NO

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

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