theorem Th78:
  op0(x) is Element of U & op1(x) is Element of U &
  op2(x) is Element of U
  proof
    thus op0(x) is Element of U;
    [: {x},{x} :] is Element of U;
    hence op1(x) is Element of U;
    [: [: {x},{x} :] , {x} :] is Element of U;
    hence op2(x) is Element of U by CLASSES4:13;
  end;
