NO

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

Proof:
 Nonconfluence Processor:
  terms: h(h(c(),c()),f(h(f(f(b())),c()))) *<- h(h(c(),c()),b()) ->* c()
  Qed
  
  first automaton:
   final states: {11}
   transitions:
    h(19,18) -> 20*
    h(15,12) -> 16*
    h(20,17) -> 11*
    b() -> 13*
    f(16) -> 13,17
    f(13) -> 14*
    f(14) -> 15*
    c() -> 19,18,12
  
  second automaton:
   final states: {21}
   transitions:
    c() -> 21*