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
   interpretation:
    [f#](x0) = -1x0 + 0,
    
    [g](x0) = -3x0 + 0,
    
    [f](x0) = 4x0 + 5
   orientation:
    f#(f(X)) = 3X + 4 >= X + 1 = f#(g(f(X)))
    
    f#(f(X)) = 3X + 4 >= 1X + 2 = f#(g(f(g(f(X)))))
    
    f#(g(f(X))) = X + 1 >= -4X + 0 = f#(g(X))
    
    f(f(X)) = 8X + 9 >= 6X + 7 = f(g(f(g(f(X)))))
    
    f(g(f(X))) = 5X + 6 >= 1X + 5 = f(g(X))
   problem:
    DPs:
     
    TRS:
     f(f(X)) -> f(g(f(g(f(X)))))
     f(g(f(X))) -> f(g(X))
   Qed