y,c') Î Aff' for all c' Î C\D.  Thus C Ç Top' must be empty, establishing the 
"only if" clause.  Next we prove the "if" clause:  Assume C Ç Top is not empty.  
We must show C Ç Top' is not empty.  Since C Ç Top is not empty, there must 
exist c Î C that is not second in any pair in Aff.  This means (y,c) Ï Aff for all 
y Î A\{c}.  Thus (y,c) Ï Aff for all y Î A\C.  By 1.13.1, (y,c') Ï Aff for all  
y Î A\C and all c' Î C.  By 1.16, (y,c') Ï Aff' for all y Î A\C and all c' Î C\D.  
By theorem "Consistency Maintained (2)" (temporarily relabeling A = A', R = R', 
etc.), Aff' must be internally consistent.  Thus Pairs(C\D) Ç Aff' must also 
be internally consistent.  By theorem "At Least One Top (1)" there must exist 
c' Î C\D that is not second in any pair in Pairs(C\D) Ç Aff'.  Since (y,c') Ï Aff'  
for all y Î A\C and c' is not second in any pair in Pairs(C\D) Ç Aff', it follows 
that c' is not second in any pair in Aff'.  This means c' Î Top', which means 
C Ç Top' is not empty.  Thus the "if" clause has been established.  Since both 
the "if" and "only if" clauses have been established, 1.18 is established.          QED 

Resuming the proof of proposition 1.5...  First we establish the "only if" clause of 1.5.  
Assume s Î W_x. (This means MAM = x.)  We aim to show MAM' = x.  
Since MAM = x, this implies x Î Top.  By 1.17, x Î Top'.  Since MAM = x, 
by the definition of MAM() the following condition must hold: 

(1.5.1)  RVH ranks x over all a Î Top\{x}. 

By 1.9, 1.17 and 1.5.1, the following condition must hold: 

(1.5.2)  RVH' ranks x over all x' Î (A\C)\{x} Ç Top'.

There are two cases to consider: 

Case 1.5.3:  C Ç Top is empty.  By 1.18, C Ç Top' must also be empty.  
Thus, by 1.5.2 RVH' ranks x over all a Î Top'\{x}.  Thus MAM' = x.

Case 1.5.4:  C Ç Top is not empty.  By 1.5.1, RVH ranks x over 
all c Î C Ç Top.  By 1.6, RVH ranks x over all c Î C.  By 1.9, 
RVH' ranks x over all c Î C\D.  Thus, by 1.5.2 RVH' ranks x  
over all a Î Top'\{x}.  Thus MAM' = x.

Since MAM' = x in both cases, s' Î W'_x, establishing the "only if" clause of 1.5.  
Next we establish the "if" clause.  Assume s' Î W'_x.  This means MAM' = x.  
We must show MAM = x.  Since MAM' = x, this implies x Î Top'.  By 1.17, 
x Î Top.  Since MAM' = x, by the definition of MAM() the following must hold: 

(1.5.5)  RVH' ranks x over all a Î Top'\{x}. 

By 1.9, 1.17 and 1.5.5, the following condition must hold: 

(1.5.6)  RVH ranks x over all x' Î (A\C)\{x} Ç Top.

There are two cases to consider: 

Case 1.5.7:  C Ç Top' is empty.  By 1.18, C Ç Top must also be empty.  
Thus, by 1.5.6 RVH must rank x over all a Î Top\{x}.  Thus MAM = x.

Case 1.5.8:  C Ç Top' is not empty.  By 1.5.5, RVH' ranks x over 
all c Î C Ç Top'.  By 1.9, RVH ranks x over all c Î C Ç Top'.  
By 1.6, RVH ranks x over all c Î C.  Thus, by 1.5.6 RVH ranks x  
over all a Î Top\{x}.  It follows that MAM = x.

