
theorem
  for k being Element of NAT holds PFBrt (1,k) = { PFArt (1,k), PFCrt (1 ,k) }
proof
  let k be Element of NAT;
  thus PFBrt (1,k) c= { PFArt (1,k), PFCrt (1,k) }
  proof
    let x be object;
    assume
A1: x in PFBrt (1,k);
    per cases by A1,Def4;
    suppose
      ex m being non zero Element of NAT st m <= 1 & x = PFArt (m,k);
      then consider m being non zero Element of NAT such that
A2:   m <= 1 and
A3:   x = PFArt (m,k);
      m >= 1 by NAT_1:14;
      then m = 1 by A2,XXREAL_0:1;
      hence thesis by A3,TARSKI:def 2;
    end;
    suppose
      x = PFCrt (1,k);
      hence thesis by TARSKI:def 2;
    end;
  end;
  let x be object;
  assume
A4: x in { PFArt (1,k), PFCrt (1,k) };
  per cases by A4,TARSKI:def 2;
  suppose
    x = PFArt (1,k);
    hence thesis by Def4;
  end;
  suppose
    x = PFCrt (1,k);
    hence thesis by Def4;
  end;
end;
