NO

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

Proof:
 Nonconfluence Processor:
  terms: f(f3(),b()) *<- f(f3(),a()) ->* a()
  Qed
  
  first automaton:
   final states: {11}
   transitions:
    f3() -> 13*
    b() -> 12*
    f(13,12) -> 11*
  
  second automaton:
   final states: {6}
   transitions:
    a() -> 6*
    b() -> 6*
    f(6,6) -> 6*