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;
reserve T,T1 for DecoratedTree,
  P for AntiChain_of_Prefixes of dom T,
  t for Element of dom T,
  p1, p2, r1, r2 for FinSequence of NAT;

theorem Th14:
  for q st q in dom tree(T,P,T1) &
  q in {p^s where p is Element of dom T, s is Element of dom T1 : p in P}
  holds ex p9 being Element of dom T, r being Element of dom T1 st
  q = p9^r & p9 in P & tree(T,P,T1).q = T1.r
proof
  let q;
  assume that
A1: q in dom tree(T,P,T1) and
A2: q in {p^s where p is Element of dom T, s is Element of dom T1 : p in P};
  per cases by A1,Th10;
  suppose
A3: for p st p in P holds not p is_a_prefix_of q & tree(T,P,T1).q = T.q;
    consider p9 being Element of dom T, r being Element of dom T1 such that
A4: q= p9^r and
A5: p9 in P by A2;
    tree(T,P,T1).q = T1.r by A4,TREES_1:1,A3,A5;
   hence thesis by A4,A5;
  end;
  suppose
     ex p,r st p in P & r in dom T1 & q = p^r & tree(T,P,T1).q = T1.r;
    then consider p,r such that
A6: p in P and
A7: r in dom T1 and
A8: q = p^r and
A9: tree(T,P,T1).q = T1.r;
    consider p9 being Element of dom T, r9 being Element of dom T1 such that
A10: q = p9^r9 and
A11: p9 in P by A2;
 now
      assume
A12:  p <> p9;
      p,p9 are_c=-comparable by A8,A10,Th1;
      hence contradiction by A6,A11,A12,TREES_1:def 10;
    end;
    hence thesis by A6,A7,A8,A9;
  end;
end;
