YES
Trs:
 {f(a(), b()) -> f(a(), c()),
  f(c(), d()) -> f(b(), d())}
 Comment:
  We consider a non-duplicating trs.
  BOUND:
   Automaton:
    {      a_1() -> q6,
           a_0() -> q5,
           c_1() -> q7,
           c_0() -> q5,
           d_1() -> q7,
           d_0() -> q5,
           b_1() -> q6,
           b_0() -> q5,
     f_1(q6, q7) -> q5,
     f_0(q5, q5) -> q5}
   Strict:
   {}
   Qed