NO

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

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