theorem
  for B1,B2 being finite natural-membered set st
     B1 <N< B2 holds
Sum (SubXFinS(cF,B1\/B2))=Sum (SubXFinS(cF,B1))+Sum(SubXFinS(cF,B2))
proof
  let B1,B2 be finite natural-membered set such that A1: B1 <N< B2;
  set B12=B1\/B2;
  set B12L=B12/\len cF;
  set B1L=B1/\len cF;
  set B2L=B2/\len cF;
  B1L\/B2L=B12L by XBOOLE_1:23;
  then A3:Sgm0(B12L) = Sgm0(B1L) ^ Sgm0(B2L) by Th35,A1,Th25;
rng Sgm0(B1L) = B1L & rng Sgm0(B2L) = B2L by Def4;
  then rng Sgm0(B1L) c= dom cF & rng Sgm0(B2L) c= dom cF by XBOOLE_1:17;
  then SubXFinS (cF,B1) ^ SubXFinS (cF,B2) = SubXFinS (cF,B12) by A3,Th69;
hence thesis by Th54;
end;
