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 Th2:
  P <> {} implies tree(T,P,T1) = {t1 where t1 is Element of T :
  for p st p in P holds not p is_a_proper_prefix_of t1}
  \/ { p^s where p is Element of T, s is Element of T1 : p in P }
proof
  assume
A1: P <> {};
  thus tree(T,P,T1) c=
  {t : for p st p in P holds not p is_a_proper_prefix_of t} \/
  { p^s where p is Element of T, s is Element of T1 : p in P }
  proof
    let x be object;
    assume
A2: x in tree(T,P,T1);
    then reconsider q = x as FinSequence of NAT by TREES_1:19;
A3: now
      given p,r such that
A4:   p in P & r in T1 & q = p^r;
      P c= T by TREES_1:def 11;
      hence x in { p9^s where p9 is Element of T,
      s is Element of T1 : p9 in P } by A4;
    end;
    q in T & (for p st p in P holds not p is_a_proper_prefix_of q) implies
    x in { t : for p st p in P holds not p is_a_proper_prefix_of t };
    hence thesis by A1,A2,A3,Def1,XBOOLE_0:def 3;
  end;
  let x be object such that
A5: x in { t : for p st p in P holds not p is_a_proper_prefix_of t } \/
  { p^s where p is Element of T, s is Element of T1 : p in P };
A6: now
    assume x in { p^s where p is Element of T, s is Element of T1 : p in P };
    then ex p being Element of T st
      ex s being Element of T1 st x = p^s & p in P;
    hence thesis by Def1;
  end;
 now
    assume x in { t : for p st p in P holds not p is_a_proper_prefix_of t };
    then ex t st x = t & for p st p in P holds not p is_a_proper_prefix_of t;
    hence thesis by A1,Def1;
  end;
  hence thesis by A5,A6,XBOOLE_0:def 3;
end;
