NO

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

Proof:
 Nonconfluence Processor:
  terms: h(f(c()),f(h(c(),a()))) *<- c() ->* f(b())
  Qed
  
  first automaton:
   final states: {3}
   transitions:
    c() -> 8,5
    b() -> 48*
    f(8) -> 9*
    f(6) -> 7*
    f(48) -> 48,5,8
    h(5,4) -> 6*
    h(9,7) -> 5,8,3
    a() -> 4*
  
  second automaton:
   final states: {1}
   transitions:
    b() -> 2*
    f(2) -> 2,1