Theorem: "MAM₂ satisfies LIIA": 
        Some definitions:  For all r Î L(A) and all B Í A, let r|_B denote 
        the restriction of r to B. (That is, the strict ranking of B such that, 
        for all x,y Î B, r|_B ranks x over y if and only if r ranks x over y.)  
        For all r Î L(A), let Contiguous(r) denote {C Í A such that, 
        for all c,c' Î C and all x Î A\C, r does not rank x between c and c'}. 
        (That is, all subsets of alternatives that are contiguous in r.)  
        For all r Î L(A), all non-empty C Î Contiguous(r) and all r' Î L(C), 
        let substitute(r,r') denote the strict ranking of A that is the same as r  
        except the alternatives in C are deleted and replaced with r'.  
        That is, all three of the following conditions hold: 
            (i) For all x Î A\C and all y Î A, substitute(r,r') ranks x over y  
                  if and only if r ranks x over y. 
            (ii) For all x Î A and all y Î A\C, substitute(r,r') ranks x over y  
                  if and only if r ranks x over y. 
            (iii) For all x,y Î C, substitute(r,r') ranks x over y  
                   if and only if r' ranks x over y.  
(1) For all r Î L(A), all non-empty C Î Contiguous(r) and all r' Î L(C), 
      DistinctAgreed(r,substitute(r,r'),A,R) = DistinctAgreed(r|_C,r',C,R|_C). 
(2) For all s Î L(R,A), all r Î L(A), all non-empty C Î Contiguous(r) 
      and all r' Î UndominatedRankings(C,R|_C,s|_C), 
      r does not  dominate substitute(r,r') given (R,s).  
(3) For all s Î L(R,A), all r Î UndominatedRankings(A,R,s), all non-empty 
      C Î Contiguous(r) and all r' Î UndominatedRankings(C,R|_C,s|_C), 
      substitute(r,r') Î UndominatedRankings(A,R,s). 
(4) For all s Î L(R,A), all r Î UndominatedRankings(A,R,s) 
      and all non-empty C Î Contiguous(r), 
      r|_C Î UndominatedRankings(C,R|_C,s|_C). 
(5) For all s Î L(R,A) and all non-empty C Î Contiguous(R_social2(A,R,s)), 
      R_social2(A,R,s)|_C = R_social2(C,R|_C,s|_C). 

Proof of (1):  Pick any r Î L(A), any non-empty C Î Contiguous(r) and any r' Î L(C).  
Abbreviate r" = substitute(r,r').  Make the following abbreviations: 
        Maj = Majorities(A,R).
        Agreed(r) = Agreed(r,R).  
        Agreed(r") = Agreed(r",R).  
        Distinct(r,r") = DistinctAgreed(r,r",A,R).  
        Maj' = Majorities(C,R|_C). 
        Agreed(r|_C) = Agreed(r|_C,R|_C).  
        Agreed(r') = Agreed(r',R|_C).  
        Distinct(r|_C,r') = DistinctAgreed(r|_C,r',C,R|_C).  
We must show Distinct(r,r") = Distinct(r|_C,r').  By the definition of DistinctAgreed(), 
both of the following statements hold:  
        (1.1)  Distinct(r,r") = (Agreed(r) È Agreed(r"))\(Agreed(r) Ç Agreed(r")). 
        (1.2)  Distinct(r|_C,r') = (Agreed(r|_C) È Agreed(r'))\(Agreed(r|_C) Ç Agreed(r')). 
By construction and the definition of Agreed(), both of the following statements hold: 
        (1.3)  For all (x,y) Î Pairs(A)\Pairs(C), 
                  (x,y) Î Agreed(r) if and only if (x,y) Î Agreed(r").  
By 1.3, the following statement holds: 
        (1.4)  Pairs(A)\(Agreed(r) Ç Agreed(r")) = Pairs(C).  
Since (Agreed(r) È Agreed(r")) Í Pairs(A), by 1.1 and 1.4 the following statement holds: 
        (1.5)  Distinct(r,r") = (Agreed(r) È Agreed(r")) Ç Pairs(C). 
By theorem "Majorities are Independent of Irrelevant Alternatives", Maj' = Maj Ç Pairs(C).  
Thus, by construction and the definition of Agreed(), both of the following statements hold: 
        (1.6)  For all (x,y) Î Pairs(C), 
                  (x,y) Î Agreed(r) if and only if (x,y) Î Agreed(r|_C).  
        (1.7)  For all (x,y) Î Pairs(C), 
                  (x,y) Î Agreed(r") if and only if (x,y) Î Agreed(r').  
By 1.5, 1.6, 1.7 and 1.2, Distinct(r,r") = Distinct(r|_C,r') Ç Pairs(C).  
Since Distinct(r|_C,r') Í Pairs(C), this implies Distinct(r,r") = Distinct(r',r|_C), 
establishing 1.  

