NO

Problem:
 F(x) -> A()
 F(x) -> G(F(x))
 G(F(x)) -> F(H(x))
 G(F(x)) -> B()

Proof:
 Nonconfluence Processor:
  terms: G(A()) *<- G(F(f5())) ->* B()
  Qed
  
  first automaton:
   final states: {13}
   transitions:
    A() -> 14*
    G(14) -> 13*
  
  second automaton:
   final states: {1}
   transitions:
    B() -> 1*