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))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [f#](x0) = x0,
    
    [h](x0) = 0,
    
    [f](x0) = x0,
    
    [g](x0) = 1x0 + 0
   orientation:
    f#(g(X)) = 1X + 0 >= X = f#(X)
    
    f#(g(X)) = 1X + 0 >= X = f#(f(X))
    
    f(g(X)) = 1X + 0 >= 1X + 0 = g(f(f(X)))
    
    f(h(X)) = 0 >= 0 = h(g(X))
   problem:
    DPs:
     
    TRS:
     f(g(X)) -> g(f(f(X)))
     f(h(X)) -> h(g(X))
   Qed