reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th50:
  for x, y, z being Element of L holds (((x | y) | (x | y)) | ((z
  | ((x | y) | z)) | (x | y))) | (x | x) = (z | ((x | y) | z)) | (x | x)
proof
  let x, y, z be Element of L;
  set Y = z;
  set X = x | y;
  (Y | (X | Y)) | X = X | X by Th36;
  hence thesis by Th48;
end;
