YES

Problem:
 f(X) -> g(n__h(n__f(X)))
 h(X) -> n__h(X)
 f(X) -> n__f(X)
 activate(n__h(X)) -> h(activate(X))
 activate(n__f(X)) -> f(activate(X))
 activate(X) -> X

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
                    [0 0]  
   [activate](x0) = [0 0]x0,
   
             [0  -&]  
   [h](x0) = [0  0 ]x0,
   
             [0  -&]  
   [g](x0) = [0  -&]x0,
   
                [0  -&]  
   [n__h](x0) = [0  0 ]x0,
   
                [0 0]  
   [n__f](x0) = [2 2]x0,
   
             [2 1]  
   [f](x0) = [2 2]x0
  orientation:
          [2 1]     [0 0]                    
   f(X) = [2 2]X >= [0 0]X = g(n__h(n__f(X)))
   
          [0  -&]     [0  -&]           
   h(X) = [0  0 ]X >= [0  0 ]X = n__h(X)
   
          [2 1]     [0 0]           
   f(X) = [2 2]X >= [2 2]X = n__f(X)
   
                       [0 0]     [0 0]                  
   activate(n__h(X)) = [0 0]X >= [0 0]X = h(activate(X))
   
                       [2 2]     [2 2]                  
   activate(n__f(X)) = [2 2]X >= [2 2]X = f(activate(X))
   
                 [0 0]          
   activate(X) = [0 0]X >= X = X
  problem:
   h(X) -> n__h(X)
   f(X) -> n__f(X)
   activate(n__h(X)) -> h(activate(X))
   activate(n__f(X)) -> f(activate(X))
   activate(X) -> X
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
                     [3  -&]  
    [activate](x0) = [0  1 ]x0,
    
              [3  2 ]  
    [h](x0) = [-& 3 ]x0,
    
                 [3  1 ]  
    [n__h](x0) = [-& 3 ]x0,
    
                 [3  0 ]  
    [n__f](x0) = [-& 3 ]x0,
    
              [3  0 ]  
    [f](x0) = [-& 3 ]x0
   orientation:
           [3  2 ]     [3  1 ]           
    h(X) = [-& 3 ]X >= [-& 3 ]X = n__h(X)
    
           [3  0 ]     [3  0 ]           
    f(X) = [-& 3 ]X >= [-& 3 ]X = n__f(X)
    
                        [6 4]     [6 3]                  
    activate(n__h(X)) = [3 4]X >= [3 4]X = h(activate(X))
    
                        [6 3]     [6 1]                  
    activate(n__f(X)) = [3 4]X >= [3 4]X = f(activate(X))
    
                  [3  -&]          
    activate(X) = [0  1 ]X >= X = X
   problem:
    h(X) -> n__h(X)
    f(X) -> n__f(X)
    activate(n__h(X)) -> h(activate(X))
    activate(n__f(X)) -> f(activate(X))
   KBO Processor:
    weight function:
     w0 = 1
     w(activate) = w(h) = w(n__h) = w(n__f) = w(f) = 1
    precedence:
     activate > h ~ f > n__h ~ n__f
    problem:
     
    Qed