YES

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

Proof:
 DP Processor:
  DPs:
   f#(f(X)) -> f#(g(f(X)))
   f#(f(X)) -> f#(g(f(g(f(X)))))
   f#(g(f(X))) -> f#(g(X))
  TRS:
   f(f(X)) -> f(g(f(g(f(X)))))
   f(g(f(X))) -> f(g(X))
  Arctic Interpretation Processor:
   dimension: 1
   usable rules:
    f(f(X)) -> f(g(f(g(f(X)))))
    f(g(f(X))) -> f(g(X))
   interpretation:
    [f#](x0) = x0 + 1,
    
    [g](x0) = -8x0 + 0,
    
    [f](x0) = 9x0 + 10
   orientation:
    f#(f(X)) = 9X + 10 >= 1X + 2 = f#(g(f(X)))
    
    f#(f(X)) = 9X + 10 >= 2X + 3 = f#(g(f(g(f(X)))))
    
    f#(g(f(X))) = 1X + 2 >= -8X + 1 = f#(g(X))
    
    f(f(X)) = 18X + 19 >= 11X + 12 = f(g(f(g(f(X)))))
    
    f(g(f(X))) = 10X + 11 >= 1X + 10 = f(g(X))
   problem:
    DPs:
     
    TRS:
     f(f(X)) -> f(g(f(g(f(X)))))
     f(g(f(X))) -> f(g(X))
   Qed