We consider the following Problem:

  Strict Trs:
    {  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}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  Arguments of following rules are not normal-forms:
  {f(f(X)) -> c(n__f(n__g(n__f(X))))}
  
  All above mentioned rules can be savely removed.
  
  We consider the following Problem:
  
    Strict Trs:
      {  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}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component:
      {  h(X) -> c(n__d(X))
       , f(X) -> n__f(X)
       , d(X) -> n__d(X)}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(f) = {1}, Uargs(c) = {}, Uargs(n__f) = {}, Uargs(n__g) = {},
        Uargs(d) = {1}, Uargs(activate) = {}, Uargs(h) = {},
        Uargs(n__d) = {}, Uargs(g) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       f(x1) = [1 0] x1 + [1]
               [0 0]      [1]
       c(x1) = [1 0] x1 + [1]
               [1 0]      [1]
       n__f(x1) = [1 0] x1 + [0]
                  [0 0]      [0]
       n__g(x1) = [0 0] x1 + [0]
                  [0 0]      [0]
       d(x1) = [1 0] x1 + [1]
               [1 0]      [1]
       activate(x1) = [1 0] x1 + [0]
                      [0 0]      [1]
       h(x1) = [1 0] x1 + [2]
               [1 0]      [1]
       n__d(x1) = [1 0] x1 + [0]
                  [0 0]      [0]
       g(x1) = [0 0] x1 + [0]
               [0 0]      [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  c(X) -> d(activate(X))
         , g(X) -> n__g(X)
         , activate(n__f(X)) -> f(activate(X))
         , activate(n__g(X)) -> g(X)
         , activate(n__d(X)) -> d(X)
         , activate(X) -> X}
      Weak Trs:
        {  h(X) -> c(n__d(X))
         , f(X) -> n__f(X)
         , d(X) -> n__d(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(c) = {}, Uargs(n__f) = {}, Uargs(n__g) = {},
          Uargs(d) = {1}, Uargs(activate) = {}, Uargs(h) = {},
          Uargs(n__d) = {}, Uargs(g) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         f(x1) = [1 0] x1 + [1]
                 [0 0]      [1]
         c(x1) = [1 0] x1 + [1]
                 [1 0]      [1]
         n__f(x1) = [1 0] x1 + [0]
                    [0 0]      [0]
         n__g(x1) = [0 0] x1 + [0]
                    [0 0]      [0]
         d(x1) = [1 0] x1 + [1]
                 [1 0]      [1]
         activate(x1) = [1 0] x1 + [0]
                        [0 0]      [1]
         h(x1) = [1 0] x1 + [2]
                 [1 0]      [2]
         n__d(x1) = [1 0] x1 + [0]
                    [0 0]      [0]
         g(x1) = [0 0] x1 + [2]
                 [0 0]      [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  c(X) -> d(activate(X))
           , activate(n__f(X)) -> f(activate(X))
           , activate(n__g(X)) -> g(X)
           , activate(n__d(X)) -> d(X)
           , activate(X) -> X}
        Weak Trs:
          {  g(X) -> n__g(X)
           , h(X) -> c(n__d(X))
           , f(X) -> n__f(X)
           , d(X) -> n__d(X)}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {c(X) -> d(activate(X))}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(f) = {1}, Uargs(c) = {}, Uargs(n__f) = {}, Uargs(n__g) = {},
            Uargs(d) = {1}, Uargs(activate) = {}, Uargs(h) = {},
            Uargs(n__d) = {}, Uargs(g) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           f(x1) = [1 0] x1 + [1]
                   [0 0]      [1]
           c(x1) = [1 0] x1 + [3]
                   [1 0]      [1]
           n__f(x1) = [1 0] x1 + [0]
                      [0 0]      [0]
           n__g(x1) = [0 0] x1 + [0]
                      [0 0]      [0]
           d(x1) = [1 0] x1 + [1]
                   [1 0]      [1]
           activate(x1) = [1 0] x1 + [0]
                          [0 0]      [1]
           h(x1) = [1 0] x1 + [3]
                   [1 0]      [2]
           n__d(x1) = [1 0] x1 + [0]
                      [0 0]      [0]
           g(x1) = [0 0] x1 + [0]
                   [0 0]      [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs:
            {  activate(n__f(X)) -> f(activate(X))
             , activate(n__g(X)) -> g(X)
             , activate(n__d(X)) -> d(X)
             , activate(X) -> X}
          Weak Trs:
            {  c(X) -> d(activate(X))
             , g(X) -> n__g(X)
             , h(X) -> c(n__d(X))
             , f(X) -> n__f(X)
             , d(X) -> n__d(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(n__g(X)) -> g(X)
             , activate(n__d(X)) -> d(X)}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(f) = {1}, Uargs(c) = {}, Uargs(n__f) = {}, Uargs(n__g) = {},
              Uargs(d) = {1}, Uargs(activate) = {}, Uargs(h) = {},
              Uargs(n__d) = {}, Uargs(g) = {}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             f(x1) = [1 0] x1 + [0]
                     [0 0]      [1]
             c(x1) = [1 0] x1 + [1]
                     [0 0]      [1]
             n__f(x1) = [1 0] x1 + [0]
                        [0 0]      [0]
             n__g(x1) = [0 0] x1 + [0]
                        [0 0]      [0]
             d(x1) = [1 0] x1 + [0]
                     [0 0]      [1]
             activate(x1) = [1 0] x1 + [1]
                            [1 0]      [1]
             h(x1) = [1 0] x1 + [2]
                     [0 0]      [2]
             n__d(x1) = [1 0] x1 + [0]
                        [0 0]      [0]
             g(x1) = [0 0] x1 + [0]
                     [0 0]      [0]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Strict Trs:
              {  activate(n__f(X)) -> f(activate(X))
               , activate(X) -> X}
            Weak Trs:
              {  activate(n__g(X)) -> g(X)
               , activate(n__d(X)) -> d(X)
               , c(X) -> d(activate(X))
               , g(X) -> n__g(X)
               , h(X) -> c(n__d(X))
               , f(X) -> n__f(X)
               , d(X) -> n__d(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(c) = {}, Uargs(n__f) = {}, Uargs(n__g) = {},
                Uargs(d) = {1}, Uargs(activate) = {}, Uargs(h) = {},
                Uargs(n__d) = {}, Uargs(g) = {}
              We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
              Interpretation Functions:
               f(x1) = [1 0] x1 + [0]
                       [0 0]      [1]
               c(x1) = [1 0] x1 + [1]
                       [0 1]      [1]
               n__f(x1) = [1 0] x1 + [0]
                          [0 0]      [0]
               n__g(x1) = [0 0] x1 + [0]
                          [0 0]      [0]
               d(x1) = [1 0] x1 + [0]
                       [0 0]      [1]
               activate(x1) = [1 0] x1 + [1]
                              [0 1]      [1]
               h(x1) = [1 0] x1 + [1]
                       [0 0]      [2]
               n__d(x1) = [1 0] x1 + [0]
                          [0 0]      [0]
               g(x1) = [0 0] x1 + [0]
                       [0 0]      [0]
            
