
theorem Th6:
  for n being Ordinal, T being admissible TermOrder of n, b1,b2,b3,
  b4 being bag of n holds b1 <= b2,T & b3 < b4,T implies b1 + b3 < b2 + b4,T
proof
  let n be Ordinal, T be admissible TermOrder of n, b1,b2,b3,b4 be bag of n;
  assume that
A1: b1 <= b2,T and
A2: b3 < b4,T;
  b3 <= b4,T by A2,TERMORD:def 3;
  then [b3,b4] in T by TERMORD:def 2;
  then [b2+b3,b2+b4] in T by BAGORDER:def 5;
  then
A3: b2+b3 <= b2+b4,T by TERMORD:def 2;
  [b1,b2] in T by A1,TERMORD:def 2;
  then [b1+b3,b2+b3] in T by BAGORDER:def 5;
  then
A4: b1+b3 <= b2+b3,T by TERMORD:def 2;
A5: now
    assume b1+b3 = b2+b4;
    then
A6: b2+b4 = b2 +b3 by A4,A3,TERMORD:7;
    b4 = b4 + b2 -' b2 & b3 = b3 + b2 -' b2 by PRE_POLY:48;
    hence contradiction by A2,A6,TERMORD:def 3;
  end;
  b1+b3 <= b2+b4,T by A4,A3,TERMORD:8;
  hence thesis by A5,TERMORD:def 3;
end;
