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

theorem Th6:
  for L being satisfying_DN_1 non empty ComplLLattStr, x, y being
  Element of L holds ((x + y)` + (x` + y)`)` = y
proof
  let L be satisfying_DN_1 non empty ComplLLattStr;
  let x, y be Element of L;
  set Z = (x + x`)`;
  ((x + y)` + ((Z + x)` + y)`)` = y by Th5;
  hence thesis by Th3;
end;
