reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th3:
  for x, y, z being Element of L holds (x | ((y | x) | x)) | (y | (
  z | ((x | z) | z))) = y
proof
  let x, y, z be Element of L;
  set X = (x | z);
  set Y = (z | ((x | z) | z));
  (X | ((Y | X) | X)) | x = Y by Th2;
  hence thesis by Def1;
end;