            The strictly oriented rules are moved into the weak component.
            
            We consider the following Problem:
            
              Strict Trs: {activate(n__f(X)) -> f(activate(X))}
              Weak Trs:
                {  activate(X) -> X
                 , activate(n__g(X)) -> g(X)
                 , activate(n__d(X)) -> d(X)
                 , c(X) -> d(activate(X))
                 , g(X) -> n__g(X)
                 , h(X) -> c(n__d(X))
                 , f(X) -> n__f(X)
                 , d(X) -> n__d(X)}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(?,O(n^1))
            
            Proof:
              We consider the following Problem:
              
                Strict Trs: {activate(n__f(X)) -> f(activate(X))}
                Weak Trs:
                  {  activate(X) -> X
                   , activate(n__g(X)) -> g(X)
                   , activate(n__d(X)) -> d(X)
                   , c(X) -> d(activate(X))
                   , g(X) -> n__g(X)
                   , h(X) -> c(n__d(X))
                   , f(X) -> n__f(X)
                   , d(X) -> n__d(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) -> 1
                 , f_1(3) -> 1
                 , f_1(3) -> 3
                 , c_0(2) -> 1
                 , c_1(4) -> 1
                 , n__f_0(2) -> 1
                 , n__f_0(2) -> 2
                 , n__f_0(2) -> 3
                 , n__f_1(3) -> 1
                 , n__f_1(3) -> 3
                 , n__g_0(2) -> 1
                 , n__g_0(2) -> 2
                 , n__g_0(2) -> 3
                 , n__g_1(2) -> 3
                 , d_0(1) -> 1
                 , d_0(2) -> 1
                 , d_1(2) -> 3
                 , d_1(3) -> 1
                 , activate_0(2) -> 1
                 , activate_1(2) -> 3
                 , activate_1(4) -> 3
                 , h_0(2) -> 1
                 , n__d_0(1) -> 1
                 , n__d_0(2) -> 1
                 , n__d_0(2) -> 2
                 , n__d_0(2) -> 3
                 , n__d_1(2) -> 3
                 , n__d_1(2) -> 4
                 , n__d_1(3) -> 1
                 , g_0(2) -> 1
                 , g_1(2) -> 3}

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