theorem
  n <> m implies not <*n*> is_a_prefix_of <*m*>^s
proof
  assume
A1: n <> m;
  assume <*n*> is_a_prefix_of <*m*>^s;
then A2: ex a be FinSequence st <*m*>^s = <*n*>^a by Th1;
 m = (<*m*>^s).1 by FINSEQ_1:41
    .= n by A2,FINSEQ_1:41;
  hence contradiction by A1;
end;