Since MAM = x in both cases, s Î W_x, establishing the "if" clause of 1.5.  
Since both the "if" and "only if" of 1.5 clauses have been established, 1.5 is 
established.  It follows that proposition 1.3 holds, and thus that proposition 1 holds.  
Thus MAM is completely independent of clone alternatives.        QED

It is instructive to provide a second proof that MAM satisfies ICA.  
Relying on theorem "MAM₂ = MAM", we prove the following proposition:

Proposition (2):  MAM₂ satisfies ICA.  

Assume C Í A is a finite non-empty set of clones within A given R.  Assume D is 
a strict subset of C.  Abbreviate A' = A\D.  Let R' denote the restriction of R to A'.  
For all s Î L(R,A), let s' denote the restriction of s to A'. (Thus s' Î L(R',A')).  
By 1.2 and 1.4, proposition 2 will follow immediately if we establish the following: 

Proposition (2.1):  For all x Î A\C and all s Î L(R,A), 
                             MAM₂(A,R,s) = x if and only if MAM₂(A',R',s') = x. 

Pick any s Î L(R,A).  Refer to the definitions comprising MAM₂.  
Make the following abbreviations: 

Maj = Majorities(A,R).
For all r Î L(A), Agr(r) = Agreed(r,A,R). 
For all r,r' Î L(A), Dist(r,r') = DistinctAgreed(r,r',A,R).  
For all x,y Î A, Precede(x,y) = Precede(x,y,A,R,s).
For all r,r' Î L(A) such that Dist(r,r') is not empty, 
        LDA(r,r') = LargestDistinctAgreed(r,r',A,R,s). 
Undom = UndominatedRankings(A,R,s). 
Top2 = Top₂(A,R,s). 
Maj' = Majorities(A',R').
For all r Î L(A'), Agr'(r) = Agreed(r,A',R'). 
For all r,r' Î L(A'), Dist'(r,r') = DistinctAgreed(r,r',A',R').  
For all x,y Î A', Precede'(x,y) = Precede(x,y,A',R',s').
For all r,r' Î L(A') such that Dist'(r,r') is not empty, 
        LDA'(r,r') = LargestDistinctAgreed(r,r',A',R',s'). 
Undom' = UndominatedRankings(A',R',s'). 
Top2' = Top₂(A',R',s').  

The following four propositions will facilitate the proof of proposition 2.1: 

Proposition (2.2):  For all r Î Undom, #R(x,c) = #R(c,x) for all c Î C  
and all x Î A\C such that r ranks x over at least one alternative in C and 
below at least one alternative in C. (In other words, the number of votes 
that rank x over c equals the number of votes that rank c over x.)

Proposition (2.3):  For all r Î Undom, there exists r' Î Undom such that 
both of the following conditions hold: 
        (2.3.1)  The alternative that tops r also tops r'. 
        (2.3.2)  For all x Î A\C, either r' ranks x over every alternative in C 
                    or r' ranks every alternative in C over x. (That is, r' ranks 
                    the clones together, no non-clones between clones.) 

Proposition (2.4):  A\C Ç Top2 = A\C Ç Top2'. 

Proposition (2.5):  C Ç Top2 is empty if and only if C Ç Top2' is empty.

Proof of 2.2:  Pick any r Î Undom.  By inspection of the definition of 
UndominatedRankings(), r must be a strict ordering of A, so we can 
let c₀ denote the alternative in C that r ranks over all other alternatives in C, 
and we can let c₁ denote the alternative in C that r ranks below all other 
alternatives in C.  Let B denote {x Î A\C such that r ranks c₀ over x 
and x over c₁}. (That is, the "non-clones" ranked between clones by r.)  
Let B' denote {x Î B such that #R(x,c) ¹ #R(c,x) for some c Î C}.  
We aim to show B' is empty.  Suppose the contrary.  By condition 1 
of the definition of clones, R(b,c) = R(b,c') and R(c,b) = R(c',b) for 
all b Î B and all c,c' Î C.  Thus, by the definition of Majorities(), 
the following two statements hold: 

(2.2.1)  For all b Î B and all c,c' Î C, (b,c) Î Maj if and only if (b,c') Î Maj. 
(2.2.2)  For all b Î B and all c,c' Î C, (c,b) Î Maj if and only if (c',b) Î Maj. 

