We consider the following Problem:

  Strict Trs:
    {  active(f(f(X))) -> mark(c(f(g(f(X)))))
     , active(c(X)) -> mark(d(X))
     , active(h(X)) -> mark(c(d(X)))
     , mark(f(X)) -> active(f(mark(X)))
     , mark(c(X)) -> active(c(X))
     , mark(g(X)) -> active(g(X))
     , mark(d(X)) -> active(d(X))
     , mark(h(X)) -> active(h(mark(X)))
     , f(mark(X)) -> f(X)
     , f(active(X)) -> f(X)
     , c(mark(X)) -> c(X)
     , c(active(X)) -> c(X)
     , g(mark(X)) -> g(X)
     , g(active(X)) -> g(X)
     , d(mark(X)) -> d(X)
     , d(active(X)) -> d(X)
     , h(mark(X)) -> h(X)
     , h(active(X)) -> h(X)}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  active(f(f(X))) -> mark(c(f(g(f(X)))))
       , active(c(X)) -> mark(d(X))
       , active(h(X)) -> mark(c(d(X)))
       , mark(f(X)) -> active(f(mark(X)))
       , mark(c(X)) -> active(c(X))
       , mark(g(X)) -> active(g(X))
       , mark(d(X)) -> active(d(X))
       , mark(h(X)) -> active(h(mark(X)))
       , f(mark(X)) -> f(X)
       , f(active(X)) -> f(X)
       , c(mark(X)) -> c(X)
       , c(active(X)) -> c(X)
       , g(mark(X)) -> g(X)
       , g(active(X)) -> g(X)
       , d(mark(X)) -> d(X)
       , d(active(X)) -> d(X)
       , h(mark(X)) -> h(X)
       , h(active(X)) -> h(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(mark(X)) -> f(X)
       , f(active(X)) -> f(X)
       , c(mark(X)) -> c(X)
       , c(active(X)) -> c(X)
       , h(mark(X)) -> h(X)
       , h(active(X)) -> h(X)}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {1},
        Uargs(c) = {1}, Uargs(g) = {}, Uargs(d) = {}, Uargs(h) = {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) = [1 0] x1 + [0]
               [0 0]      [1]
       mark(x1) = [1 0] x1 + [1]
                  [0 0]      [1]
       c(x1) = [1 0] x1 + [0]
               [0 0]      [1]
       g(x1) = [0 0] x1 + [0]
               [0 0]      [1]
       d(x1) = [0 0] x1 + [0]
               [0 0]      [1]
       h(x1) = [1 0] x1 + [0]
               [0 0]      [1]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  active(f(f(X))) -> mark(c(f(g(f(X)))))
         , active(c(X)) -> mark(d(X))
         , active(h(X)) -> mark(c(d(X)))
         , mark(f(X)) -> active(f(mark(X)))
         , mark(c(X)) -> active(c(X))
         , mark(g(X)) -> active(g(X))
         , mark(d(X)) -> active(d(X))
         , mark(h(X)) -> active(h(mark(X)))
         , g(mark(X)) -> g(X)
         , g(active(X)) -> g(X)
         , d(mark(X)) -> d(X)
         , d(active(X)) -> d(X)}
      Weak Trs:
        {  f(mark(X)) -> f(X)
         , f(active(X)) -> f(X)
         , c(mark(X)) -> c(X)
         , c(active(X)) -> c(X)
         , h(mark(X)) -> h(X)
         , h(active(X)) -> h(X)}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {active(h(X)) -> mark(c(d(X)))}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {1},
          Uargs(c) = {1}, Uargs(g) = {}, Uargs(d) = {}, Uargs(h) = {1}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         active(x1) = [1 3] x1 + [1]
                      [0 0]      [1]
         f(x1) = [1 0] x1 + [0]
                 [0 0]      [0]
         mark(x1) = [1 0] x1 + [1]
                    [0 0]      [1]
         c(x1) = [1 0] x1 + [0]
                 [0 0]      [0]
         g(x1) = [0 0] x1 + [0]
                 [0 0]      [0]
         d(x1) = [0 0] x1 + [0]
                 [0 0]      [0]
         h(x1) = [1 0] x1 + [0]
                 [0 0]      [3]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  active(f(f(X))) -> mark(c(f(g(f(X)))))
           , active(c(X)) -> mark(d(X))
           , mark(f(X)) -> active(f(mark(X)))
           , mark(c(X)) -> active(c(X))
           , mark(g(X)) -> active(g(X))
