NO

Problem:
 a() -> a()
 f(f(x,a()),a()) -> b()

Proof:
 Nonconfluence Processor:
  terms: f(b(),a()) *<- f(f(f(f4(),a()),a()),a()) ->* b()
  Qed
  
  first automaton:
   final states: {1}
   transitions:
    a() -> 2*
    f(3,2) -> 1*
    b() -> 3*
  
  second automaton:
   final states: {4}
   transitions:
    b() -> 4*