"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).

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:

 65% 35% x y y z z x

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 xICA 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:

 50% 50% x z y y z x

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 xICA 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 anonymousneutral 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 MAMx 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 MAMx = 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 Wx = {s Î L(R,A) such that
MAM(A,R,s) = x}.  Thus MAMx = #Wx/#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)  #Wx/#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):  #Wx = #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' = (sR,sA)|A' = (sR|A',sA|A') where sR|A' is the same as sR except that
all alternatives in D are deleted, and sA|A' is the same as sA 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 Wx 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 Rsyc = Rsyc'. (That is, the vote in R(yc) ranked highest
by sR is the same as the vote in R(yc') ranked highest by sR.)  Furthermore,
condition 1 of the definition of clones implies Rsyc ranks y relative to c
the same as Rsyc' 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 Affab = 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) Î Q1.  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 Affac.  This
means there exist alternatives z1,z2,...,zk Î A such that z1 = c and zk = a
and {(z1,z2),(z2,z3),...,(zk-1,zk)} Í Affac.  Since z1 Î C, we can let i denote
the largest integer in {1,2,...,k} such that zi Î C.  Since zk = a Î A\C, i < k.
Since (zk,c) Î Maj, (c,zk) Ï Maj.  By 1.11.4, (c',zk) Ï Maj for all c' Î C.
Since (zk-1,zk) Î Maj, this means zk-1 Ï C, so i < k-1.  Thus zi+1 Î A\C and
zi+2 Î A\C.  Thus we can let P denote {(zi+1,zi+2),...,(zk-1,zk)} È {(zk,b)}.
Since P Í Aff, (b,zi+1) is inconsistent with Aff.  By theorem "Consistency
Maintained"
, Aff is internally consistent, so (b,zi+1) Ï Aff.  Since (zi,zi+1) Î Aff
and (b,zi+1) Ï Aff, (zi,zi+1) Î Q3.  Since (zi,zi+1) Î Precede(a,c), by 1.12.4
(zi,zi+1) Î Precede(a,b).  Thus (zi,zi+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) Î Q2.  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 Affab.  This
means there exist alternatives z1,z2,...,zk Î A such that z1 = b and zk = a and
{(z1,z2),(z2,z3),...,(zk-1,zk)} Í Affab.  Since z1 Î C, we can let i denote the
largest integer in {1,2,...,k} such that zi Î C.  Since zk = a Î A\C, i < k.
Since (zk,c) Î Maj, (c,zk) Ï Maj.  By 1.11.4, (c',zk) Ï Maj for all c' Î C.
Since (zk-1,zk) Î Maj, this means zk-1 Ï C, so i < k-1.  Thus zi+1 Î A\C and
zi+2 Î A\C.  Thus we can let P denote {(zi+1,zi+2),...,(zk-1,zk)} È {(zk,c)}.
Since P Í Aff, (c,zi+1) is inconsistent with Aff.  By theorem "Consistency
Maintained"
, Aff is internally consistent, so (c,zi+1) Ï Aff.  Since (zi,zi+1) Î Aff
and (c,zi+1) Ï Aff, (zi,zi+1) Î Q3.  Thus (zi,zi+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) Î Q3.  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 Affcb.  This
means there exist alternatives z1,z2,...,zk Î A such that z1 = b and zk = c and
{(z1,z2),(z2,z3),...,(zk-1,zk)} Í Affcb.  Since zk Î C, we can let i denote the
smallest integer in {1,2,...,k} such that zi 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 zi-1 Î A\C and
zi-2 Î A\C.  Thus we can let P denote {(a,z1)} È {(z1,z2),...,(zi-2,zi-1)}.
Since P Í Aff, (zi-1,a) is inconsistent with Aff.  By theorem "Consistency
Maintained"
, Aff is internally consistent, so (zi-1,a) Ï Aff.  Since (zi-1,zi) Î Aff
and (zi-1,a) Ï Aff, (zi-1,zi) Î Q1.  Since (zi-1,zi) Î Precede(c,b), by 1.12.3
(zi-1,zi) Î Precede(a,b).  Thus (zi-1,zi) 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) Î Q4.  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 Affab.  This
means there exist alternatives z1,z2,...,zk Î A such that z1 = b and zk = a and
{(z1,z2),(z2,z3),...,(zk-1,zk)} Í Affab.  Since zk Î C, we can let i denote the
smallest integer in {1,2,...,k} such that zi 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 zi-1 Î A\C and
zi-2 Î A\C.  Thus we can let P denote {(c,z1)} È {(z1,z2),...,(zi-2,zi-1)}.
Since P Í Aff, (zi-1,c) is inconsistent with Aff.  By theorem "Consistency
Maintained"
, Aff is internally consistent, so (zi-1,c) Ï Aff.  Since (zi-1,zi) Î Aff
and (zi-1,c) Ï Aff, (zi-1,zi) Î Q1.  Thus (zi-1,zi) 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 AR' 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 Affab = Aff Ç Precede(a,b).
Pick any (a,b) Î XPairs and any P Í Affab.  Assume (a,b) is inconsistent
with P and consistent with all  subsets of Affab smaller than P.  This means
there exist alternatives z1,z2,...,zk Î A such that z1 = b and zk = a and
P = {(z1,z2),(z2,z3),...,(zk-1,zk)}.  We must show that at most one alternative
in C is in {z1,z2,...,zk}.  Suppose the contrary.  This implies means we can
let s denote the smallest integer in {1,2,...,k} such that zi Î C and we can
let l denote the largest integer in {1,2,...,k} such that zi 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 (zs-1,zs) Î Affab, it follows by 1.12.2 and  1.13.1 that
(zs-1,a) Î Affab.  But since {(zi,zi+1) Î P such that 1 £ i < s - 1} È {(zs-1,a)
is a subset of Affab 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 (zl,zl+1) Î Affab, it follows by 1.12.1 and 1.13.2 that
(b,zl+1) Î Affab.  But since {(b,zl+1)} È {(zi,zi+1) Î P such that l + 1 £ i < k
is a subset of Affab 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 (zs-1,zs) Î Affab
it follows by 1.12.2 and  1.13.1 that (zs-1,zl) Î Affab.  But since {(zi,zi+1) Î P
such that 1 £ i < s - 1} È {(zs-1,zl)È {(zi,zi+1) Î P such that l + 1 £ i < k
is a subset of Affab 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,bÇ 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,bÇ 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,bÇ 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),...,(ak-1,ak)
where a1 = b and ak = a and there exists at least one j Î {1,2,...,k} such
that aj = c.  Since (a,b) Î XPairs', a1 Ï D and ak Ï D.  Thus we can
let i denote the smallest integer in {2,3,...,k-1} such that ai = c and
we can let j denote the largest integer in {2,3,...,k-1} such that aj = c.
Clearly 1 < i £ j < k.  By construction, ai-1 Î A\C and aj+1 Î A\C.
Pick any c' Î C\D.  Since (ai-1,ai) Î Aff, by 1.13.1 (ai-1,c') Î Aff.  Since
(aj,aj+1) Î Aff, by 1.13.2 (c',aj+1) Î Aff.  Since (ai-1,ai) Î Precede(a,b),
by 1.12.2 (ai-1,c') Î Precede(a,b).  Since (aj,aj+1) Î Precede(a,b),
by 1.12.1 (c',aj+1) Î Precede(a,b).  Let P denote the following set:
{(a1,a2),(a2,a3),...,(ai-1,c'),(c',aj+1),(aj+1,aj+2),...,(ak-1,ak)
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 Î Top'.  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 Î Top.  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 Wx. (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 Wx, 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 cL denote z and let bL
denote z'. (Thus (cL,bL) = LDA(r,r').)  By theorem "Precedence is a Strict
Ordering"
and the definition of LargestDistinctAgreed(), the following
statement must hold:

