NO

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

Proof:
 Nonconfluence Processor:
  terms: f(f(f(h(f(f(c())),f(f(h(f(f(c())),h(h(b(),h(c(),c())),f(c()))))))))) *<- 
  f(f(f(f(a())))) ->* a()
  Qed