NO

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

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