NO

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

Proof:
 Nonconfluence Processor:
  terms: h(h(c(),h(a(),b())),b()) *<- h(h(c(),a()),b()) ->* b()
  Qed
  
  first automaton:
   final states: {11}
   transitions:
    b() -> 13,12
    c() -> 16*
    a() -> 14*
    h(16,15) -> 17*
    h(14,13) -> 15*
    h(17,12) -> 11*
    h(14,12) -> 14*
  
  second automaton:
   final states: {10}
   transitions:
    b() -> 10*