reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th35:
  for x, y, z being Element of L holds (x | (y | z)) | ((y |x) | x
  ) = (x | (y | z)) | (x | (y | z))
proof
  let x, y, z be Element of L;
  set X = x | (y | z);
  X | (y | x) = x by Th25;
  hence thesis by Th32;
end;
