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