NO

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

Proof:
 Nonconfluence Processor:
  terms: h(b(),c()) *<- h(b(),b()) ->* a()
  Qed
  
  first automaton:
   final states: {5}
   transitions:
    b() -> 7*
    c() -> 5,7,6
    h(7,6) -> 5*
  
  second automaton:
   final states: {8}
   transitions:
    a() -> 8*