
theorem Th7:
  for X being set, S, T being finite Subset of X, n being Element
  of NAT st S misses T holds (S \/ T, n)-bag = (S,n)-bag + (T,n)-bag
proof
  let X be set, S, T be finite Subset of X, n be Element of NAT;
  assume S misses T;
  then
A1: S /\ T = {} by XBOOLE_0:def 7;
  now
    let i be object such that
    i in X;
    per cases by XBOOLE_0:def 3;
    suppose
A2:   not i in S \/ T;
      then
A3:   not i in T by XBOOLE_0:def 3;
A4:   not i in S by A2,XBOOLE_0:def 3;
      thus (S \/ T, n)-bag.i = 0 by A2,Th3
        .= (S,n)-bag.i + (0 qua Nat) by A4,Th3
        .= (S,n)-bag.i + (T,n)-bag.i by A3,Th3
        .= ((S,n)-bag + (T,n)-bag).i by PRE_POLY:def 5;
    end;
    suppose
A5:   i in S;
      then
A6:   not i in T by A1,XBOOLE_0:def 4;
      i in S \/ T by A5,XBOOLE_0:def 3;
      hence (S \/ T, n)-bag.i = n by Th4
        .= (S,n)-bag.i + (0 qua Nat) by A5,Th4
        .= (S,n)-bag.i +(T,n)-bag.i by A6,Th3
        .= ((S,n)-bag + (T,n)-bag).i by PRE_POLY:def 5;
    end;
    suppose
A7:   i in T;
      then
A8:   not i in S by A1,XBOOLE_0:def 4;
      i in S \/ T by A7,XBOOLE_0:def 3;
      hence (S \/ T, n)-bag.i = n by Th4
        .= (T,n)-bag.i + (0 qua Nat) by A7,Th4
        .= (S,n)-bag.i + (T,n)-bag.i by A8,Th3
        .= ((S,n)-bag + (T,n)-bag).i by PRE_POLY:def 5;
    end;
  end;
  hence thesis by PBOOLE:3;
end;
