YES

Problem:
 f(f(X)) -> c(n__f(n__g(n__f(X))))
 c(X) -> d(activate(X))
 h(X) -> c(n__d(X))
 f(X) -> n__f(X)
 g(X) -> n__g(X)
 d(X) -> n__d(X)
 activate(n__f(X)) -> f(activate(X))
 activate(n__g(X)) -> g(X)
 activate(n__d(X)) -> d(X)
 activate(X) -> X

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [activate](x0) = x0,
   
   [g](x0) = x0,
   
   [n__f](x0) = 4x0 + 4,
   
   [c](x0) = x0,
   
   [d](x0) = x0,
   
   [f](x0) = 4x0 + 4,
   
   [n__d](x0) = x0,
   
   [h](x0) = 4x0 + 1,
   
   [n__g](x0) = x0
  orientation:
   f(f(X)) = 16X + 20 >= 16X + 20 = c(n__f(n__g(n__f(X))))
   
   c(X) = X >= X = d(activate(X))
   
   h(X) = 4X + 1 >= X = c(n__d(X))
   
   f(X) = 4X + 4 >= 4X + 4 = n__f(X)
   
   g(X) = X >= X = n__g(X)
   
   d(X) = X >= X = n__d(X)
   
   activate(n__f(X)) = 4X + 4 >= 4X + 4 = f(activate(X))
   
   activate(n__g(X)) = X >= X = g(X)
   
   activate(n__d(X)) = X >= X = d(X)
   
   activate(X) = X >= X = X
  problem:
   f(f(X)) -> c(n__f(n__g(n__f(X))))
   c(X) -> d(activate(X))
   f(X) -> n__f(X)
   g(X) -> n__g(X)
   d(X) -> n__d(X)
   activate(n__f(X)) -> f(activate(X))
   activate(n__g(X)) -> g(X)
   activate(n__d(X)) -> d(X)
   activate(X) -> X
  String Reversal Processor:
   f(f(X)) -> n__f(n__g(n__f(c(X))))
   c(X) -> activate(d(X))
   f(X) -> n__f(X)
   g(X) -> n__g(X)
   d(X) -> n__d(X)
   n__f(activate(X)) -> activate(f(X))
   n__g(activate(X)) -> g(X)
   n__d(activate(X)) -> d(X)
   activate(X) -> X
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                           [0]
     [activate](x0) = x0 + [1]
                           [0],
     
               [1 0 0]  
     [g](x0) = [0 0 0]x0
               [0 1 0]  ,
     
                  [1 1 0]     [0]
     [n__f](x0) = [0 1 1]x0 + [1]
                  [0 1 1]     [0],
     
               [1 1 0]     [0]
     [c](x0) = [0 1 1]x0 + [1]
               [0 1 1]     [0],
     
               [1 1 0]  
     [d](x0) = [0 0 0]x0
               [0 0 1]  ,
     
               [1 1 0]     [0]
     [f](x0) = [0 1 1]x0 + [1]
               [0 1 1]     [1],
     
                  [1 1 0]  
     [n__d](x0) = [0 0 0]x0
                  [0 0 1]  ,
     
                  [1 0 0]  
     [n__g](x0) = [0 0 0]x0
                  [0 1 0]  
    orientation:
               [1 2 1]    [1]    [1 2 1]    [1]                         
     f(f(X)) = [0 2 2]X + [3] >= [0 2 2]X + [3] = n__f(n__g(n__f(c(X))))
               [0 2 2]    [3]    [0 2 2]    [2]                         
     
            [1 1 0]    [0]    [1 1 0]    [0]                 
     c(X) = [0 1 1]X + [1] >= [0 0 0]X + [1] = activate(d(X))
            [0 1 1]    [0]    [0 0 1]    [0]                 
     
            [1 1 0]    [0]    [1 1 0]    [0]          
     f(X) = [0 1 1]X + [1] >= [0 1 1]X + [1] = n__f(X)
            [0 1 1]    [1]    [0 1 1]    [0]          
     
            [1 0 0]     [1 0 0]           
     g(X) = [0 0 0]X >= [0 0 0]X = n__g(X)
            [0 1 0]     [0 1 0]           
     
            [1 1 0]     [1 1 0]           
     d(X) = [0 0 0]X >= [0 0 0]X = n__d(X)
            [0 0 1]     [0 0 1]           
     
                         [1 1 0]    [1]    [1 1 0]    [0]                 
     n__f(activate(X)) = [0 1 1]X + [2] >= [0 1 1]X + [2] = activate(f(X))
                         [0 1 1]    [1]    [0 1 1]    [1]                 
     
                         [1 0 0]    [0]    [1 0 0]        
     n__g(activate(X)) = [0 0 0]X + [0] >= [0 0 0]X = g(X)
                         [0 1 0]    [1]    [0 1 0]        
     
                         [1 1 0]    [1]    [1 1 0]        
     n__d(activate(X)) = [0 0 0]X + [0] >= [0 0 0]X = d(X)
                         [0 0 1]    [0]    [0 0 1]        
     
                       [0]         
     activate(X) = X + [1] >= X = X
                       [0]         
    problem:
     f(f(X)) -> n__f(n__g(n__f(c(X))))
     c(X) -> activate(d(X))
     f(X) -> n__f(X)
     g(X) -> n__g(X)
     d(X) -> n__d(X)
     n__g(activate(X)) -> g(X)
     activate(X) -> X
    String Reversal Processor:
     f(f(X)) -> c(n__f(n__g(n__f(X))))
     c(X) -> d(activate(X))
     f(X) -> n__f(X)
     g(X) -> n__g(X)
     d(X) -> n__d(X)
     activate(n__g(X)) -> g(X)
     activate(X) -> X
     Bounds Processor:
      bound: 2
      enrichment: match
      automaton:
       final states: {2,10,9,8,3,6,1}
       transitions:
        d1(12) -> 13*
        activate1(11) -> 12*
        d0(7) -> 6*
        n__d0(2) -> 9*
        n__g0(3) -> 4*
        n__g0(2) -> 8*
        n__f0(2) -> 3*
        n__f0(4) -> 5*
        c0(5) -> 1*
        f90() -> 2*
        g0(2) -> 10*
        n__d1(21) -> 22*
        n__g1(19) -> 20*
        n__d2(25) -> 26*
        activate0(2) -> 7*
        7 -> 21*
        12 -> 25*
        11 -> 12*
        2 -> 7,19
        22 -> 6*
        26 -> 13,1
        13 -> 1*
        20 -> 10*
        5 -> 11*
        10 -> 7,21
      problem:
       
      Qed