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

theorem Th38:
  for L being satisfying_DN_1 non empty ComplLLattStr, x, y, z
  being Element of L holds x + (y + ((z + y)` + x)`)` = (z + y)` + x
proof
  let L be satisfying_DN_1 non empty ComplLLattStr;
  let x, y, z be Element of L;
  set Y = (z + y)`, Z = y`;
  x + ((Y + Z)` + (Y + x)`)` = Y + x by Th37;
  hence thesis by Th19;
end;
