NO

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

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