
theorem Th13e:
for n,m being Ordinal
for b1,b2 being bag of n
st support b1 = {m} & support b2 = {m} holds b1 <=' b2 iff b1.m <= b2.m
proof
let n,m be Ordinal, b1,b2 be bag of n;
assume H0: support b1 = {m} & support b2 = {m};
H1: now let l be Ordinal;
    assume l <> m;
    then H2: not l in support b1 & not l in support b2 by H0,TARSKI:def 1;
    hence b1.l = 0 by PRE_POLY:def 7;
    thus b2.l = 0 by H2,PRE_POLY:def 7;
    end;
A: now assume A1: b1 <=' b2;
   per cases;
   suppose b1 = b2;
     hence b1.m <= b2.m;
     end;
   suppose b1 <> b2; then
     consider k being Ordinal such that
     A3: b1.k < b2.k & for l being  Ordinal st l in k holds b1.l = b2.l
         by A1,PRE_POLY:def 10,PRE_POLY:def 9;
     k = m by H1,A3;
     hence b1.m <= b2.m by A3;
     end;
   end;
now assume A1: b1.m <= b2.m;
   per cases;
   suppose b1 = b2;
     hence b1 <=' b2;
     end;
   suppose A4: b1 <> b2;
     A2: now assume A5: b1.m = b2.m;
         now let o be object;
           assume A6: o in n;
           per cases;
           suppose o = m;
             hence b1.o = b2.o by A5;
             end;
           suppose A7: o <> m;
             hence b1.o = 0 by H1,A6 .= b2.o by A6,A7,H1;
             end;
           end;
         hence contradiction by A4,PBOOLE:def 10;
         end;
     ex k being Ordinal st b1.k < b2.k &
          for l being Ordinal st l in k holds b1.l = b2.l
       proof
       take k = m;
       thus b1.k < b2.k by A2,A1,XXREAL_0:1;
       now let l be Ordinal;
         assume l in k;
         then A3: l <> m;
         hence b1.l = 0 by H1 .= b2.l by A3,H1;
         end;
       hence thesis;
       end;
     hence b1 <=' b2 by PRE_POLY:def 9,PRE_POLY:def 10;
     end;
   end;
hence thesis by A;
end;
