NO

Problem:
 b() -> f(b())
 c() -> b()
 f(f(f(c()))) -> a()
 f(h(b(),b())) -> h(h(c(),a()),b())

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