We consider the following Problem:

  Strict Trs:
    {  a__f(f(a())) -> a__f(g(f(a())))
     , mark(f(X)) -> a__f(X)
     , mark(a()) -> a()
     , mark(g(X)) -> g(mark(X))
     , a__f(X) -> f(X)}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  a__f(f(a())) -> a__f(g(f(a())))
       , mark(f(X)) -> a__f(X)
       , mark(a()) -> a()
       , mark(g(X)) -> g(mark(X))
       , a__f(X) -> f(X)}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component:
      {  mark(a()) -> a()
       , a__f(X) -> f(X)}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(a__f) = {}, Uargs(f) = {}, Uargs(g) = {1}, Uargs(mark) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       a__f(x1) = [0 0] x1 + [1]
                  [0 0]      [1]
       f(x1) = [0 0] x1 + [0]
               [0 0]      [0]
       a() = [0]
             [0]
       g(x1) = [1 0] x1 + [0]
               [0 0]      [1]
       mark(x1) = [0 0] x1 + [1]
                  [0 0]      [1]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  a__f(f(a())) -> a__f(g(f(a())))
         , mark(f(X)) -> a__f(X)
         , mark(g(X)) -> g(mark(X))}
      Weak Trs:
        {  mark(a()) -> a()
         , a__f(X) -> f(X)}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {mark(f(X)) -> a__f(X)}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(a__f) = {}, Uargs(f) = {}, Uargs(g) = {1}, Uargs(mark) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         a__f(x1) = [0 0] x1 + [1]
                    [0 0]      [1]
         f(x1) = [0 0] x1 + [0]
                 [0 0]      [0]
         a() = [0]
               [0]
         g(x1) = [1 2] x1 + [0]
                 [0 0]      [1]
         mark(x1) = [0 0] x1 + [3]
                    [0 0]      [3]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  a__f(f(a())) -> a__f(g(f(a())))
           , mark(g(X)) -> g(mark(X))}
        Weak Trs:
          {  mark(f(X)) -> a__f(X)
           , mark(a()) -> a()
           , a__f(X) -> f(X)}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {mark(g(X)) -> g(mark(X))}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(a__f) = {}, Uargs(f) = {}, Uargs(g) = {1}, Uargs(mark) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           a__f(x1) = [0 0] x1 + [0]
                      [0 0]      [0]
           f(x1) = [0 0] x1 + [0]
                   [0 0]      [0]
           a() = [0]
                 [0]
           g(x1) = [1 0] x1 + [0]
                   [0 1]      [2]
           mark(x1) = [0 2] x1 + [0]
                      [0 1]      [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs: {a__f(f(a())) -> a__f(g(f(a())))}
          Weak Trs:
            {  mark(g(X)) -> g(mark(X))
             , mark(f(X)) -> a__f(X)
             , mark(a()) -> a()
             , a__f(X) -> f(X)}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          The weightgap principle applies, where following rules are oriented strictly:
          
          TRS Component: {a__f(f(a())) -> a__f(g(f(a())))}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(a__f) = {}, Uargs(f) = {}, Uargs(g) = {1}, Uargs(mark) = {}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             a__f(x1) = [0 1] x1 + [2]
                        [0 1]      [1]
             f(x1) = [0 1] x1 + [2]
                     [0 1]      [0]
             a() = [2]
                   [2]
             g(x1) = [1 0] x1 + [0]
                     [0 0]      [0]
             mark(x1) = [1 0] x1 + [1]
                        [1 0]      [1]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Weak Trs:
              {  a__f(f(a())) -> a__f(g(f(a())))
               , mark(g(X)) -> g(mark(X))
               , mark(f(X)) -> a__f(X)
               , mark(a()) -> a()
               , a__f(X) -> f(X)}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(O(1),O(1))
          
          Proof:
            We consider the following Problem:
            
              Weak Trs:
                {  a__f(f(a())) -> a__f(g(f(a())))
                 , mark(g(X)) -> g(mark(X))
                 , mark(f(X)) -> a__f(X)
                 , mark(a()) -> a()
                 , a__f(X) -> f(X)}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(O(1),O(1))
            
            Proof:
              Empty rules are trivially bounded

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