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
  p in P implies T1 = tree(T,P,T1)|p
proof
  assume
A1: p in P;
 ex q,r st q in P & r in T1 & p = q^r
  proof
    consider q such that
A2: q = p;
    consider r such that
A3: r = <*> NAT;
A4: r in T1 by A3,TREES_1:22;
 p = q^r by A2,A3,FINSEQ_1:34;
    hence thesis by A1,A2,A4;
  end;
then A5: p in tree(T,P,T1) by Def1;
  let x be FinSequence of NAT;
  thus x in T1 implies x in tree(T,P,T1)|p
  proof
    assume x in T1;
    then p^x in tree(T,P,T1) by A1,Def1;
    hence thesis by A5,TREES_1:def 6;
  end;
  thus x in tree(T,P,T1)|p implies x in T1
  proof
    assume x in tree(T,P,T1)|p;
then A6: p^x in tree(T,P,T1) by A5,TREES_1:def 6;
A7: now
      assume that p^x in T and
A8:  for r st r in P holds not r is_a_proper_prefix_of p^x;
A9:  not p is_a_proper_prefix_of p^x by A1,A8;
  p is_a_prefix_of p^x by TREES_1:1;
then   p^x = p by A9
        .= p^<*>NAT by FINSEQ_1:34;
then   x = {} by FINSEQ_1:33;
      hence thesis by TREES_1:22;
    end;
 now
      given s,r such that
A10:  s in P and
A11:  r in T1 and
A12:  p^x = s^r;
      now
        assume s <> p;
        then not s,p are_c=-comparable by A1,A10,TREES_1:def 10;
        hence contradiction by A12,Th1;
      end;
      hence thesis by A11,A12,FINSEQ_1:33;
    end;
    hence thesis by A1,A6,A7,Def1;
  end;
end;
