We consider the following Problem:

  Strict Trs:
    {  a__f(a(), X, X) -> a__f(X, a__b(), b())
     , a__b() -> a()
     , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3)
     , mark(b()) -> a__b()
     , mark(a()) -> a()
     , a__f(X1, X2, X3) -> f(X1, X2, X3)
     , a__b() -> b()}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  a__f(a(), X, X) -> a__f(X, a__b(), b())
       , a__b() -> a()
       , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3)
       , mark(b()) -> a__b()
       , mark(a()) -> a()
       , a__f(X1, X2, X3) -> f(X1, X2, X3)
       , a__b() -> b()}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component:
      {  mark(b()) -> a__b()
       , mark(a()) -> a()
       , a__f(X1, X2, X3) -> f(X1, X2, X3)}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(a__f) = {2}, Uargs(mark) = {}, Uargs(f) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       a__f(x1, x2, x3) = [0 0] x1 + [1 1] x2 + [0 0] x3 + [1]
                          [1 1]      [0 0]      [1 1]      [1]
       a() = [0]
             [0]
       a__b() = [0]
                [0]
       b() = [0]
             [0]
       mark(x1) = [0 0] x1 + [1]
                  [0 0]      [0]
       f(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
                       [0 0]      [0 0]      [0 0]      [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  a__f(a(), X, X) -> a__f(X, a__b(), b())
         , a__b() -> a()
         , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3)
         , a__b() -> b()}
      Weak Trs:
        {  mark(b()) -> a__b()
         , mark(a()) -> a()
         , a__f(X1, X2, X3) -> f(X1, X2, X3)}
      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() -> a()
         , a__b() -> b()}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(a__f) = {2}, Uargs(mark) = {}, Uargs(f) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         a__f(x1, x2, x3) = [0 0] x1 + [1 1] x2 + [0 0] x3 + [1]
                            [1 1]      [0 0]      [1 1]      [1]
         a() = [0]
               [0]
         a__b() = [2]
                  [2]
         b() = [0]
               [0]
         mark(x1) = [0 0] x1 + [3]
                    [0 0]      [2]
         f(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
                         [0 0]      [0 0]      [0 0]      [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  a__f(a(), X, X) -> a__f(X, a__b(), b())
           , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3)}
        Weak Trs:
          {  a__b() -> a()
           , a__b() -> b()
           , mark(b()) -> a__b()
           , mark(a()) -> a()
           , a__f(X1, X2, X3) -> f(X1, X2, X3)}
        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(X1, X2, X3)) -> a__f(X1, mark(X2), X3)}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(a__f) = {2}, Uargs(mark) = {}, Uargs(f) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           a__f(x1, x2, x3) = [0 0] x1 + [1 0] x2 + [0 1] x3 + [0]
                              [0 0]      [0 1]      [0 1]      [3]
           a() = [0]
                 [0]
           a__b() = [0]
                    [3]
           b() = [0]
                 [1]
           mark(x1) = [0 1] x1 + [1]
                      [0 1]      [2]
           f(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
                           [0 0]      [0 1]      [0 1]      [3]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs: {a__f(a(), X, X) -> a__f(X, a__b(), b())}
          Weak Trs:
            {  mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3)
             , a__b() -> a()
             , a__b() -> b()
             , mark(b()) -> a__b()
             , mark(a()) -> a()
             , a__f(X1, X2, X3) -> f(X1, X2, X3)}
          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(a(), X, X) -> a__f(X, a__b(), b())}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(a__f) = {2}, Uargs(mark) = {}, Uargs(f) = {}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             a__f(x1, x2, x3) = [0 1] x1 + [1 0] x2 + [0 1] x3 + [0]
                                [0 0]      [1 0]      [0 1]      [1]
             a() = [0]
                   [1]
             a__b() = [0]
                      [2]
             b() = [0]
                   [0]
             mark(x1) = [1 0] x1 + [1]
                        [1 0]      [3]
             f(x1, x2, x3) = [0 1] x1 + [1 0] x2 + [0 1] x3 + [0]
                             [0 0]      [0 0]      [0 0]      [0]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Weak Trs:
              {  a__f(a(), X, X) -> a__f(X, a__b(), b())
               , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3)
               , a__b() -> a()
               , a__b() -> b()
               , mark(b()) -> a__b()
               , mark(a()) -> a()
               , a__f(X1, X2, X3) -> f(X1, X2, X3)}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(O(1),O(1))
          
          Proof:
            We consider the following Problem:
            
              Weak Trs:
                {  a__f(a(), X, X) -> a__f(X, a__b(), b())
                 , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3)
                 , a__b() -> a()
                 , a__b() -> b()
                 , mark(b()) -> a__b()
                 , mark(a()) -> a()
                 , a__f(X1, X2, X3) -> f(X1, X2, X3)}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(O(1),O(1))
            
            Proof:
              Empty rules are trivially bounded

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