reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th46:
  for x, y, z being Element of L holds (x | (y | z)) | (x | (y | y
  )) = (x | (y | z)) | (x | (y | z))
proof
  let x, y, z be Element of L;
  (x | (y | z)) | ((y |x) | x) = (x | (y | z)) | (x | (y | y)) by Th43;
  hence thesis by Th35;
end;
