reserve D for non empty set,
  i,j,k for Nat,
  n,m for Nat,
  r for Real,
  e for real-valued FinSequence;

theorem Th52:
  for n being non zero Nat ex P being FinSequence of REAL st len
  P = n & (for i st i in dom P holds P.i >= 0) & Sum P = 1
proof
  let n be non zero Nat;
  reconsider n as non zero Nat;
  consider e being FinSequence of REAL such that
A1: len e = n and
A2: for i be Nat st i in Seg n holds (i in Seg 1 implies e.i = jj) & (
  not i in Seg 1 implies e.i = zz) by Th2;
A3: n >= 1 by NAT_1:14;
A4: Sum e = 1
  proof
    consider f be Real_Sequence such that
A5: f.1 = e.1 and
A6: for n be Nat st 0 <> n & n < len e holds f.(n+1) = f.n+e.(n+1) and
A7: Sum e = f.len e by A1,NAT_1:14,PROB_3:63;
    for n st n <> 0 & n <= len e holds f.n = 1
    proof
      defpred p[Nat] means $1 <> 0 & $1 <= len e implies f.$1 = 1;
A8:   now
        let k be Nat such that
A9:     p[k];
        now
          assume that
          k+1 <> 0 and
A10:      k+1 <= len e;
          per cases by A10,NAT_1:13;
          suppose
A11:        k = 0 & k < len e;
            1 in Seg 1 & 1 in Seg n by A3;
            hence f.(k+1) = 1 by A2,A5,A11;
          end;
          suppose
A12:        k > 0 & k < len e;
            then k >= 1 by NAT_1:14;
            then k + 1 > 1 by NAT_1:13;
            then
A13:        k + 1 in Seg n & not k + 1 in Seg 1 by A1,A10,FINSEQ_1:1;
            thus f.(k+1) =1 + e.(k+1) by A6,A9,A12
              .= 1 + 0 by A2,A13
              .= 1;
          end;
        end;
        hence p[k+1];
      end;
A14:  p[0];
      for n holds p[n] from NAT_1:sch 2(A14,A8);
      hence thesis;
    end;
    hence thesis by A1,A7;
  end;
  take e;
  for i st i in dom e holds e.i >= 0
  proof
    let i;
    assume i in dom e;
    then i in Seg n by A1,FINSEQ_1:def 3;
    hence thesis by A2;
  end;
  hence thesis by A1,A4;
end;
