NO

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

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