NO

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

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