NO

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

Proof:
 Nonconfluence Processor:
  terms: f(h(a(),f(a()))) *<- f(h(b(),f(a()))) ->* f(c())
  Qed
  
  first automaton:
   final states: {7}
   transitions:
    h(10,9) -> 11*
    a() -> 10,8
    f(11) -> 7*
    f(8) -> 9*
  
  second automaton:
   final states: {12}
   transitions:
    h(13,13) -> 13*
    c() -> 13*
    f(13) -> 12*