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 Th10:
  Election(A,n,B,k) = Choose(Seg (n+k),n,A,B)
proof
  thus Election(A,n,B,k) c= Choose(Seg (n+k),n,A,B)
  proof
    let x be object;
    assume
A1:   x in Election(A,n,B,k);
    then reconsider v=x as Element of (n+k)-tuples_on {A,B};
A2:   rng v c= {A,B};
    len v = n+k by CARD_1:def 7;
    then dom v = Seg (n+k) by FINSEQ_1:def 3;
    then reconsider V=v as Function of Seg (n+k),{A,B} by FUNCT_2:2,A2;
    card (v"{A}) = n by A1,Def1;
    then V in Choose(Seg (n+k),n,A,B) by CARD_FIN:def 1;
    hence thesis;
  end;
  let x be object;
  assume x in Choose(Seg (n+k),n,A,B);
  then consider f be Function of Seg (n+k),{A,B} such that
A3: f = x and
A4: card (f"{A})=n by CARD_FIN:def 1;
A5: rng f c= {A,B};
A6: dom f = Seg (n+k) by FUNCT_2:def 1;
  then reconsider f as FinSequence by FINSEQ_1:def 2;
  reconsider f as FinSequence of {A,B} by A5,FINSEQ_1:def 4;
  n+k in NAT by ORDINAL1:def 12;
  then len f = n+k by FINSEQ_1:def 3,A6;
  then f is Element of (n+k)-tuples_on {A,B} by FINSEQ_2:92;
  hence x in Election(A,n,B,k) by A3,A4,Def1;
end;