           , mark(d(X)) -> active(d(X))
           , mark(h(X)) -> active(h(mark(X)))
           , g(mark(X)) -> g(X)
           , g(active(X)) -> g(X)
           , d(mark(X)) -> d(X)
           , d(active(X)) -> d(X)}
        Weak Trs:
          {  active(h(X)) -> mark(c(d(X)))
           , f(mark(X)) -> f(X)
           , f(active(X)) -> f(X)
           , c(mark(X)) -> c(X)
           , c(active(X)) -> c(X)
           , h(mark(X)) -> h(X)
           , h(active(X)) -> h(X)}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {active(c(X)) -> mark(d(X))}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {1},
            Uargs(c) = {1}, Uargs(g) = {}, Uargs(d) = {}, Uargs(h) = {1}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           active(x1) = [1 3] x1 + [1]
                        [0 0]      [1]
           f(x1) = [1 0] x1 + [0]
                   [0 0]      [0]
           mark(x1) = [1 0] x1 + [1]
                      [0 0]      [1]
           c(x1) = [1 0] x1 + [0]
                   [0 0]      [2]
           g(x1) = [0 0] x1 + [0]
                   [0 0]      [0]
           d(x1) = [0 0] x1 + [0]
                   [0 0]      [0]
           h(x1) = [1 0] x1 + [0]
                   [0 0]      [1]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs:
            {  active(f(f(X))) -> mark(c(f(g(f(X)))))
             , mark(f(X)) -> active(f(mark(X)))
             , mark(c(X)) -> active(c(X))
             , mark(g(X)) -> active(g(X))
             , mark(d(X)) -> active(d(X))
             , mark(h(X)) -> active(h(mark(X)))
             , g(mark(X)) -> g(X)
             , g(active(X)) -> g(X)
             , d(mark(X)) -> d(X)
             , d(active(X)) -> d(X)}
          Weak Trs:
            {  active(c(X)) -> mark(d(X))
             , active(h(X)) -> mark(c(d(X)))
             , f(mark(X)) -> f(X)
             , f(active(X)) -> f(X)
             , c(mark(X)) -> c(X)
             , c(active(X)) -> c(X)
             , h(mark(X)) -> h(X)
             , h(active(X)) -> h(X)}
          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(f(X))) -> mark(c(f(g(f(X)))))}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {1},
              Uargs(c) = {1}, Uargs(g) = {}, Uargs(d) = {}, Uargs(h) = {1}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             active(x1) = [1 0] x1 + [1]
                          [0 1]      [1]
             f(x1) = [1 0] x1 + [0]
                     [0 1]      [2]
             mark(x1) = [1 0] x1 + [0]
                        [0 1]      [1]
             c(x1) = [1 0] x1 + [0]
                     [0 0]      [0]
             g(x1) = [0 0] x1 + [0]
                     [0 0]      [0]
             d(x1) = [0 0] x1 + [0]
                     [0 0]      [0]
             h(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:
              {  mark(f(X)) -> active(f(mark(X)))
               , mark(c(X)) -> active(c(X))
               , mark(g(X)) -> active(g(X))
               , mark(d(X)) -> active(d(X))
               , mark(h(X)) -> active(h(mark(X)))
               , g(mark(X)) -> g(X)
               , g(active(X)) -> g(X)
               , d(mark(X)) -> d(X)
               , d(active(X)) -> d(X)}
            Weak Trs:
              {  active(f(f(X))) -> mark(c(f(g(f(X)))))
               , active(c(X)) -> mark(d(X))
               , active(h(X)) -> mark(c(d(X)))
               , f(mark(X)) -> f(X)
               , f(active(X)) -> f(X)
               , c(mark(X)) -> c(X)
               , c(active(X)) -> c(X)
               , h(mark(X)) -> h(X)
               , h(active(X)) -> h(X)}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(?,O(n^1))
          
