NO

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

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