reserve a,b,c,d,e,x,r for Real,
  A for non empty closed_interval Subset of REAL,
  f,g for PartFunc of REAL,REAL;

theorem Th11:
  A c= dom f & A c= dom g & f is_integrable_on A & f|A is bounded
  & g is_integrable_on A & g|A is bounded implies f+g is_integrable_on A & f-g
is_integrable_on A & integral(f+g,A) = integral(f,A) + integral(g,A) & integral
  (f-g,A) = integral(f,A) - integral(g,A)
proof
  assume that
A1: A c= dom f & A c= dom g and
A2: f is_integrable_on A and
A3: f|A is bounded and
A4: g is_integrable_on A and
A5: g|A is bounded;
A6: f||A|A is bounded & g||A|A is bounded by A3,A5;
A7: f||A is Function of A,REAL & g||A is Function of A,REAL by A1,Lm1;
A8: f||A + g||A = (f+g)||A & f||A - g||A = (f-g)||A by RFUNCT_1:44,47;
A9: f||A is integrable & g||A is integrable by A2,A4;
  then f||A + g||A is integrable & f||A - g||A is integrable by A6,A7,
INTEGRA1:57,INTEGRA2:33;
  hence f+g is_integrable_on A & f-g is_integrable_on A by A8;
  thus thesis by A9,A6,A7,A8,INTEGRA1:57,INTEGRA2:33;
end;
