reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th59:
  for x, y being Element of L holds x | (y | (y | y)) = x | x
proof
  let x, y be Element of L;
  set Y = y | (x | y);
  Y | (x | y) = y by Th25;
  hence thesis by Th58;
end;
