NO

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

Proof:
 Nonconfluence Processor:
  terms: f(g(g(f(g(x8))))) *<- f(g(f(g(f(x8))))) ->* g(f(g(g(f(x8)))))
  Qed