theorem Th32:
  i in dom (p1-p2) & a1 = p1.i & a2 = p2.i implies (p1-p2).i = a1 - a2
proof
  assume i in dom (p1-p2) & a1 = p1.i & a2 = p2.i;
  then (p1 - p2).i = (diffield(K)).(a1,a2) by FUNCOP_1:22;
  hence thesis by Th11;
end;
