We consider the following Problem:

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

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  f(n__f(n__a())) -> f(n__g(n__f(n__a())))
       , f(X) -> n__f(X)
       , a() -> n__a()
       , g(X) -> n__g(X)
       , activate(n__f(X)) -> f(X)
       , activate(n__a()) -> a()
       , 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:
      {  f(X) -> n__f(X)
       , activate(n__a()) -> a()}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(f) = {}, Uargs(n__f) = {}, Uargs(n__g) = {}, Uargs(g) = {1},
        Uargs(activate) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       f(x1) = [0 0] x1 + [1]
               [0 0]      [1]
       n__f(x1) = [0 0] x1 + [0]
                  [0 0]      [0]
       n__a() = [0]
                [0]
       n__g(x1) = [1 0] x1 + [0]
                  [0 0]      [0]
       a() = [0]
             [0]
       g(x1) = [1 0] x1 + [0]
               [0 0]      [1]
       activate(x1) = [1 0] x1 + [1]
                      [0 0]      [1]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  f(n__f(n__a())) -> f(n__g(n__f(n__a())))
         , a() -> n__a()
         , g(X) -> n__g(X)
         , activate(n__f(X)) -> f(X)
         , activate(n__g(X)) -> g(activate(X))
         , activate(X) -> X}
      Weak Trs:
        {  f(X) -> n__f(X)
         , activate(n__a()) -> a()}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {a() -> n__a()}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(f) = {}, Uargs(n__f) = {}, Uargs(n__g) = {}, Uargs(g) = {1},
          Uargs(activate) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         f(x1) = [0 0] x1 + [1]
                 [0 0]      [1]
         n__f(x1) = [0 0] x1 + [0]
                    [0 0]      [0]
         n__a() = [0]
                  [0]
         n__g(x1) = [1 0] x1 + [0]
                    [0 0]      [0]
         a() = [1]
               [0]
         g(x1) = [1 0] x1 + [0]
                 [0 0]      [1]
         activate(x1) = [1 0] x1 + [1]
                        [0 0]      [1]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  f(n__f(n__a())) -> f(n__g(n__f(n__a())))
           , g(X) -> n__g(X)
           , activate(n__f(X)) -> f(X)
           , activate(n__g(X)) -> g(activate(X))
           , activate(X) -> X}
        Weak Trs:
          {  a() -> n__a()
           , f(X) -> n__f(X)
           , activate(n__a()) -> a()}
        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) = {}, Uargs(n__f) = {}, Uargs(n__g) = {}, Uargs(g) = {1},
            Uargs(activate) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           f(x1) = [0 0] x1 + [1]
                   [0 0]      [1]
           n__f(x1) = [0 0] x1 + [0]
                      [0 0]      [0]
           n__a() = [0]
                    [0]
           n__g(x1) = [1 0] x1 + [0]
                      [0 0]      [0]
           a() = [0]
                 [0]
           g(x1) = [1 0] x1 + [2]
                   [0 0]      [1]
           activate(x1) = [1 0] x1 + [1]
                          [0 0]      [1]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs:
            {  f(n__f(n__a())) -> f(n__g(n__f(n__a())))
             , activate(n__f(X)) -> f(X)
             , activate(n__g(X)) -> g(activate(X))
             , activate(X) -> X}
          Weak Trs:
            {  g(X) -> n__g(X)
             , a() -> n__a()
             , f(X) -> n__f(X)
             , activate(n__a()) -> a()}
          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__f(X)) -> f(X)}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(f) = {}, Uargs(n__f) = {}, Uargs(n__g) = {}, Uargs(g) = {1},
              Uargs(activate) = {}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             f(x1) = [0 0] x1 + [1]
                     [0 0]      [1]
             n__f(x1) = [0 0] x1 + [0]
                        [0 0]      [0]
             n__a() = [0]
                      [0]
             n__g(x1) = [1 0] x1 + [0]
                        [0 0]      [0]
             a() = [0]
                   [0]
             g(x1) = [1 2] x1 + [0]
                     [0 0]      [1]
             activate(x1) = [1 0] x1 + [3]
                            [0 0]      [1]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Strict Trs:
              {  f(n__f(n__a())) -> f(n__g(n__f(n__a())))
               , activate(n__g(X)) -> g(activate(X))
               , activate(X) -> X}
            Weak Trs:
              {  activate(n__f(X)) -> f(X)
               , g(X) -> n__g(X)
               , a() -> n__a()
               , f(X) -> n__f(X)
               , activate(n__a()) -> a()}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(?,O(n^1))
          
