
theorem Th25:
  for n, k being Element of NAT holds PFBrt (n,k) is Element of
  SubstPoset (NAT, {k})
proof
  let n, k be Element of NAT;
A1: for s, t being Element of PFuncs (NAT,{k}) holds ( s in PFBrt (n,k) & t
  in PFBrt (n,k) & s c= t implies s = t )
  proof
    let s, t be Element of PFuncs (NAT,{k});
    assume that
A2: s in PFBrt (n,k) and
A3: t in PFBrt (n,k) and
A4: s c= t;
A5: ( ex m being non zero Element of NAT st m <= n & s = PFArt (m,k) ) or
    s = PFCrt (n,k) by A2,Def4;
A6: ( ex m being non zero Element of NAT st m <= n & t = PFArt (m,k) ) or
    t = PFCrt (n,k) by A3,Def4;
    per cases by A5,A6;
    suppose
      ( ex m being Element of NAT st m <= n & s = PFArt (m,k) ) & ex m
      being Element of NAT st m <= n & t = PFArt (m,k);
      hence thesis by A4,Th23;
    end;
    suppose
      ( ex m being Element of NAT st m <= n & s = PFArt (m,k) ) & t =
      PFCrt (n,k);
      hence thesis by A4,Th18;
    end;
    suppose
      s = PFCrt (n,k) & ex m being Element of NAT st m <= n & t = PFArt (m,k);
      hence thesis by A4,Th24;
    end;
    suppose
      s = PFCrt (n,k) & t = PFCrt (n,k);
      hence thesis;
    end;
  end;
  for u being set st u in PFBrt (n,k) holds u is finite by Lm8;
  then SubstitutionSet (NAT, {k}) = the carrier of SubstPoset (NAT, {k}) &
  PFBrt (n,k) in { A where A is Element of Fin PFuncs (NAT,{k}) : ( for u being
set st u in A holds u is finite ) & for s, t being Element of PFuncs (NAT,{k})
  holds ( s in A & t in A & s c= t implies s = t ) } by A1,SUBSTLAT:def 4;
  hence thesis by SUBSTLAT:def 1;
end;
