We consider the following Problem:

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

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  active(f(a(), X, X)) -> mark(f(X, b(), b()))
       , active(b()) -> mark(a())
       , mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
       , mark(a()) -> active(a())
       , mark(b()) -> active(b())
       , f(mark(X1), X2, X3) -> f(X1, X2, X3)
       , f(X1, mark(X2), X3) -> f(X1, X2, X3)
       , f(X1, X2, mark(X3)) -> f(X1, X2, X3)
       , f(active(X1), X2, X3) -> f(X1, X2, X3)
       , f(X1, active(X2), X3) -> f(X1, X2, X3)
       , f(X1, X2, active(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:
      {  f(X1, mark(X2), X3) -> f(X1, X2, X3)
       , f(X1, active(X2), X3) -> f(X1, X2, X3)}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(active) = {1}, Uargs(f) = {2}, Uargs(mark) = {1}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       active(x1) = [1 0] x1 + [1]
                    [0 0]      [1]
       f(x1, x2, x3) = [0 0] x1 + [1 0] x2 + [0 0] x3 + [0]
                       [0 0]      [0 0]      [0 0]      [1]
       a() = [0]
             [0]
       mark(x1) = [1 0] x1 + [1]
                  [0 0]      [1]
       b() = [0]
             [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  active(f(a(), X, X)) -> mark(f(X, b(), b()))
         , active(b()) -> mark(a())
         , mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
         , mark(a()) -> active(a())
         , mark(b()) -> active(b())
         , f(mark(X1), X2, X3) -> f(X1, X2, X3)
         , f(X1, X2, mark(X3)) -> f(X1, X2, X3)
         , f(active(X1), X2, X3) -> f(X1, X2, X3)
         , f(X1, X2, active(X3)) -> f(X1, X2, X3)}
      Weak Trs:
        {  f(X1, mark(X2), X3) -> f(X1, X2, X3)
         , f(X1, active(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: {active(b()) -> mark(a())}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(active) = {1}, Uargs(f) = {2}, Uargs(mark) = {1}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         active(x1) = [1 1] x1 + [1]
                      [0 0]      [1]
         f(x1, x2, x3) = [0 0] x1 + [1 0] x2 + [0 0] x3 + [0]
                         [0 0]      [0 0]      [0 0]      [0]
         a() = [0]
               [0]
         mark(x1) = [1 0] x1 + [1]
                    [0 0]      [1]
         b() = [0]
               [2]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  active(f(a(), X, X)) -> mark(f(X, b(), b()))
           , mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
           , mark(a()) -> active(a())
           , mark(b()) -> active(b())
           , f(mark(X1), X2, X3) -> f(X1, X2, X3)
           , f(X1, X2, mark(X3)) -> f(X1, X2, X3)
           , f(active(X1), X2, X3) -> f(X1, X2, X3)
           , f(X1, X2, active(X3)) -> f(X1, X2, X3)}
        Weak Trs:
          {  active(b()) -> mark(a())
           , f(X1, mark(X2), X3) -> f(X1, X2, X3)
           , f(X1, active(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(a()) -> active(a())}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(active) = {1}, Uargs(f) = {2}, Uargs(mark) = {1}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           active(x1) = [1 2] x1 + [1]
                        [0 0]      [1]
           f(x1, x2, x3) = [0 0] x1 + [1 0] x2 + [0 0] x3 + [0]
                           [0 0]      [0 0]      [0 0]      [0]
           a() = [0]
                 [0]
           mark(x1) = [1 0] x1 + [3]
                      [0 0]      [1]
           b() = [0]
                 [1]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs:
            {  active(f(a(), X, X)) -> mark(f(X, b(), b()))
             , mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
             , mark(b()) -> active(b())
             , f(mark(X1), X2, X3) -> f(X1, X2, X3)
             , f(X1, X2, mark(X3)) -> f(X1, X2, X3)
             , f(active(X1), X2, X3) -> f(X1, X2, X3)
             , f(X1, X2, active(X3)) -> f(X1, X2, X3)}
          Weak Trs:
            {  mark(a()) -> active(a())
             , active(b()) -> mark(a())
             , f(X1, mark(X2), X3) -> f(X1, X2, X3)
             , f(X1, active(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(b()) -> active(b())}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(active) = {1}, Uargs(f) = {2}, Uargs(mark) = {1}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             active(x1) = [1 0] x1 + [1]
                          [0 0]      [1]
             f(x1, x2, x3) = [0 0] x1 + [1 0] x2 + [0 0] x3 + [0]
                             [0 0]      [0 0]      [0 0]      [1]
             a() = [0]
                   [0]
             mark(x1) = [1 0] x1 + [2]
                        [0 0]      [1]
             b() = [2]
                   [0]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Strict Trs:
              {  active(f(a(), X, X)) -> mark(f(X, b(), b()))
               , mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
               , f(mark(X1), X2, X3) -> f(X1, X2, X3)
               , f(X1, X2, mark(X3)) -> f(X1, X2, X3)
               , f(active(X1), X2, X3) -> f(X1, X2, X3)
               , f(X1, X2, active(X3)) -> f(X1, X2, X3)}
            Weak Trs:
              {  mark(b()) -> active(b())
               , mark(a()) -> active(a())
               , active(b()) -> mark(a())
               , f(X1, mark(X2), X3) -> f(X1, X2, X3)
               , f(X1, active(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: {active(f(a(), X, X)) -> mark(f(X, b(), b()))}
            
