theorem Th20:
  f|A is bounded implies
  for D,D1 ex D2 st D<=D2 & D1<=D2 & rng D2=rng D1 \/ rng D &
    0 <= lower_sum(f,D2)-lower_sum(f,D) & 0<=lower_sum(f,D2)-lower_sum(f,D1)
  proof
    assume
A1: f|A is bounded;
    for D,D1 ex D2 st D<=D2 & D1<=D2 & rng D2=rng D1 \/ rng D &
    0 <= lower_sum(f,D2)-lower_sum(f,D) & 0<=lower_sum(f,D2)-lower_sum(f,D1)
    proof
      let D,D1;
      consider D2 such that
A6:   D<=D2 and
A7:   D1<=D2 and
A8:   rng D2=rng D1 \/ rng D by Th4;
A9:   lower_sum(f,D2)-lower_sum(f,D1)>=0 by A1,A7,INTEGRA1:46,XREAL_1:48;
      lower_sum(f,D2)-lower_sum(f,D)>=0 by A1,A6,INTEGRA1:46,XREAL_1:48;
      hence thesis by A6,A7,A8,A9;
    end;
    hence thesis;
  end;
