theorem  :: (2.40)
  -(a "/\" b) = ((-a) "\/" (-b))
  proof
    (a "/\" b)` = (a`` "/\" b)` by ROBBINS3:def 6
               .= (a`` "/\" b ``)` by ROBBINS3:def 6
               .= (a` "\/" b`)`` by Th1
               .= (a` "\/" b`) by ROBBINS3:def 6;
    hence thesis;
  end;
