NO

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

Proof:
 Nonconfluence Processor:
  terms: h(h(h(b(),h(h(f(a()),c()),f(a()))),f(f(a()))),f(f(f(a())))) *<- 
  h(h(h(b(),b()),f(f(a()))),f(f(f(a())))) ->* a()
  Qed
  
  first automaton:
   final states: {1}
   transitions:
    h(17,8) -> 18*
    h(14,10) -> 15*
    h(16,15) -> 17*
    h(14,3) -> 16*
    h(13,11) -> 14*
    h(18,5) -> 1*
    c() -> 11*
    f(12) -> 13*
    f(9) -> 10*
    f(4) -> 5*
    f(7) -> 8*
    f(11) -> 48*
    f(6) -> 7*
    f(3) -> 4*
    f(2) -> 3*
    f(48) -> 11*
    a() -> 12,9,6,2
    b() -> 16*
  
  second automaton:
   final states: {19}
   transitions:
    a() -> 19*