
theorem
  for L being non empty reflexive antisymmetric RelStr holds L is up-complete
  iff for X being non empty directed Subset of L holds ex_sup_of X,L
proof
  let L be non empty reflexive antisymmetric RelStr;
  hereby
    assume
A1: L is up-complete;
    let X be non empty directed Subset of L;
    consider x being Element of L such that
A2: x is_>=_than X and
A3: for y being Element of L st y is_>=_than X holds x <= y by A1;
    thus ex_sup_of X,L
    proof
      take x;
      thus X is_<=_than x &
      for b being Element of L st X is_<=_than b holds b >= x by A2,A3;
      let c be Element of L;
      assume that
A4:   X is_<=_than c and
A5:   for b being Element of L st X is_<=_than b holds b >= c;
A6:   c <= x by A2,A5;
      c >= x by A3,A4;
      hence thesis by A6,ORDERS_2:2;
    end;
  end;
  assume
A7: for X being non empty directed Subset of L holds ex_sup_of X,L;
  let X be non empty directed Subset of L;
  ex_sup_of X,L by A7;
  then ex a being Element of L st X is_<=_than a &
  (for b being Element of L st X is_<=_than b holds b >= a) &
  for c being Element of L st X is_<=_than c &
  for b being Element of L st X is_<=_than b holds b >= c
  holds c = a;
  hence thesis;
end;
