reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th68:
  for L being non empty ShefferStr st L is satisfying_Sh_1 holds L
  is satisfying_Sheffer_2
proof
  let L be non empty ShefferStr;
  assume L is satisfying_Sh_1;
  then for x, y being Element of L holds x | (y | (y | y)) = x | x by Th59;
  hence thesis by SHEFFER1:def 14;
end;
