NO

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

Proof:
 Nonconfluence Processor:
  terms: f(f(c())) *<- f(f(a())) ->* b()
  Qed
  
  first automaton:
   final states: {1}
   transitions:
    h(2,2) -> 7*
    h(9,8) -> 1*
    h(7,6) -> 8*
    a() -> 9*
    c() -> 9,2
    b() -> 6*
    f(6) -> 25*
    f(2) -> 3*
    f(3) -> 1*
    f(25) -> 6*
  
  second automaton:
   final states: {4}
   transitions:
    b() -> 4*
    f(4) -> 10*
    f(10) -> 4*