reserve
  X,x,y,z for set,
  k,n,m for Nat ,
  f for Function,
  p,q,r for FinSequence of NAT;
reserve p1,p2,p3 for FinSequence;

theorem Th6:
  p1^<*x*> is_a_prefix_of p2 implies p1 is_a_proper_prefix_of p2
proof
  assume p1^<*x*> is_a_prefix_of p2;
  then consider p3 such that
A1: p2 = p1^<*x*>^p3 by Th1;
 p2 = p1^(<*x*>^p3) by A1,FINSEQ_1:32;
  hence p1 is_a_prefix_of p2 by Th1;
  assume p1 = p2;
then  len p1 = len(p1^<*x*>) + len p3 by A1,FINSEQ_1:22
    .= len p1 + len <*x*> + len p3 by FINSEQ_1:22
    .= len p1 + 1 + len p3 by FINSEQ_1:39;
then  len p1 + 1 <= len p1 by NAT_1:11;
  hence contradiction by NAT_1:13;
end;
