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

theorem Th40:
  for L being satisfying_DN_1 non empty ComplLLattStr, x, y, z
  being Element of L holds ((x + y)` + ((x + y)` + (x + z)`)`)` + y = y
proof
  let L be satisfying_DN_1 non empty ComplLLattStr;
  let x, y, z be Element of L;
  set Y = ((x + y)` + (x + z)`)`;
  ((y + Y)` + Y)`+ y = y by Th28;
  hence thesis by Th37;
end;