          Proof:
            The weightgap principle applies, where following rules are oriented strictly:
            
            TRS Component: {g(active(X)) -> g(X)}
            
            Interpretation of nonconstant growth:
            -------------------------------------
              The following argument positions are usable:
                Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {1},
                Uargs(c) = {1}, Uargs(g) = {}, Uargs(d) = {}, Uargs(h) = {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) = [1 0] x1 + [0]
                       [0 0]      [0]
               mark(x1) = [1 0] x1 + [0]
                          [0 1]      [0]
               c(x1) = [1 0] x1 + [0]
                       [0 0]      [0]
               g(x1) = [0 2] x1 + [0]
                       [0 0]      [1]
               d(x1) = [0 0] x1 + [0]
                       [0 0]      [0]
               h(x1) = [1 0] x1 + [1]
                       [0 0]      [0]
            
            The strictly oriented rules are moved into the weak component.
            
            We consider the following Problem:
            
              Strict Trs:
                {  mark(f(X)) -> active(f(mark(X)))
                 , mark(c(X)) -> active(c(X))
                 , mark(g(X)) -> active(g(X))
                 , mark(d(X)) -> active(d(X))
                 , mark(h(X)) -> active(h(mark(X)))
                 , g(mark(X)) -> g(X)
                 , d(mark(X)) -> d(X)
                 , d(active(X)) -> d(X)}
              Weak Trs:
                {  g(active(X)) -> g(X)
                 , active(f(f(X))) -> mark(c(f(g(f(X)))))
                 , active(c(X)) -> mark(d(X))
                 , active(h(X)) -> mark(c(d(X)))
                 , f(mark(X)) -> f(X)
                 , f(active(X)) -> f(X)
                 , c(mark(X)) -> c(X)
                 , c(active(X)) -> c(X)
                 , h(mark(X)) -> h(X)
                 , h(active(X)) -> h(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(c(X)) -> active(c(X))
                 , mark(g(X)) -> active(g(X))
                 , mark(d(X)) -> active(d(X))}
              
              Interpretation of nonconstant growth:
              -------------------------------------
                The following argument positions are usable:
                  Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {1},
                  Uargs(c) = {1}, Uargs(g) = {}, Uargs(d) = {}, Uargs(h) = {1}
                We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                Interpretation Functions:
                 active(x1) = [1 0] x1 + [0]
                              [0 0]      [0]
                 f(x1) = [1 0] x1 + [2]
                         [0 0]      [0]
                 mark(x1) = [1 0] x1 + [1]
                            [0 0]      [0]
                 c(x1) = [1 0] x1 + [1]
                         [0 0]      [0]
                 g(x1) = [0 0] x1 + [0]
                         [0 0]      [0]
                 d(x1) = [0 0] x1 + [0]
                         [0 0]      [1]
                 h(x1) = [1 0] x1 + [2]
                         [0 0]      [0]
              
              The strictly oriented rules are moved into the weak component.
              
              We consider the following Problem:
              
                Strict Trs:
                  {  mark(f(X)) -> active(f(mark(X)))
                   , mark(h(X)) -> active(h(mark(X)))
                   , g(mark(X)) -> g(X)
                   , d(mark(X)) -> d(X)
                   , d(active(X)) -> d(X)}
                Weak Trs:
                  {  mark(c(X)) -> active(c(X))
                   , mark(g(X)) -> active(g(X))
                   , mark(d(X)) -> active(d(X))
                   , g(active(X)) -> g(X)
                   , active(f(f(X))) -> mark(c(f(g(f(X)))))
                   , active(c(X)) -> mark(d(X))
                   , active(h(X)) -> mark(c(d(X)))
                   , f(mark(X)) -> f(X)
                   , f(active(X)) -> f(X)
                   , c(mark(X)) -> c(X)
                   , c(active(X)) -> c(X)
                   , h(mark(X)) -> h(X)
                   , h(active(X)) -> h(X)}
                StartTerms: basic terms
                Strategy: innermost
              
              Certificate: YES(?,O(n^1))
              
              Proof:
                The weightgap principle applies, where following rules are oriented strictly:
                
                TRS Component:
                  {  d(mark(X)) -> d(X)
                   , d(active(X)) -> d(X)}
                
                Interpretation of nonconstant growth:
                -------------------------------------
                  The following argument positions are usable:
                    Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {1},
                    Uargs(c) = {1}, Uargs(g) = {}, Uargs(d) = {}, Uargs(h) = {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) = [1 0] x1 + [0]
                           [0 0]      [0]
                   mark(x1) = [1 0] x1 + [0]
                              [0 1]      [2]
                   c(x1) = [1 2] x1 + [0]
                           [0 0]      [0]
                   g(x1) = [0 0] x1 + [0]
                           [0 0]      [1]
                   d(x1) = [0 2] x1 + [0]
                           [0 0]      [0]
                   h(x1) = [1 2] x1 + [1]
                           [0 0]      [0]
                
                The strictly oriented rules are moved into the weak component.
                
