We consider the following Problem:

  Strict Trs:
    {  f(g(X), Y) -> f(X, n__f(n__g(X), activate(Y)))
     , f(X1, X2) -> n__f(X1, X2)
     , g(X) -> n__g(X)
     , activate(n__f(X1, X2)) -> f(activate(X1), X2)
     , activate(n__g(X)) -> g(activate(X))
     , activate(X) -> X}
  StartTerms: basic terms
  Strategy: innermost

Certificate: YES(?,O(n^1))

Proof:
  Arguments of following rules are not normal-forms:
  {f(g(X), Y) -> f(X, n__f(n__g(X), activate(Y)))}
  
  All above mentioned rules can be savely removed.
  
  We consider the following Problem:
  
    Strict Trs:
      {  f(X1, X2) -> n__f(X1, X2)
       , g(X) -> n__g(X)
       , activate(n__f(X1, X2)) -> f(activate(X1), X2)
       , activate(n__g(X)) -> g(activate(X))
       , activate(X) -> X}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {g(X) -> n__g(X)}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(f) = {1}, Uargs(g) = {1}, Uargs(n__f) = {}, Uargs(n__g) = {},
        Uargs(activate) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       f(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                   [0 0]      [0 0]      [1]
       g(x1) = [1 0] x1 + [2]
               [0 0]      [1]
       n__f(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                      [0 0]      [0 0]      [0]
       n__g(x1) = [1 0] x1 + [0]
                  [0 0]      [0]
       activate(x1) = [1 0] x1 + [1]
                      [1 0]      [1]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  f(X1, X2) -> n__f(X1, X2)
         , activate(n__f(X1, X2)) -> f(activate(X1), X2)
         , activate(n__g(X)) -> g(activate(X))
         , activate(X) -> X}
      Weak Trs: {g(X) -> n__g(X)}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {f(X1, X2) -> n__f(X1, X2)}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(f) = {1}, Uargs(g) = {1}, Uargs(n__f) = {}, Uargs(n__g) = {},
          Uargs(activate) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         f(x1, x2) = [1 0] x1 + [0 0] x2 + [2]
                     [0 0]      [0 0]      [0]
         g(x1) = [1 0] x1 + [0]
                 [0 0]      [1]
         n__f(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                        [0 0]      [0 0]      [0]
         n__g(x1) = [1 0] x1 + [0]
                    [0 0]      [0]
         activate(x1) = [1 0] x1 + [1]
                        [1 0]      [1]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  activate(n__f(X1, X2)) -> f(activate(X1), X2)
           , activate(n__g(X)) -> g(activate(X))
           , activate(X) -> X}
        Weak Trs:
          {  f(X1, X2) -> n__f(X1, X2)
           , g(X) -> n__g(X)}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {activate(X) -> X}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(f) = {1}, Uargs(g) = {1}, Uargs(n__f) = {}, Uargs(n__g) = {},
            Uargs(activate) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           f(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                       [0 0]      [0 1]      [1]
           g(x1) = [1 0] x1 + [3]
                   [0 0]      [1]
           n__f(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                          [0 0]      [0 0]      [1]
           n__g(x1) = [1 0] x1 + [0]
                      [0 0]      [0]
           activate(x1) = [1 0] x1 + [1]
                          [0 1]      [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs:
            {  activate(n__f(X1, X2)) -> f(activate(X1), X2)
             , activate(n__g(X)) -> g(activate(X))}
          Weak Trs:
            {  activate(X) -> X
             , f(X1, X2) -> n__f(X1, X2)
             , g(X) -> n__g(X)}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          We consider the following Problem:
          
            Strict Trs:
              {  activate(n__f(X1, X2)) -> f(activate(X1), X2)
               , activate(n__g(X)) -> g(activate(X))}
            Weak Trs:
              {  activate(X) -> X
               , f(X1, X2) -> n__f(X1, X2)
               , g(X) -> n__g(X)}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(?,O(n^1))
          
          Proof:
            The problem is match-bounded by 1.
            The enriched problem is compatible with the following automaton:
            {  f_0(2, 2) -> 1
             , f_1(3, 2) -> 1
             , f_1(3, 2) -> 3
             , g_0(2) -> 1
             , g_1(3) -> 1
             , g_1(3) -> 3
             , n__f_0(2, 2) -> 1
             , n__f_0(2, 2) -> 2
             , n__f_0(2, 2) -> 3
             , n__f_1(3, 2) -> 1
             , n__f_1(3, 2) -> 3
             , n__g_0(2) -> 1
             , n__g_0(2) -> 2
             , n__g_0(2) -> 3
             , n__g_1(3) -> 1
             , n__g_1(3) -> 3
             , activate_0(2) -> 1
             , activate_1(2) -> 3}

Hurray, we answered YES(?,O(n^1))