theorem Th8:
  i in Seg (Sum f) implies min(f^g,i)=min(f,i)
proof
  reconsider fg=f^g as FinSequence of NAT;
  assume
A1: i in Seg Sum f;
  then
A2: i<=Sum(f|min(f,i)) by Def1;
  Sum fg=Sum f+Sum g by RVSUM_1:75;
  then Sum fg>=Sum f+0 by XREAL_1:6;
  then
A3: Seg Sum f c= Seg Sum fg by FINSEQ_1:5;
  min(f,i) in dom f by A1,Def1;
  then
A4: min(f,i)<=len f by FINSEQ_3:25;
  then fg|min(f,i)=f|min(f,i) by FINSEQ_5:22;
  then
A5: min(fg,i)<=min(f,i) by A1,A3,A2,Def1;
  then fg|min(fg,i)=f|min(fg,i) by A4,FINSEQ_5:22,XXREAL_0:2;
  then i<=Sum(f|min(fg,i)) by A1,A3,Def1;
  then min(f,i)<=min(fg,i) by A1,Def1;
  hence thesis by A5,XXREAL_0:1;
end;
