YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [0  -&]  
   [g](x0) = [-& -&]x0,
   
             [0 0]  
   [f](x0) = [1 2]x0
  orientation:
             [1 2]     [0 0]                    
   f(f(X)) = [3 4]X >= [1 1]X = f(g(f(g(f(X)))))
   
                [0 0]     [0  -&]           
   f(g(f(X))) = [1 1]X >= [1  -&]X = f(g(X))
  problem:
   f(g(f(X))) -> f(g(X))
  Arctic Interpretation Processor:
   dimension: 3
   interpretation:
              [1  -& 1 ]  
    [g](x0) = [1  1  0 ]x0
              [0  0  1 ]  ,
    
              [1  1  1 ]  
    [f](x0) = [-& 1  0 ]x0
              [1  0  1 ]  
   orientation:
                 [3 3 3]     [2 2 2]           
    f(g(f(X))) = [3 3 3]X >= [2 2 1]X = f(g(X))
                 [3 3 3]     [2 1 2]           
   problem:
    
   Qed