reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th31:
  for x, y, z, u being Element of L holds (x | (y | z)) | (x | (u
  | (y | x))) = x
proof
  let x, y, z be Element of L;
  (x | (y | z)) | (y | x) = x by Th25;
  hence thesis by Th29;
end;
