YES
Trs:
 {f(x, x) -> f(a(), b()),
      b() -> c()}
 Comment:
  We consider a non-duplicating trs.
  BOUND:
   Automaton:
    {      a_1() -> q7 | q3 | q2,
           a_0() -> q2,
     f_1(q7, q8) -> q2,
     f_1(q7, q6) -> q2,
     f_1(q7, q4) -> q2,
     f_1(q3, q8) -> q2,
     f_1(q3, q6) -> q2,
     f_1(q3, q4) -> q2,
     f_0(q8, q8) -> q2,
     f_0(q8, q7) -> q2,
     f_0(q8, q6) -> q2,
     f_0(q8, q2) -> q2,
     f_0(q7, q8) -> q2,
     f_0(q7, q7) -> q2,
     f_0(q7, q6) -> q2,
     f_0(q7, q2) -> q2,
     f_0(q6, q8) -> q2,
     f_0(q6, q7) -> q2,
     f_0(q6, q6) -> q2,
     f_0(q6, q2) -> q2,
     f_0(q2, q8) -> q2,
     f_0(q2, q7) -> q2,
     f_0(q2, q6) -> q2,
     f_0(q2, q2) -> q2,
           b_1() -> q6 | q4 | q2,
           b_0() -> q2,
           c_2() -> q8 | q6 | q2,
           c_1() -> q2,
           c_0() -> q2}
   Strict:
   {}
   Qed