We consider the following Problem:

  Strict Trs:
    {  f(X) -> if(X, c(), n__f(n__true()))
     , if(true(), X, Y) -> X
     , if(false(), X, Y) -> activate(Y)
     , f(X) -> n__f(X)
     , true() -> n__true()
     , activate(n__f(X)) -> f(activate(X))
     , activate(n__true()) -> true()
     , activate(X) -> X}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  Arguments of following rules are not normal-forms:
  {if(true(), X, Y) -> X}
  
  All above mentioned rules can be savely removed.
  
  We consider the following Problem:
  
    Strict Trs:
      {  f(X) -> if(X, c(), n__f(n__true()))
       , if(false(), X, Y) -> activate(Y)
       , f(X) -> n__f(X)
       , true() -> n__true()
       , activate(n__f(X)) -> f(activate(X))
       , activate(n__true()) -> true()
       , 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: {true() -> n__true()}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(f) = {1}, Uargs(if) = {}, Uargs(n__f) = {},
        Uargs(activate) = {}
      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]
       if(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [0 0] x3 + [1]
                        [1 0]      [0 0]      [0 0]      [1]
       c() = [0]
             [0]
       n__f(x1) = [1 0] x1 + [0]
                  [0 0]      [0]
       n__true() = [0]
                   [0]
       true() = [2]
                [0]
       false() = [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(X) -> if(X, c(), n__f(n__true()))
         , if(false(), X, Y) -> activate(Y)
         , f(X) -> n__f(X)
         , activate(n__f(X)) -> f(activate(X))
         , activate(n__true()) -> true()
         , activate(X) -> X}
      Weak Trs: {true() -> n__true()}
      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__true()) -> true()}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(f) = {1}, Uargs(if) = {}, Uargs(n__f) = {},
          Uargs(activate) = {}
        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]
         if(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [0 0] x3 + [1]
                          [1 0]      [0 0]      [0 0]      [1]
         c() = [0]
               [0]
         n__f(x1) = [1 0] x1 + [0]
                    [0 0]      [0]
         n__true() = [0]
                     [0]
         true() = [0]
                  [0]
         false() = [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(X) -> if(X, c(), n__f(n__true()))
           , if(false(), X, Y) -> activate(Y)
           , f(X) -> n__f(X)
           , activate(n__f(X)) -> f(activate(X))
           , activate(X) -> X}
        Weak Trs:
          {  activate(n__true()) -> true()
           , true() -> n__true()}
        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) -> if(X, c(), n__f(n__true()))
           , f(X) -> n__f(X)}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(f) = {1}, Uargs(if) = {}, Uargs(n__f) = {},
            Uargs(activate) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           f(x1) = [1 0] x1 + [2]
                   [0 0]      [1]
           if(x1, x2, x3) = [0 0] x1 + [1 0] x2 + [0 0] x3 + [1]
                            [0 0]      [1 0]      [0 0]      [1]
           c() = [0]
                 [0]
           n__f(x1) = [1 0] x1 + [0]
                      [0 0]      [0]
           n__true() = [0]
                       [0]
           true() = [0]
                    [0]
           false() = [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:
            {  if(false(), X, Y) -> activate(Y)
             , activate(n__f(X)) -> f(activate(X))
             , activate(X) -> X}
          Weak Trs:
            {  f(X) -> if(X, c(), n__f(n__true()))
             , f(X) -> n__f(X)
             , activate(n__true()) -> true()
             , true() -> n__true()}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          The weightgap principle applies, where following rules are oriented strictly:
          
          TRS Component: {if(false(), X, Y) -> activate(Y)}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(f) = {1}, Uargs(if) = {}, Uargs(n__f) = {},
              Uargs(activate) = {}
            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]
             if(x1, x2, x3) = [0 0] x1 + [1 0] x2 + [1 0] x3 + [1]
                              [0 0]      [1 0]      [1 0]      [1]
             c() = [0]
                   [0]
             n__f(x1) = [1 0] x1 + [0]
                        [0 0]      [0]
             n__true() = [0]
                         [0]
             true() = [0]
                      [0]
             false() = [0]
                       [0]
             activate(x1) = [1 0] x1 + [0]
                            [1 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:
              {  if(false(), X, Y) -> activate(Y)
               , f(X) -> if(X, c(), n__f(n__true()))
               , f(X) -> n__f(X)
               , activate(n__true()) -> true()
               , true() -> n__true()}
            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(if) = {}, Uargs(n__f) = {},
                Uargs(activate) = {}
              We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
              Interpretation Functions:
               f(x1) = [1 2] x1 + [0]
                       [0 0]      [3]
               if(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [0]
                                [0 0]      [0 1]      [0 1]      [3]
               c() = [0]
                     [0]
               n__f(x1) = [1 2] x1 + [0]
                          [0 0]      [0]
               n__true() = [0]
                           [0]
               true() = [0]
                        [0]
               false() = [2]
                         [0]
               activate(x1) = [1 0] x1 + [1]
                              [0 1]      [2]
            
            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
                 , if(false(), X, Y) -> activate(Y)
                 , f(X) -> if(X, c(), n__f(n__true()))
                 , f(X) -> n__f(X)
                 , activate(n__true()) -> true()
                 , true() -> n__true()}
              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
                   , if(false(), X, Y) -> activate(Y)
                   , f(X) -> if(X, c(), n__f(n__true()))
                   , f(X) -> n__f(X)
                   , activate(n__true()) -> true()
                   , true() -> n__true()}
                StartTerms: basic terms
                Strategy: innermost
              
              Certificate: YES(?,O(n^1))
              
              Proof:
                The problem is match-bounded by 2.
                The enriched problem is compatible with the following automaton:
                {  f_0(2) -> 1
                 , f_1(3) -> 1
                 , f_1(3) -> 3
                 , f_2(7) -> 1
                 , f_2(7) -> 3
                 , if_0(2, 2, 2) -> 1
                 , if_1(2, 4, 5) -> 1
                 , if_1(3, 4, 5) -> 1
                 , if_1(3, 4, 5) -> 3
                 , if_2(3, 8, 9) -> 1
                 , if_2(3, 8, 9) -> 3
                 , if_2(7, 8, 9) -> 1
                 , if_2(7, 8, 9) -> 3
                 , c_0() -> 1
                 , c_0() -> 2
                 , c_0() -> 3
                 , c_1() -> 4
                 , c_2() -> 8
                 , 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__f_1(6) -> 1
                 , n__f_1(6) -> 3
                 , n__f_1(6) -> 5
                 , n__f_2(7) -> 1
                 , n__f_2(7) -> 3
                 , n__f_2(10) -> 1
                 , n__f_2(10) -> 3
                 , n__f_2(10) -> 9
                 , n__true_0() -> 1
                 , n__true_0() -> 2
                 , n__true_0() -> 3
                 , n__true_1() -> 3
                 , n__true_1() -> 6
                 , n__true_1() -> 7
                 , n__true_2() -> 7
                 , n__true_2() -> 10
                 , true_0() -> 1
                 , true_1() -> 3
                 , true_2() -> 7
                 , false_0() -> 1
                 , false_0() -> 2
                 , false_0() -> 3
                 , activate_0(2) -> 1
                 , activate_1(2) -> 3
                 , activate_1(5) -> 1
                 , activate_1(5) -> 3
                 , activate_1(9) -> 1
                 , activate_1(9) -> 3
                 , activate_2(6) -> 7
                 , activate_2(10) -> 7}

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