            Interpretation of nonconstant growth:
            -------------------------------------
              The following argument positions are usable:
                Uargs(active) = {1}, Uargs(f) = {2}, Uargs(mark) = {1}
              We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
              Interpretation Functions:
               active(x1) = [1 2] x1 + [0]
                            [0 0]      [1]
               f(x1, x2, x3) = [0 0] x1 + [1 0] x2 + [0 0] x3 + [0]
                               [0 0]      [0 0]      [0 0]      [2]
               a() = [0]
                     [0]
               mark(x1) = [1 0] x1 + [0]
                          [0 0]      [1]
               b() = [0]
                     [0]
            
            The strictly oriented rules are moved into the weak component.
            
            We consider the following Problem:
            
              Strict Trs:
                {  mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
                 , f(mark(X1), X2, X3) -> f(X1, X2, X3)
                 , f(X1, X2, mark(X3)) -> f(X1, X2, X3)
                 , f(active(X1), X2, X3) -> f(X1, X2, X3)
                 , f(X1, X2, active(X3)) -> f(X1, X2, X3)}
              Weak Trs:
                {  active(f(a(), X, X)) -> mark(f(X, b(), b()))
                 , mark(b()) -> active(b())
                 , mark(a()) -> active(a())
                 , active(b()) -> mark(a())
                 , f(X1, mark(X2), X3) -> f(X1, X2, X3)
                 , f(X1, active(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: {f(X1, X2, mark(X3)) -> f(X1, X2, X3)}
              
              Interpretation of nonconstant growth:
              -------------------------------------
                The following argument positions are usable:
                  Uargs(active) = {1}, Uargs(f) = {2}, Uargs(mark) = {1}
                We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                Interpretation Functions:
                 active(x1) = [1 3] x1 + [0]
                              [0 0]      [0]
                 f(x1, x2, x3) = [0 0] x1 + [1 0] x2 + [1 0] x3 + [0]
                                 [0 0]      [0 0]      [0 0]      [2]
                 a() = [1]
                       [0]
                 mark(x1) = [1 0] x1 + [3]
                            [0 0]      [0]
                 b() = [1]
                       [1]
              
              The strictly oriented rules are moved into the weak component.
              
              We consider the following Problem:
              
                Strict Trs:
                  {  mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
                   , f(mark(X1), X2, X3) -> f(X1, X2, X3)
                   , f(active(X1), X2, X3) -> f(X1, X2, X3)
                   , f(X1, X2, active(X3)) -> f(X1, X2, X3)}
                Weak Trs:
                  {  f(X1, X2, mark(X3)) -> f(X1, X2, X3)
                   , active(f(a(), X, X)) -> mark(f(X, b(), b()))
                   , mark(b()) -> active(b())
                   , mark(a()) -> active(a())
                   , active(b()) -> mark(a())
                   , f(X1, mark(X2), X3) -> f(X1, X2, X3)
                   , f(X1, active(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: {f(X1, X2, active(X3)) -> f(X1, X2, X3)}
                
                Interpretation of nonconstant growth:
                -------------------------------------
                  The following argument positions are usable:
                    Uargs(active) = {1}, Uargs(f) = {2}, Uargs(mark) = {1}
                  We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                  Interpretation Functions:
                   active(x1) = [1 0] x1 + [0]
                                [0 1]      [2]
                   f(x1, x2, x3) = [0 0] x1 + [1 0] x2 + [0 2] x3 + [0]
                                   [0 0]      [0 0]      [0 0]      [0]
                   a() = [0]
                         [0]
                   mark(x1) = [1 0] x1 + [0]
                              [0 1]      [2]
                   b() = [0]
                         [0]
                
                The strictly oriented rules are moved into the weak component.
                
