NO

Problem:
 c() -> f(b())
 c() -> b()

Proof:
 Nonconfluence Processor:
  terms: b() *<- c() ->* f(b())
  Qed
  
  first automaton:
   final states: {1}
   transitions:
    b() -> 1*
  
  second automaton:
   final states: {2}
   transitions:
    b() -> 3*
    f(3) -> 2*