          Proof:
            The weightgap principle applies, where following rules are oriented strictly:
            
            TRS Component: {f(n__f(n__a())) -> f(n__g(n__f(n__a())))}
            
            Interpretation of nonconstant growth:
            -------------------------------------
              The following argument positions are usable:
                Uargs(f) = {}, Uargs(n__f) = {}, Uargs(n__g) = {}, Uargs(g) = {1},
                Uargs(activate) = {}
              We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
              Interpretation Functions:
               f(x1) = [0 1] x1 + [0]
                       [0 0]      [2]
               n__f(x1) = [0 1] x1 + [0]
                          [0 0]      [2]
               n__a() = [0]
                        [0]
               n__g(x1) = [1 0] x1 + [1]
                          [0 0]      [0]
               a() = [0]
                     [0]
               g(x1) = [1 0] x1 + [1]
                       [1 0]      [1]
               activate(x1) = [1 0] x1 + [0]
                              [0 0]      [2]
            
            The strictly oriented rules are moved into the weak component.
            
            We consider the following Problem:
            
              Strict Trs:
                {  activate(n__g(X)) -> g(activate(X))
                 , activate(X) -> X}
              Weak Trs:
                {  f(n__f(n__a())) -> f(n__g(n__f(n__a())))
                 , activate(n__f(X)) -> f(X)
                 , g(X) -> n__g(X)
                 , a() -> n__a()
                 , f(X) -> n__f(X)
                 , activate(n__a()) -> a()}
              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) = {}, Uargs(n__f) = {}, Uargs(n__g) = {}, Uargs(g) = {1},
                  Uargs(activate) = {}
                We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                Interpretation Functions:
                 f(x1) = [0 0] x1 + [1]
                         [0 0]      [1]
                 n__f(x1) = [0 0] x1 + [0]
                            [0 0]      [0]
                 n__a() = [0]
                          [0]
                 n__g(x1) = [1 0] x1 + [0]
                            [0 0]      [0]
                 a() = [0]
                       [0]
                 g(x1) = [1 0] x1 + [0]
                         [0 0]      [1]
                 activate(x1) = [1 0] x1 + [1]
                                [0 1]      [1]
              
              The strictly oriented rules are moved into the weak component.
              
              We consider the following Problem:
              
                Strict Trs: {activate(n__g(X)) -> g(activate(X))}
                Weak Trs:
                  {  activate(X) -> X
                   , f(n__f(n__a())) -> f(n__g(n__f(n__a())))
                   , activate(n__f(X)) -> f(X)
                   , g(X) -> n__g(X)
                   , a() -> n__a()
                   , f(X) -> n__f(X)
                   , activate(n__a()) -> a()}
                StartTerms: basic terms
                Strategy: innermost
              
              Certificate: YES(?,O(n^1))
              
              Proof:
                We consider the following Problem:
                
                  Strict Trs: {activate(n__g(X)) -> g(activate(X))}
                  Weak Trs:
                    {  activate(X) -> X
                     , f(n__f(n__a())) -> f(n__g(n__f(n__a())))
                     , activate(n__f(X)) -> f(X)
                     , g(X) -> n__g(X)
                     , a() -> n__a()
                     , f(X) -> n__f(X)
                     , activate(n__a()) -> a()}
                  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(2) -> 3
                   , f_1(4) -> 1
                   , f_1(4) -> 3
                   , n__f_0(2) -> 1
                   , n__f_0(2) -> 2
                   , n__f_0(2) -> 3
                   , n__f_1(2) -> 3
                   , n__f_1(4) -> 1
                   , n__f_1(4) -> 3
                   , n__f_1(6) -> 5
                   , n__a_0() -> 1
                   , n__a_0() -> 2
                   , n__a_0() -> 3
                   , n__a_1() -> 3
                   , n__a_1() -> 6
                   , 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
                   , n__g_1(5) -> 4
                   , a_0() -> 1
                   , a_1() -> 3
                   , g_0(2) -> 1
                   , g_1(3) -> 1
                   , g_1(3) -> 3
                   , activate_0(2) -> 1
                   , activate_1(2) -> 3}

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