NO

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

Proof:
 Nonconfluence Processor:
  terms: f(a(),b()) *<- f(a(),f(a(),f(a(),f3()))) ->* b()
  Qed
  
  first automaton:
   final states: {1}
   transitions:
    a() -> 3*
    f(3,2) -> 1*
    b() -> 2*
  
  second automaton:
   final states: {4}
   transitions:
    b() -> 4*