YES
Trs:
 {f(f(x, a()), a()) -> f(f(f(a(), f(a(), a())), a()), x)}
 Comment:
  We consider a non-duplicating trs.
  BOUND:
   Automaton:
    {      a_1() -> q3,
           a_0() -> q2,
     f_1(q6, q6) -> q2,
     f_1(q6, q5) -> q2,
     f_1(q6, q2) -> q2,
     f_1(q5, q3) -> q6,
     f_1(q3, q4) -> q5,
     f_1(q3, q3) -> q4,
     f_0(q2, q2) -> q2}
   Strict:
   {}
   Qed