Furthermore, by the definition of B', the following two statements hold: 

(2.2.3)  For all b Î B' and all c Î C, (b,c) Î Maj if and only if (c,b) Ï Maj. 
(2.2.4)  For all b Î B\B' and all c Î C, (b,c) Ï Maj and (c,b) Ï Maj. 

Let r' denote the strict ordering of A that is the same as r except the alternatives 
in B are raised just over c₀, and let r" denote the strict ordering of A that is the 
same as r except the alternatives in B are lowered just below c₁.  By construction 
and by 2.2.3 and 2.2.4, all six of the following statements must hold:

(2.2.5)  For all (x,y) Î Dist(r,r') Ç Agr(r), x Î C and y Î B'. 
(2.2.6)  For all (x,y) Î Dist(r,r') Ç Agr(r'), x Î B' and y Î C. 
(2.2.7)  For all (x,y) Î Dist(r,r") Ç Agr(r), x Î B' and y Î C. 
(2.2.8)  For all (x,y) Î Dist(r,r") Ç Agr(r"), x Î C and y Î B'.  
(2.2.9)  For all b Î B', (c₀,b) Î Dist(r,r') Ç Agr(r) 
             if and only if (c₁,b) Î Dist(r,r") Ç Agr(r").  
(2.2.10)  For all b Î B', (b,c₀) Î Dist(r,r') Ç Agr(r') 
              if and only if (b,c₁) Î Dist(r,r") Ç Agr(r).  

Since r Î Undom, r' does not dominate r given (R,s).  Since B' is not empty, 
Dist(r,r') is not empty.  By theorem "Precedence is a Strict Ordering" and 
the definition of LargestDistinctAgreed(), we can let (z,z') denote the 
unique ordered pair LDA(r,r').  Since r' does not dominate r, this implies 
(z,z') Ï Agr(r'), which implies (z,z') Î Agr(r).  Thus r dominates r' given (R,s).  
By 2.2.5, z Î C and z' Î B'.  For greater clarity, let c_L denote z and let b_L 
denote z'. (Thus (c_L,b_L) = LDA(r,r').)  By theorem "Precedence is a Strict 
Ordering" and the definition of LargestDistinctAgreed(), the following 
statement must hold: 

(2.2.11)  (c_L,b_L) Î Precede(x,y) for all (x,y) Î Dist(r,r') Ç Agr(r').  

By 1.12.1, 2.2.6 and 2.2.11, the following statement must hold: 

(2.2.12)  (c,b_L) Î Precede(x,y) for all c Î C and all (x,y) Î Dist(r,r') Ç Agr(r').  

By 2.2.9, (c₁,b_L) Î Dist(r,r") Ç Agr(r").  Therefore, by 1.12.1, 2.2.12, 
2.2.6 and 2.2.10 (c₁,b_L) Î Precede(x,y) for all (x,y) Î Dist(r,r") Ç Agr(r).  
This implies LDA(r,r") Ï Agr(r), which implies LDA(r,r") Î Agr(r").  
Thus r" dominates r given (R,s).  But this contradicts r Î Undom.  
Thus the contrary assumption that B' is not empty cannot hold.  
Since B' is empty, proposition 2.2 follows immediately.         QED

Proof of 2.3:  Pick any r Î Undom.  There are two cases to consider:

Case 2.3.3:  For all x Î A\C, either r ranks x over every alternative in C 
or r ranks every alternative in C over x.  In this case we can simply 
let r' = r and clearly both conditions 2.3.1 and 2.3.2 hold.

