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

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