NO

Problem:
 g(x) -> h(k(x))
 g(x) -> x
 h(k(x)) -> f(x)
 f(x) -> x
 k(c()) -> c()
 f(c()) -> g(c())

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