reserve T, T1 for Tree,
  P for AntiChain_of_Prefixes of T,
  p1 for FinSequence,
  p, q, r, s, p9 for FinSequence of NAT,
  x, Z for set,
  t for Element of T,
  k, n for Nat;

theorem Th3:
  {t1 where t1 is Element of T :
  for p st p in P holds not p is_a_prefix_of t1} c=
  {t1 where t1 is Element of T :
  for p st p in P holds not p is_a_proper_prefix_of t1}
proof
  let x be object;
  assume x in {t1 where t1 is Element of T :
  for p st p in P holds not p is_a_prefix_of t1};
  then consider t9 being Element of T such that
A1: x = t9 and
A2: for p st p in P holds not p is_a_prefix_of t9;
   for p st p in P holds not p is_a_proper_prefix_of t9 by A2;
  hence thesis by A1;
end;
