reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th62:
  for x, y, z being Element of L holds x | (y | ((z | z) | x)) = x | (y | z)
proof
  let x, y, z be Element of L;
  set X = z | z;
  X | (z | z) = z by Th21;
  hence thesis by Th61;
end;
