reserve D,D1,D2 for non empty set,
        d,d1,d2 for XFinSequence of D,
        n,k,i,j for Nat;
reserve A,B for object,
        v for Element of (n+k)-tuples_on {A,B},
        f,g for FinSequence;

theorem Th15:
  A <> B & n > 0 implies n|->A is A,n,B,0-dominated-election
proof
  set nA=n|->A;
  assume that
A1:   A<>B and
A2:   n>0;
  A is Element of {A,B} by TARSKI:def 2;
  then reconsider nA as  Element of (n+0)-tuples_on {A,B} by FINSEQ_2:112;
  nA"rng nA c= nA"{A} by FUNCOP_1:13,RELAT_1:143;
  then  dom nA = nA"{A} by RELAT_1:132, RELAT_1:134;
  then card (nA"{A}) = card Seg len nA by FINSEQ_1:def 3
                    .= len nA by FINSEQ_1:57
                    .= n by CARD_1:def 7;
  hence n|->A in Election(A,n,B,0) by Def1;
  let i such that
A3: i>0;
  set nAi=nA|i;
A4: nAi = (Seg n /\ Seg i)-->A by FUNCOP_1:12;
A5: not A in {B} by A1,TARSKI:def 1;
  per cases;
    suppose i<= n;
      then (Seg i) /\(Seg n) = Seg i by FINSEQ_1:5,XBOOLE_1:28;
      then
A6:     nAi"{A} = Seg i by A4,FUNCOP_1:15;
      nAi"{B} = {} by A5,FUNCOP_1:16,A4;
      hence thesis by A6,A3;
    end;
    suppose i > n;
      then (Seg i) /\(Seg n) = Seg n by FINSEQ_1:5,XBOOLE_1:28;
      then
A7:   nAi"{A} = Seg n by A4,FUNCOP_1:15;
      nAi"{B} = {} by A5,FUNCOP_1:16,A4;
      hence thesis by A2,A7;
    end;
end;
