NO

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

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