NO

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

Proof:
 Nonconfluence Processor:
  terms: f(c()) *<- f(a()) ->* b()
  Qed
  
  first automaton:
   final states: {1}
   transitions:
    h(62,2) -> 2*
    b() -> 61*
    f(61) -> 62*
    f(2) -> 2,1
    c() -> 2*
  
  second automaton:
   final states: {3}
   transitions:
    b() -> 3*