
theorem bag2b:
for Z being non empty set
for B being bag of Z
for o1,o2 being object st B.o1 = card B & o2 <> o1 holds B.o2 = 0
proof
let Z be non empty set, B be bag of Z, o1,o2 be object;
assume H: B.o1 = card B & o2 <> o1;
per cases;
suppose B = EmptyBag Z;
  hence thesis by PBOOLE:5;
  end;
suppose I: B <> EmptyBag Z;
  consider f being FinSequence of NAT such that
  A: degree B = Sum f & f = B*canFS(support B) by UPROOTS:def 4;
  set cS = canFS(support B);
  now assume AS: B.o2 <> 0;
    then D1: o2 in support B by PRE_POLY:def 7;
    then o2 in rng cS by FUNCT_2:def 3;
    then consider i being Nat such that
    B: i in dom cS & cS.i = o2 by FINSEQ_2:10;
    reconsider i as Element of NAT by ORDINAL1:def 12;
    G1: f.i = B.o2 by A,B,FUNCT_1:13;
    card B <> 0 by I,UPROOTS:12;
    then D2: o1 in support B by H,PRE_POLY:def 7;
    then o1 in rng cS by FUNCT_2:def 3;
    then consider j being Nat such that
    B2: j in dom cS & cS.j = o1 by FINSEQ_2:10;
    reconsider j as Element of NAT by ORDINAL1:def 12;
    B.o1 + B.o2 > B.o1 + 0 by AS,XREAL_1:8; then
    G2: f.i + f.j > card B by H,G1,A,B2,FUNCT_1:13;
    support B c= dom B by PRE_POLY:37; then
    i in dom f & j in dom f by A,B2,B,D1,D2,FUNCT_1:11;
    hence contradiction by B2,B,A,H,G2,lembag2b;
    end;
  hence thesis;
  end;
end;
