reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th29:
  for x, y, z, u being Element of L holds (x | (y | z))| (x | (u |
  (y | x))) = (x | (y | z)) | (y | x)
proof
  let x, y, z, u be Element of L;
  set X = x | (y | z);
  set Y = y | x;
  X | Y = x by Th25;
  hence thesis by Th17;
end;
