NO

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

Proof:
 Nonconfluence Processor:
  terms: h(h(c(),b()),f(f(h(h(c(),c()),c())))) *<- h(h(b(),b()),f(f(h(h(c(),c()),c())))) ->* 
  c()
  Qed
  
  first automaton:
   final states: {16}
   transitions:
    f(22) -> 23*
    f(21) -> 22*
    b() -> 24*
    c() -> 24,25,19,18,17
    h(20,17) -> 21*
    h(25,24) -> 26*
    h(19,18) -> 20*
    h(26,23) -> 16*
  
  second automaton:
   final states: {15}
   transitions:
    c() -> 15*