
theorem Th5:
  for L being non empty reflexive RelStr, x being Element of L
  holds {x} is directed filtered
proof
  let L be non empty reflexive RelStr;
  let x be Element of L;
  set X = {x};
  hereby
    let z,y be Element of L;
    assume that
A1: z in X and
A2: y in X;
    take x;
    thus x in X & z <= x & y <= x by A1,A2,TARSKI:def 1;
  end;
  hereby
    let z,y be Element of L;
    assume that
A3: z in X and
A4: y in X;
    take x;
    thus x in X & x <= z & x <= y by A3,A4,TARSKI:def 1;
  end;
end;
