reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th53:
  for x, y, z being Element of L holds x | ((y | ((x | z) | y)) |
  x) = y | ((x | z) | y)
proof
  let x, y, z be Element of L;
  (y | ((x | z) | y)) | (x | x) = x by Th52;
  hence thesis by Th16;
end;