Proof of (2):  Pick any s Î L(R,A).  Make the following abbreviations: 
        Maj = Majorities(A,R).
        For all r Î L(A), Agreed(r) = Agreed(r,R).  
        For all r,r" Î L(A), Distinct(r,r") = DistinctAgreed(r,r",A,R).  
        For all r,r" Î L(A) such that Distinct(r,r") is not empty, 
                LDA(r,r") = LargestDistinctAgreed(r,r",A,R,s). 
        Undom = UndominatedRankings(A,R,s). 
Pick any r Î L(A) and any non-empty C Î Contiguous(r).  
Make the following additional abbreviations: 
        Maj' = Majorities(C,R|_C). 
        For all r' Î L(C), Agreed(r') = Agreed(r',R|_C).  
        For all r',r" Î L(C), Distinct(r',r") = DistinctAgreed(r',r",C,R|_C).  
        For all r',r" Î L(C) such that Distinct(r',r") is not empty, 
                LDA(r',r") = LargestDistinctAgreed(r',r",C,R|_C,s|_C). 
        Undom' = UndominatedRankings(C,R|_C,s|_C).  
Pick any r' Î Undom'.  Abbreviate r" = substitute(r,r').  We must show r does 
not dominate r" given (R,s).  If r' = r|_C, then r" = r, so by theorem "Dominance 
is an Ordering" dominance is irreflexive and we are done.  On the other hand, 
suppose r' ¹ r|_C.  Since r' Î Undom', r|_C does not dominate r' given (R|_C,s|_C).  
There are two cases to consider:  

Case 2.1:  r' does not dominate r|_C given (R|_C,s|_C).  Since neither 
dominates the other, by the definition of dominance Distinct(r',r|_C) must 
be empty.  By 1, Distinct(r,r") is empty.  Thus r does not dominate r" 
given (R,s) in case 2.1.

Case 2.2:  r' dominates r|_C given (R|_C,s|_C).  By the definition of dominance, 
this implies Distinct(r',r|_C) is not empty.  Thus, by theorem "Precedence is a Strict Ordering" and the definition of LargestDistinctAgreed(), we can 
let (x,y) denote the unique ordered pair LDA(r',r|_C).  Since r' dominates r|_C, 
(x,y) Î Agreed(r') and (x,y) Ï Agreed(r|_C).  By 1, Distinct(r,r") = Distinct(r',r|_C), 
so (x,y) Î Agreed(r") and (x,y) Ï Agreed(r).  By theorem "Precedence is 
Independent of Irrelevant Alternatives", LDA(r,r") = (x,y).  It follows that 
r" dominates r given (R,s).  By theorem "Dominance is an Ordering", 
dominance is asymmetric, so r does not dominate r" given (R,s) in case 2.2. 

Since in both cases r does not dominate r" given (R,s), (2) is established.  

Proof of 3:  Pick any s Î L(R,A).  Make the same abbreviations as in the 
proof of (2).  Pick any r Î Undom, any non-empty C Î Contiguous(r) 
and any r' Î Undom'.  Abbreviate r" = substitute(r,r').  By 2, r does not 
dominate r" given (R,s).  Pick any r₀ Î L(A).  Since r Î Undom, r₀ does not 
dominate r given (R,s).  By theorem "Dominance is an Ordering", dominance 
is negatively transitive, so r₀ does not dominate r" given (R,s).  Since r₀ was 
picked arbitrarily, it follows that no ranking in L(A) dominates r" given (R,s).  
Thus r" Î Undom, establishing 3.  

Proof of 4:  Pick any s Î L(R,A).  Make the same abbreviations as in the 
proof of (2).  Pick any r Î Undom and any non-empty C Î Contiguous(r).  
We must show r|_C Î Undom'.  Suppose the contrary.  This means there 
exists r' Î L(C) such that r' dominates r|_C given (R|_C,s|_C).  This implies 
Distinct(r',r|_C) is not empty.  By theorem "Precedence is a Strict Ordering" 
and the definition of LargestDistinctAgreed(), we can let (x,y) denote 
the unique ordered pair LDA(r',r|_C).  Since r' dominates r|_C, this implies 
(x,y) Î Agreed(r') and (x,y) Ï Agreed(r|_C).  Abbreviate r" = substitute(r,r').  
We aim to show r" dominates r given (R,s), a contradiction.  By 1, 
Distinct(r",r) = Distinct(r',r|_C).  This implies (x,y) Î Agreed(r") and 
(x,y) Ï Agreed(r).  By theorem "Precedence is Independent of Irrelevant 
Alternatives", LDA(r",r) = (x,y).  It follows that r" dominates r given (R,s).  
But since r Î Undom, this is a contradiction.  Thus the contrary assumption 
cannot hold, so r|_C Î Undom', establishing 4.

Proof of 5:  Pick any s Î L(R,A).  Make the same abbreviations as in the proof 
of (2).  Abbreviate R_s = R_social2(A,R,s).  Pick any non-empty C Î Contiguous(R_s).  
Abbreviate R_C = R_social2(C,R|_C,s|_C).  We must show R_s|_C = R_C.  Suppose the 
contrary.  By theorem "Properties of RSocial (5)", R_s Î Undom and R_C Î Undom'.  
By 4, R_s|_C Î Undom'.  

QED