Case 2.3.4:  There exists x Î A\C such that r ranks x over at least one 
alternative in C and below at least one alternative in C.  Define c₁, B and B' 
the same as in the proof of proposition 2.2.  Clearly B is not empty in this 
case.  By 2.3, B' must be empty.  Let r' denote the strict ordering of A 
that is the same as r except all alternatives in B are lowered below c₁.  
Since B' is empty, Dist(r,r') must be empty.  This implies r does not 
dominate r' given (R,s).  By theorem "Dominance is an Ordering", 
r' must also be undominated given (R,s), so r' Î Undom.  
Clearly both conditions 2.3.1 and 2.3.2 hold. 

Since in both cases there exists r' Î Undom such that conditions 2.3.1 and 2.3.2 
hold, and since r was picked arbitrarily, proposition 2.3 is established.   QED 

Proof of 2.4:  Make the following additional abbreviations:

Undom_C = UndominatedRankings(C,R|_C,s|_C). 
Undom'_C = UndominatedRankings(C\D,R|_C\D,s|_C\D).  

By theorem "Dominance is an Ordering", Undom cannot be empty, Undom' cannot 
be empty, Undom_C cannot be empty, and Undom'_C cannot be empty.  Thus we can 
pick r_C Î Undom_C and r'_C Î Undom'_C. (In words, r_C is some "best" ranking of the 
clones, and r'_C is some "best" ranking of the clones that remain when D is deleted.)  

First we show (A\C Ç Top2) Í (A\C Ç Top2'):  Pick any x Î A\C.  Assume x Î Top2.  
We must show x Î Top2'.  Since x Î Top2, by the definition of Top₂() there must 
exist r₀ Î Undom topped by x.  By 2.3, there must exist r₁ Î Undom topped 
by x such that, for all y Î A\C, either r₁ ranks y over every alternative in C 
or r₁ ranks every alternative in C over y.  Let r₂ denote the strict ordering of A 
that is the same as r₁ except the alternatives in C are deleted and replaced with r_C.  
Since r₁ Î Undom and r_C Î Undom_C, it follows that r₂ Î Undom.  Let r₃ denote 
the strict ordering of A' that is the same as r₂ except the alternatives in C are deleted 
and replaced with r'_C.  We aim to show r₃ Î Undom' (after which x Î Top2' will follow).  
Suppose to the contrary r₃ Ï Undom'.  This implies there exists r₄ Î Undom' such 
that r₄ dominates r₃ given (R',s').  Since C\D is a set of clones within A' given R', 
by 2.3 there must exist r₅ Î Undom' such that, for all y Î A\C, either r₅ ranks y over 
every alternative in C\D or r₅ ranks every alternative in C\D over y.  Let r₆ denote 
the strict ordering of A' that is the same as r₅ except the alternatives in C\D are 
deleted and replaced with r'_C.  Since r₅ Î Undom' and r'_C Î Undom'_C, it follows 
that r₆ Î Undom'.  By theorem "Dominance is an Ordering", r₆ must also dominate r₃  
given (R',s').  Let r₇ denote the strict ordering of A that is the same as r₆ except 
the alternatives in C\D are deleted and replaced with r_C.  By construction, 
Dist(r₇,r₂) = Dist'(r₆,r₃).  Therefore, since r₆ dominates r₃ given (R',s'), it follows by 
1.10 (that is, precedence is independent of irrelevant alternatives) that r₇ dominates r₂  
given (R,s).  But this contradicts r₂ Î Undom established above.  Thus the contrary 
assumption r₃ Ï Undom' cannot hold, implying r₃ Î Undom'.  Since x tops r₃ and 
r₃ Î Undom', x Î Top2'.  Thus (A\C Ç Top2) Í (A\C Ç Top2'). 

