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
  p is Tree-yielding implies (<*n*>^q in tree(p) iff n < len p & q in p.(n+ 1
  ) )
proof
  assume
A1: p is Tree-yielding;
  thus <*n*>^q in tree(p) implies n < len p & q in p.(n+1)
  proof
    assume <*n*>^q in tree(p);
    then consider k,r such that
A2: k < len p and
A3: r in p.(k+1) and
A4: <*n*>^q = <*k*>^r by A1,Def15;
A5: (<*n*>^q).1 = n by FINSEQ_1:41;
    (<*k*>^r).1 = k by FINSEQ_1:41;
    hence thesis by A2,A3,A4,A5,FINSEQ_1:33;
  end;
  thus thesis by A1,Def15;
end;
