NO

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

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