theorem Th78:
  (p1+p2)<X>p3 = (p1<X>p3)+(p2<X>p3)
proof
  (p1+p2)<X>p3 = -(p3<X>(p1+p2)) by Th74
   .= -((p3<X>p1)+(p3<X>p2)) by Th77
   .= -((p3<X>p1)-(p2<X>p3)) by Th74
   .= -(p3<X>p1)+(p2<X>p3) by RVSUM_1:36;
  hence thesis by Th74;
end;