                We consider the following Problem:
                
                  Strict Trs:
                    {  mark(f(X)) -> active(f(mark(X)))
                     , mark(h(X)) -> active(h(mark(X)))
                     , g(mark(X)) -> g(X)}
                  Weak Trs:
                    {  d(mark(X)) -> d(X)
                     , d(active(X)) -> d(X)
                     , mark(c(X)) -> active(c(X))
                     , mark(g(X)) -> active(g(X))
                     , mark(d(X)) -> active(d(X))
                     , g(active(X)) -> g(X)
                     , active(f(f(X))) -> mark(c(f(g(f(X)))))
                     , active(c(X)) -> mark(d(X))
                     , active(h(X)) -> mark(c(d(X)))
                     , f(mark(X)) -> f(X)
                     , f(active(X)) -> f(X)
                     , c(mark(X)) -> c(X)
                     , c(active(X)) -> c(X)
                     , h(mark(X)) -> h(X)
                     , h(active(X)) -> h(X)}
                  StartTerms: basic terms
                  Strategy: innermost
                
                Certificate: YES(?,O(n^1))
                
                Proof:
                  The weightgap principle applies, where following rules are oriented strictly:
                  
                  TRS Component: {g(mark(X)) -> g(X)}
                  
                  Interpretation of nonconstant growth:
                  -------------------------------------
                    The following argument positions are usable:
                      Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {1},
                      Uargs(c) = {1}, Uargs(g) = {}, Uargs(d) = {}, Uargs(h) = {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]      [2]
                     f(x1) = [1 0] x1 + [2]
                             [0 0]      [0]
                     mark(x1) = [1 0] x1 + [0]
                                [0 1]      [2]
                     c(x1) = [1 0] x1 + [0]
                             [0 0]      [0]
                     g(x1) = [1 2] x1 + [0]
                             [0 0]      [0]
                     d(x1) = [0 0] x1 + [0]
                             [0 0]      [0]
                     h(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:
                      {  mark(f(X)) -> active(f(mark(X)))
                       , mark(h(X)) -> active(h(mark(X)))}
                    Weak Trs:
                      {  g(mark(X)) -> g(X)
                       , d(mark(X)) -> d(X)
                       , d(active(X)) -> d(X)
                       , mark(c(X)) -> active(c(X))
                       , mark(g(X)) -> active(g(X))
                       , mark(d(X)) -> active(d(X))
                       , g(active(X)) -> g(X)
                       , active(f(f(X))) -> mark(c(f(g(f(X)))))
                       , active(c(X)) -> mark(d(X))
                       , active(h(X)) -> mark(c(d(X)))
                       , f(mark(X)) -> f(X)
                       , f(active(X)) -> f(X)
                       , c(mark(X)) -> c(X)
                       , c(active(X)) -> c(X)
                       , h(mark(X)) -> h(X)
                       , h(active(X)) -> h(X)}
                    StartTerms: basic terms
                    Strategy: innermost
                  
                  Certificate: YES(?,O(n^1))
                  
                  Proof:
                    We consider the following Problem:
                    
                      Strict Trs:
                        {  mark(f(X)) -> active(f(mark(X)))
                         , mark(h(X)) -> active(h(mark(X)))}
                      Weak Trs:
                        {  g(mark(X)) -> g(X)
                         , d(mark(X)) -> d(X)
                         , d(active(X)) -> d(X)
                         , mark(c(X)) -> active(c(X))
                         , mark(g(X)) -> active(g(X))
                         , mark(d(X)) -> active(d(X))
                         , g(active(X)) -> g(X)
                         , active(f(f(X))) -> mark(c(f(g(f(X)))))
                         , active(c(X)) -> mark(d(X))
                         , active(h(X)) -> mark(c(d(X)))
                         , f(mark(X)) -> f(X)
                         , f(active(X)) -> f(X)
                         , c(mark(X)) -> c(X)
                         , c(active(X)) -> c(X)
                         , h(mark(X)) -> h(X)
                         , h(active(X)) -> h(X)}
                      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) -> 1
                       , mark_0(2) -> 1
                       , c_0(2) -> 1
                       , g_0(2) -> 1
                       , d_0(2) -> 1
                       , h_0(2) -> 1}

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