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

theorem Th26:
  for L being satisfying_DN_1 non empty ComplLLattStr, x, y, z
  being Element of L holds x + ((y + z)` + (y + x)`)` = (y + x )``
proof
  let L be satisfying_DN_1 non empty ComplLLattStr;
  let x, y, z be Element of L;
  (x + ((y + z)` + (y + x)`)`)`` = (y + x)`` by Th10;
  hence thesis by Th23;
end;
