NO

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

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