NO

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

Proof:
 Nonconfluence Processor:
  terms: h(h(f(x26))) *<- f(h(x26)) ->* h(x26)
  Qed