We consider the following Problem:

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

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

Proof:
  Arguments of following rules are not normal-forms:
  {g(a(), X) -> f(b(), activate(X))}
  
  All above mentioned rules can be savely removed.
  
  We consider the following Problem:
  
    Strict Trs:
      {  h(X) -> g(X, X)
       , f(X, X) -> h(a())
       , a() -> b()
       , 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, X) -> h(a())}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(h) = {1}, Uargs(g) = {}, Uargs(f) = {}, Uargs(activate) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       h(x1) = [1 0] x1 + [1]
               [0 0]      [1]
       g(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                   [0 0]      [1 0]      [0]
       a() = [0]
             [0]
       f(x1, x2) = [0 0] x1 + [0 0] x2 + [2]
                   [0 0]      [0 0]      [2]
       b() = [0]
             [0]
       activate(x1) = [1 0] x1 + [0]
                      [0 0]      [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  h(X) -> g(X, X)
         , a() -> b()
         , activate(X) -> X}
      Weak Trs: {f(X, X) -> h(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() -> b()}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(h) = {1}, Uargs(g) = {}, Uargs(f) = {}, Uargs(activate) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         h(x1) = [1 0] x1 + [1]
                 [0 0]      [1]
         g(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                     [0 0]      [1 0]      [0]
         a() = [2]
               [0]
         f(x1, x2) = [0 0] x1 + [0 0] x2 + [3]
                     [0 0]      [0 0]      [2]
         b() = [0]
               [0]
         activate(x1) = [1 0] x1 + [0]
                        [0 0]      [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  h(X) -> g(X, X)
           , activate(X) -> X}
        Weak Trs:
          {  a() -> b()
           , f(X, X) -> h(a())}
        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) -> g(X, X)}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(h) = {1}, Uargs(g) = {}, Uargs(f) = {}, Uargs(activate) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           h(x1) = [1 0] x1 + [1]
                   [1 0]      [1]
           g(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                       [1 0]      [0 0]      [0]
           a() = [0]
                 [0]
           f(x1, x2) = [0 0] x1 + [0 0] x2 + [2]
                       [0 0]      [0 0]      [2]
           b() = [0]
                 [0]
           activate(x1) = [1 0] x1 + [0]
                          [0 0]      [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs: {activate(X) -> X}
          Weak Trs:
            {  h(X) -> g(X, X)
             , a() -> b()
             , f(X, X) -> h(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(h) = {1}, Uargs(g) = {}, Uargs(f) = {}, Uargs(activate) = {}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             h(x1) = [1 0] x1 + [1]
                     [0 1]      [1]
             g(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                         [0 0]      [0 1]      [0]
             a() = [0]
                   [0]
             f(x1, x2) = [1 0] x1 + [0 0] x2 + [2]
                         [0 1]      [0 0]      [2]
             b() = [0]
                   [0]
             activate(x1) = [1 0] x1 + [2]
                            [0 1]      [0]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Weak Trs:
              {  activate(X) -> X
               , h(X) -> g(X, X)
               , a() -> b()
               , f(X, X) -> h(a())}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(O(1),O(1))
          
          Proof:
            We consider the following Problem:
            
              Weak Trs:
                {  activate(X) -> X
                 , h(X) -> g(X, X)
                 , a() -> b()
                 , f(X, X) -> h(a())}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(O(1),O(1))
            
            Proof:
              Empty rules are trivially bounded

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