theorem Th1:
  [!f,x,x+h!] = (fdif(f,h).1.x)/h
proof
  [!f,x,x+h!] = [!f,x+h,x!] by DIFF_1:29
    .= (fD(f,h).x)/h by DIFF_1:3
    .= (fD(fdif(f,h).0,h).x)/h by DIFF_1:def 6
    .= (fdif(f,h).(0 qua Nat+1).x)/h by DIFF_1:def 6;
  hence thesis;
end;
