NO

Problem:
 a(x) -> g(g(b(x)))
 a(x) -> g(c(x))
 b(x) -> g(b(x))

Proof:
 Nonconfluence Processor:
  terms: g(g(b(f4()))) *<- a(f4()) ->* g(c(f4()))
  Qed
  
  first automaton:
   final states: {4}
   transitions:
    b(5) -> 6*
    f4() -> 5*
    g(7) -> 4*
    g(6) -> 6,7
  
  second automaton:
   final states: {1}
   transitions:
    c(2) -> 3*
    f4() -> 2*
    g(3) -> 1*