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

theorem Th21:
  for L being satisfying_DN_1 non empty ComplLLattStr, x being
  Element of L holds (x + x)` = x`
proof
  let L be satisfying_DN_1 non empty ComplLLattStr;
  let x be Element of L;
  set y = the Element of L;
  set Y = (y + x)`;
  (x + (Y + x`)`)` = x` by Th20;
  hence thesis by Th19;
end;
