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

theorem Th4:
  for L being satisfying_DN_1 non empty ComplLLattStr, x, y, z, u
being Element of L holds ((x + y)` + ((z + x)` + (((y + y`)` + y)` + (y + u)`)`
  )`)` = y
proof
  let L be satisfying_DN_1 non empty ComplLLattStr;
  let x, y, z, u be Element of L;
  ((y + y`)` + y)` = y` by Th3;
  hence thesis by Th2;
end;
