let n be Ordinal; :: thesis: for T being admissible TermOrder of n
for b1, b2, b3, b4 being bag of n st b1 <= b2,T & b3 < b4,T holds
b1 + b3 < b2 + b4,T
let T be admissible TermOrder of n; :: thesis: for b1, b2, b3, b4 being bag of n st b1 <= b2,T & b3 < b4,T holds
b1 + b3 < b2 + b4,T
let b1, b2, b3, b4 be bag of n; :: thesis: ( b1 <= b2,T & b3 < b4,T implies b1 + b3 < b2 + b4,T )
assume that
A1: b1 <= b2,T and
A2: b3 < b4,T ; :: thesis: b1 + b3 < b2 + 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 :: thesis: not b1 + b3 = b2 + b4

assume b1 + b3 = b2 + b4 ; :: thesis: contradiction
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; :: thesis: verum

end;

b1 + b3 <= b2 + b4,T by A4, A3, TERMORD:8;
hence b1 + b3 < b2 + b4,T by A5, TERMORD:def 3; :: thesis: verum