Next we show (A\C Ç Top2') Í (A\C Ç Top2):  Pick any x Î A\C.  Assume x Î Top2'.  
We must show x Î Top2.  Since x Î Top2', by the definition of Top₂() there must 
exist r₀ Î Undom' topped by x.  Since C\D is a set of clones within A' given R', 
by 2.3 there must exist r₁ Î Undom' topped by x such that, for all y Î A\C, 
either r₁ ranks y over every alternative in C\D or r₁ ranks every alternative in C\D 
over y.  Let r₂ denote the strict ordering of A' that is the same as r₁ except the 
alternatives in C\D are deleted and replaced with r'_C.  Since r₁ Î Undom' and 
r'_C Î Undom'_C, it follows that r₂ Î Undom'.  Let r₃ denote the strict ordering 
of A that is the same as r₂ except the alternatives in C\D are deleted and replaced 
with r_C.  We aim to show r₃ Î Undom (after which x Î Top2 will follow).  
Suppose to the contrary r₃ Ï Undom.  This implies there exists r₄ Î Undom such 
that r₄ dominates r₃ given (R,s).  By 2.3, there must exist r₅ Î Undom such that, 
for all y Î A\C, either r₅ ranks y over every alternative in C or r₅ ranks every 
alternative in C over y.  Let r₆ denote the strict ordering of A that is the same as r₅  
except the alternatives in C are deleted and replaced with r_C.  Since r₅ Î Undom 
and r_C Î Undom_C, it follows that r₆ Î Undom.  By theorem "Dominance is an 
Ordering", r₆ must also dominate r₃ given (R,s).  Let r₇ denote the strict ordering 
of A' that is the same as r₆ except the alternatives in C are deleted and replaced 
with r'_C.  By construction, Dist(r₇,r₂) = Dist'(r₆,r₃).  Therefore, since r₆ dominates r₃  
given (R,s), it follows by 1.10 that r₇ dominates r₂ given (R',s').  But this contradicts 
r₂ Î Undom' established above.  Thus the contrary assumption r₃ Ï Undom cannot 
hold, implying r₃ Î Undom.  Since x tops r₃ and r₃ Î Undom, x Î Top2.  
Thus (A\C Ç Top2') Í (A\C Ç Top2).  Since (A\C Ç Top2) Í (A\C Ç Top2') 
and (A\C Ç Top2') Í (A\C Ç Top2), proposition 2.2 is established.        QED 

Proof of 2.5:  Make the same abbreviations as in the proof of 2.4.  First we establish 
the "if" clause of 2.5:  Assume there exists c Î C such that c Î Top2.  We must show 
there exists c' Î C\D such that c' Î Top2'.  Since c Î Top2, by the definition of Top₂() 
there must exist r₀ Î Undom topped by c.  By 2.3, there must exist r₁ Î Undom 
topped by c such that r₁ ranks every alternative in C over every alternative in A\C.  
Let r₂ denote the strict ordering of A that is the same as r₁ except the alternatives 
in C are deleted and replaced with r_C.  Since r₁ Î Undom and r_C Î Undom_C, 
it follows that r₂ Î Undom.  Let r₃ denote the strict ordering of A' that is the same 
as r₂ except the alternatives in C are deleted and replaced with r'_C.  We aim to 
show r₃ Î Undom' (after which it will follow that the alternative atop r'_C is in Top2').  
Suppose to the contrary r₃ Ï Undom'.  This implies there exists r₄ Î Undom' such 
that r₄ dominates r₃ given (R',s').  Since C\D is a set of clones within A' given R', 
by 2.3 there exists r₅ Î Undom' such that, for all x Î A\C, either r₅ ranks x over 
every alternative in C\D or r₅ ranks every alternative in C\D over x.  Let r₆ denote 
the strict ordering of A' that is the same as r₅ except the alternatives in C\D are 
deleted and replaced with r'_C.  Since r₅ Î Undom' and r'_C Î Undom'_C, it follows 
that r₆ Î Undom'.  By theorem "Dominance is an Ordering", r₆ must also dominate r₃  
given (R',s').  Let r₇ denote the strict ordering of A that is the same as r₆ except 
the alternatives in C\D are deleted and replaced with r_C.  By construction, 
Dist(r₇,r₂) = Dist'(r₆,r₃).  Therefore, since r₆ dominates r₃ given (R',s'), it follows 
by 1.10 that r₇ dominates r₂ given (R,s).  But this contradicts r₂ Î Undom.  
Thus the contrary assumption r₃ Ï Undom' cannot hold, implying r₃ Î Undom'.  
Since r'_C is a strict ordering of C\D, we can let c' denote the alternative in C\D that 
tops r'_C.  Since r₃ Î Undom' and r₃ is topped by c', this implies c' Î Top2'.  
Thus the "if" clause of 2.5 has been established.  

