YES

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

Proof:
 DP Processor:
  DPs:
   f#(g(X)) -> f#(X)
   f#(g(X)) -> f#(f(X))
  TRS:
   f(g(X)) -> g(f(f(X)))
   f(h(X)) -> h(g(X))
  KBO Processor:
   argument filtering:
    pi(g) = [0]
    pi(f) = 0
    pi(h) = []
    pi(f#) = 0
   weight function:
    w0 = 1
    w(f#) = w(h) = 1
    w(f) = w(g) = 0
   precedence:
    f# ~ h ~ f ~ g
   problem:
    DPs:
     
    TRS:
     f(g(X)) -> g(f(f(X)))
     f(h(X)) -> h(g(X))
   Qed