
theorem Th8:
  for L1,L2 be non empty reflexive antisymmetric RelStr st the
RelStr of L1 = the RelStr of L2 & L1 is up-complete for x1,y1 be Element of L1
  for x2,y2 be Element of L2 st x1 = x2 & y1 = y2 & x1 << y1 holds x2 << y2
proof
  let L1,L2 be non empty reflexive antisymmetric RelStr;
  assume that
A1: the RelStr of L1 = the RelStr of L2 and
A2: L1 is up-complete;
  let x1,y1 be Element of L1;
  let x2,y2 be Element of L2;
  assume that
A3: x1 = x2 and
A4: y1 = y2 and
A5: x1 << y1;
  now
    let D2 be non empty directed Subset of L2;
    reconsider D1 = D2 as Subset of L1 by A1;
    reconsider D1 as non empty directed Subset of L1 by A1,WAYBEL_0:3;
    ex_sup_of D1,L1 by A2,WAYBEL_0:75;
    then
A6: sup D1 = sup D2 by A1,YELLOW_0:26;
    assume y2 <= sup D2;
    then y1 <= sup D1 by A1,A4,A6;
    then consider d1 be Element of L1 such that
A7: d1 in D1 and
A8: x1 <= d1 by A5,WAYBEL_3:def 1;
    reconsider d2 = d1 as Element of L2 by A1;
    take d2;
    thus d2 in D2 by A7;
    thus x2 <= d2 by A1,A3,A8;
  end;
  hence thesis by WAYBEL_3:def 1;
end;
