theorem Th9:
  T misses S implies Sum(T \/ S) = Sum(T) + Sum(S)
proof
  consider F such that
A1: rng F = T \/ S and
A2: F is one-to-one & Sum(T \/ S) = Sum(F) by Def1;
  consider G such that
A3: rng G = T and
A4: G is one-to-one and
A5: Sum(T) = Sum(G) by Def1;
  consider H such that
A6: rng H = S and
A7: H is one-to-one and
A8: Sum(S) = Sum(H) by Def1;
  set I = G ^ H;
  assume T misses S;
  then
A9: I is one-to-one by A3,A4,A6,A7,FINSEQ_3:91;
  rng I = rng F by A1,A3,A6,FINSEQ_1:31;
  hence Sum(T \/ S) = Sum(I) by A2,A9,RLVECT_1:42
    .= Sum(T) + Sum(S) by A5,A8,RLVECT_1:41;
end;