(2.2.11)  (cL,bL) 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,bL) 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,bL) Î Dist(r,r") Ç Agr(r").  Therefore, by 1.12.1, 2.2.12,
2.2.6 and 2.2.10 (c1,bL) 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:

UndomC = 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, UndomC cannot be empty, and Undom'C cannot be empty.  Thus we can
pick rC Î UndomC and r'C Î Undom'C. (In words, rC 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 rC.
Since r1 Î Undom and rC Î UndomC, 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 rC.  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 rC.  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 rC.  Since r5 Î Undom
and rC Î UndomC, 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 rC.  Since r1 Î Undom and rC Î UndomC
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 rC.  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 rC.  We aim to show r3 Î Undom (after which it will follow that
the alternative atop rC 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 rC.
Since r5 Î Undom and rC Î UndomC, 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 rC is a strict ordering of C, we can let c denote
the alternative in C atop rC.  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
the alternative in Top2 ranked highest by RVH is not in C.  Let x denote
this alternative.  It follows that MAM2(A,R,s) = x.  By 1.6 ("Random
Voter Hierarchy ranks clones together"), RVH ranks x over all of C.
By 2.4, A\C Ç Top2 = A\C Ç Top2'.  Thus x Î Top2', and by 1.9
("Random Voter Hierarchy is independent of irrelevant alternatives")
RVH' ranks x over all of Top2'\{x}.  Thus MAM2(A',R',s') = x
which means MAM2(A,R,s) = MAM2(A',R',s').

Case 2.1.3:  MAM2(A,R,s) Ï C and C Ç Top2 is empty.  By 2.5,
C Ç Top2' is empty.  By 2.4, Top2 = Top2'.  Thus, by 1.9 ("Random
Voter Hierarchy is independent of irrelevant alternatives") the alternative
in Top2 ranked highest by RVH is the same as the alternative in Top2'
ranked highest by RVH'.  Thus MAM2(A,R,s) = MAM2(A',R',s').

By inspection of the three cases, proposition 2.1 follows immediately.
Thus proposition 2 is also established, which means MAM2 satisfies ICA.
By theorem "MAM2 = MAM", this implies MAM satisfies ICA.         QED 