Next we establish the "only if" clause of 2.5:  Assume C Ç Top2' is not empty.  
We must show C Ç Top2 is not empty.  Since C Ç Top2' is not empty, there must 
exist c' Î C\D such that c' Î Top2'.  By the definition of Top₂(), there must exist 
r₀ Î Undom' topped by c'.  Since C\D is a set of clones within A' given R', 
by 2.3 there exists r₁ Î Undom' topped by c' such that r₁ ranks every alternative 
in C\D over every alternative in A\C.  Let r₂ denote the strict ordering of A' that 
is the same as r₁ except the alternatives in C\D are deleted and replaced with r'_C.  
Since r₁ Î Undom' and r'_C Î Undom'_C, it follows that r₂ Î Undom'.  Let r₃ denote 
the strict ordering of A that is the same as r₂ except the alternatives in C\D are deleted 
and replaced with r_C.  We aim to show r₃ Î Undom (after which it will follow that 
the alternative atop r_C is in Top2).  Suppose to the contrary r₃ Ï Undom.  This implies 
there exists r₄ Î Undom such that r₄ dominates r₃ given (R,s).  By 2.3, there must 
exist r₅ Î Undom such that, for all x Î A\C, either r₅ ranks x over every alternative 
in C or r₅ ranks every alternative in C over x.  Let r₆ denote the strict ordering of A  
that is the same as r₅ except the alternatives in C are deleted and replaced with r_C.  
Since r₅ Î Undom and r_C Î Undom_C, it follows that r₆ Î Undom.  By theorem 
"Dominance is an Ordering", r₆ must also dominate r₃ given (R,s).  Let r₇ denote 
the strict ordering of A' that is the same as r₆ except the alternatives in C are deleted 
and replaced with r'_C.  By construction, Dist(r₇,r₂) = Dist'(r₆,r₃).  Therefore, since 
r₆ dominates r₃ given (R,s), it follows by 1.10 that r₇ dominates r₂ given (R',s').  
But this contradicts r₂ Î Undom'.  Thus the contrary assumption r₃ Ï Undom cannot 
hold, implying r₃ Î Undom.  Since r_C is a strict ordering of C, we can let c denote 
the alternative in C atop r_C.  Since r₃ Î Undom and r₃ is topped by c, c Î Top2.  
Thus C Ç Top2 is not empty, establishing the "only if" clause of 2.5.  Since both 
the "if" and "only if" clauses have been established, 2.5 is established.          QED 

Resuming the proof of 2.1...  There are three cases to consider:  

Case 2.1.1:  MAM₂(A,R,s) Î C.  This implies there exists c Î Top2 that is 
ranked by RVH over all other alternatives in Top2.  By 1.6 ("Random Voter 
Hierarchy ranks clones together"), RVH ranks all of C over all alternatives 
in A\C Ç Top2.  By 2.4, A\C Ç Top2 = A\C Ç Top2'.  Thus, by 1.9 
("Random Voter Hierarchy is independent of irrelevant alternatives") 
RVH' ranks all of C\D over all alternatives in A\C Ç Top2'.  By 2.5, 
C Ç Top2' is not empty.  Thus RVH' ranks some c' Î Top2' over 
all other alternatives in Top2'.  Thus MAM₂(A',R',s') = c', 
which means MAM₂(A',R',s') Î C.  

Case 2.1.2:  MAM₂(A,R,s) Ï C and C Ç Top2 is not empty.  This implies&nb