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 Th6:
  for T, T1, P holds
  P c= { p^s where p is Element of T, s is Element of T1 : p in P }
proof
  let T, T1, P;
 now
    let x be object;
    assume
A1: x in P;
 P c= T by TREES_1:def 11;
    then consider q being Element of T such that
A2: q = x by A1;
 <*> NAT in T1 by TREES_1:22;
    then consider s9 being Element of T1 such that
A3: s9 = <*> NAT;
 q = q^s9 by A3,FINSEQ_1:34;
    thus
    then x in { p^s where p is Element of T, s is Element of T1 : p in P }
    by A1,A2;
  end;
  hence thesis;
end;
