NO

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

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