NO

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

Proof:
 Nonconfluence Processor:
  terms: g(f(f(f4()))) *<- f(g(f(f4()))) ->* f4()
  Qed
  
  first automaton:
   final states: {8}
   transitions:
    f(10) -> 11*
    f(9) -> 10*
    g(11) -> 8*
    f4() -> 9*
  
  second automaton:
   final states: {12}
   transitions:
    f4() -> 12*