theorem Th68:
  X (\) (X (\) Y) = X (/\) Y
proof
  thus X (\) (X (\) Y) = (X (\) X) (\/) X (/\) Y by Th64
    .= EmptyMS I (\/) X (/\) Y by Th52
    .= X (/\) Y by Th22,Th43;
end;
