NO

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

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