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

theorem Th2:
  for L being satisfying_DN_1 non empty ComplLLattStr, x, y, z, u
  being Element of L holds ((x + y)` + ((z + x)` + (y` + (y + u)`)`)`)` = y
proof
  let L be satisfying_DN_1 non empty ComplLLattStr;
  let x, y, z, u be Element of L;
  set v = the Element of L;
  ((x + z)` + (((y + u)` + x)` + (z` + (z + v)`)`)`)` = z by Th1;
  hence thesis by Th1;
end;