                We consider the following Problem:
                
                  Strict Trs:
                    {  mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
                     , f(mark(X1), X2, X3) -> f(X1, X2, X3)
                     , f(active(X1), X2, X3) -> f(X1, X2, X3)}
                  Weak Trs:
                    {  f(X1, X2, active(X3)) -> f(X1, X2, X3)
                     , f(X1, X2, mark(X3)) -> f(X1, X2, X3)
                     , active(f(a(), X, X)) -> mark(f(X, b(), b()))
                     , mark(b()) -> active(b())
                     , mark(a()) -> active(a())
                     , active(b()) -> mark(a())
                     , f(X1, mark(X2), X3) -> f(X1, X2, X3)
                     , f(X1, active(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:
                    {  f(mark(X1), X2, X3) -> f(X1, X2, X3)
                     , f(active(X1), X2, X3) -> f(X1, X2, X3)}
                  
                  Interpretation of nonconstant growth:
                  -------------------------------------
                    The following argument positions are usable:
                      Uargs(active) = {1}, Uargs(f) = {2}, Uargs(mark) = {1}
                    We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                    Interpretation Functions:
                     active(x1) = [1 0] x1 + [0]
                                  [0 1]      [1]
                     f(x1, x2, x3) = [0 2] x1 + [1 0] x2 + [0 2] x3 + [0]
                                     [0 0]      [0 0]      [0 0]      [0]
                     a() = [0]
                           [0]
                     mark(x1) = [1 0] x1 + [0]
                                [0 1]      [1]
                     b() = [0]
                           [0]
                  
                  The strictly oriented rules are moved into the weak component.
                  
                  We consider the following Problem:
                  
                    Strict Trs: {mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))}
                    Weak Trs:
                      {  f(mark(X1), X2, X3) -> f(X1, X2, X3)
                       , f(active(X1), X2, X3) -> f(X1, X2, X3)
                       , f(X1, X2, active(X3)) -> f(X1, X2, X3)
                       , f(X1, X2, mark(X3)) -> f(X1, X2, X3)
                       , active(f(a(), X, X)) -> mark(f(X, b(), b()))
                       , mark(b()) -> active(b())
                       , mark(a()) -> active(a())
                       , active(b()) -> mark(a())
                       , f(X1, mark(X2), X3) -> f(X1, X2, X3)
                       , f(X1, active(X2), X3) -> f(X1, X2, X3)}
                    StartTerms: basic terms
                    Strategy: innermost
                  
                  Certificate: YES(?,O(n^1))
                  
                  Proof:
                    We consider the following Problem:
                    
                      Strict Trs: {mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))}
                      Weak Trs:
                        {  f(mark(X1), X2, X3) -> f(X1, X2, X3)
                         , f(active(X1), X2, X3) -> f(X1, X2, X3)
                         , f(X1, X2, active(X3)) -> f(X1, X2, X3)
                         , f(X1, X2, mark(X3)) -> f(X1, X2, X3)
                         , active(f(a(), X, X)) -> mark(f(X, b(), b()))
                         , mark(b()) -> active(b())
                         , mark(a()) -> active(a())
                         , active(b()) -> mark(a())
                         , f(X1, mark(X2), X3) -> f(X1, X2, X3)
                         , f(X1, active(X2), X3) -> f(X1, X2, X3)}
                      StartTerms: basic terms
                      Strategy: innermost
                    
                    Certificate: YES(?,O(n^1))
                    
                    Proof:
                      The problem is match-bounded by 0.
                      The enriched problem is compatible with the following automaton:
                      {  active_0(2) -> 1
                       , f_0(2, 2, 2) -> 1
                       , a_0() -> 2
                       , mark_0(2) -> 1
                       , b_0() -> 2}

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