YES
Trs:
 {a b -> a c b}
 Comment:
  We consider a non-duplicating trs.
  BOUND:
   Automaton:
    {b_1(q3) -> q4,
     b_0(q2) -> q2,
     c_1(q4) -> q5,
     c_0(q2) -> q2,
     a_1(q5) -> q6,
     a_0(q2) -> q2,
          q6 -> q2,
          q2 -> q3}
   Strict:
   {}
   Qed