NO

Problem:
 f(g(x)) -> g(f(f(x)))
 g(x) -> x

Proof:
 Nonconfluence Processor:
  terms: f(f2()) *<- f(g(f2())) ->* g(f(f(f2())))
  Qed
  
  first automaton:
   final states: {1}
   transitions:
    f(2) -> 1*
    f2() -> 2*
  
  second automaton:
   final states: {3}
   transitions:
    g(6) -> 3*
    f(4) -> 5*
    f(5) -> 6*
    f2() -> 4*
    6 -> 3*