reserve L for satisfying_DN_1 non empty ComplLLattStr;
reserve x, y, z for Element of L;

theorem Th44:
  for L being satisfying_DN_1 non empty ComplLLattStr, x, y being
  Element of L holds (x + y)` + x = x + y`
proof
  let L be satisfying_DN_1 non empty ComplLLattStr;
  let x, y be Element of L;
  set Z = x;
  x + (y + ((Z + y)` + x)`)` = (Z + y)` + x by Th38;
  hence thesis by Th28;
end;
