reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th43:
  for x, y being Element of L holds (x | y) | y = y | (x | x)
proof
  let x, y be Element of L;
  set X = x;
  set Y = y;
  Y | (X | Y) = (X | Y) | Y by Th20;
  hence thesis by Th42;
end;
