NO

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

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