reserve m, n for Nat;

theorem Th14:
  for n being non zero Nat holds ex k being Element of
  NAT st support ppf n c= Seg k
proof
  let n be non zero Nat;
A1: support ppf n = support pfexp n by NAT_3:def 9;
  per cases;
  suppose
A2:  support ppf n is empty;
    take 0;
    thus thesis by A2;
  end;
  suppose
    support ppf n is non empty;
    then reconsider S = support ppf n as finite non empty Subset of NAT by
XBOOLE_1:1;
    take max S;
A3: max S is Element of NAT by ORDINAL1:def 12;
    support ppf n c= Seg max S
    proof
      let x be object;
      assume
A4:   x in support ppf n;
      then reconsider m = x as Nat;
      x is Prime by A1,A4,NAT_3:34;
      then
A5:   1 <= m by INT_2:def 4;
      m <= max S by A4,XXREAL_2:def 8;
      hence thesis by A5,FINSEQ_1:1;
    end;
    hence thesis by A3;
  end;
end;
