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
  A <> B & n >= k implies prob DominatedElection(A,n,B,k) = (n-k) / (n+k)
proof
    assume
A1:   A<>B & n >= k;
    then
A2:   card Election(A,n,B,k) = (n+k) choose n by Th12;
A3: n+ k >= k+0 & (n+k)-k =n & n+k >= n by NAT_1:11;
A4: (n+k) choose n >0 by NAT_1:11,CATALAN2:26;
    per cases by A1,XXREAL_0:1;
      suppose
A5:     n=k;
        then DominatedElection(A,n,B,k) is empty by Th18;
        then card DominatedElection(A,n,B,k) = 0;
        then prob DominatedElection(A,n,B,k) = 0/((n+k) choose n)
          by A2,RPR_1:def 1;
        hence thesis by A5;
      end;
      suppose n > k; then
A6:       card DominatedElection(A,n,B,k) =((n-k)/(n+k))*((n+k) choose k)
          by A1,Th27
              .= ((n-k) / (n+k))*((n+k) choose n) by A3,NEWTON:20;
        thus prob DominatedElection(A,n,B,k)
           = ((n-k)/(n+k))*((n+k) choose n) / ((n+k) choose n)
             by A2,A6,RPR_1:def 1
          .= ((n-k)/(n+k))*(((n+k) choose n)/((n+k) choose n)) by XCMPLX_1:74
          .= ((n-k)/(n+k))*1 by A4,XCMPLX_1:60
          .= (n-k)/(n+k);
  end;
end;
