reserve x,y,z for object, X,Y for set,
  i,k,n for Nat,
  p,q,r,s for FinSequence,
  w for FinSequence of NAT,
  f for Function;

theorem
  for T1, T2 being Tree st T1-level 1 c= T2-level 1 &
  for n being Element of NAT st <*n*> in T1 holds T1|<*n*> = T2|<*n*>
   holds T1 c= T2
proof
  let T1, T2 be Tree;
  assume that
A1: T1-level 1 c= T2-level 1 and
A2: for n being Element of NAT st <*n*> in T1 holds T1|<*n*> = T2|<*n*>;
  let x be object;
  assume x in T1;
  then reconsider t = x as Element of T1;
  now
    assume t <> {};
    then consider p being FinSequence of NAT,
       n being Element of NAT such that
A3: t = <*n*>^p by FINSEQ_2:130;
A4: len <*n*> = 1 by FINSEQ_1:39;
    reconsider n as Element of NAT;
    reconsider q = <*n*> as Element of T1 by A3,TREES_1:21;
A5: q in T1-level 1 by A4;
    then
A6: q in T2-level 1 by A1;
A7: p in T1|q by A3,TREES_1:def 6;
    T1|<*n*> = T2|<*n*> by A2,A5;
    hence t in T2 by A3,A6,A7,TREES_1:def 6;
  end;
  hence thesis by TREES_1:22;
end;
