NO

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

Proof:
 Nonconfluence Processor:
  terms: f(h(c(),a())) *<- f(h(b(),a())) ->* a()
  Qed
  
  first automaton:
   final states: {11}
   transitions:
    f(14) -> 11*
    a() -> 12*
    c() -> 13*
    h(13,12) -> 14*
  
  second automaton:
   final states: {5}
   transitions:
    a() -> 5*