NO

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

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