reserve x for set,
  t,t1,t2 for DecoratedTree;

theorem
  for n being Node of t holds Subtrees (t|n) c= Subtrees t
proof
  let n be Node of t;
  let x be object;
  assume x in Subtrees (t|n);
  then consider p being Node of t|n such that
A1: x = (t|n)|p;
  dom (t|n) = (dom t)|n by TREES_2:def 10;
  then reconsider q = n^p as Node of t by TREES_1:def 6;
  x = t|q by A1,Th3;
  hence thesis;
end;
