We consider the following Problem:

  Strict Trs:
    {  g(X) -> u(h(X), h(X), X)
     , u(d(), c(Y), X) -> k(Y)
     , h(d()) -> c(a())
     , h(d()) -> c(b())
     , f(k(a()), k(b()), X) -> f(X, X, X)}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  g(X) -> u(h(X), h(X), X)
       , u(d(), c(Y), X) -> k(Y)
       , h(d()) -> c(a())
       , h(d()) -> c(b())
       , f(k(a()), k(b()), X) -> f(X, 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:
      {  g(X) -> u(h(X), h(X), X)
       , u(d(), c(Y), X) -> k(Y)}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(g) = {}, Uargs(u) = {1, 2}, Uargs(h) = {}, Uargs(c) = {},
        Uargs(k) = {}, Uargs(f) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       g(x1) = [1 1] x1 + [2]
               [0 0]      [2]
       u(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 1] x3 + [1]
                       [0 0]      [0 0]      [0 0]      [1]
       h(x1) = [0 0] x1 + [0]
               [0 0]      [1]
       d() = [0]
             [0]
       c(x1) = [0 0] x1 + [0]
               [1 1]      [1]
       k(x1) = [0 0] x1 + [0]
               [0 0]      [0]
       a() = [0]
             [0]
       b() = [0]
             [0]
       f(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [1 1] x3 + [1]
                       [0 0]      [1 1]      [0 0]      [1]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  h(d()) -> c(a())
         , h(d()) -> c(b())
         , f(k(a()), k(b()), X) -> f(X, X, X)}
      Weak Trs:
        {  g(X) -> u(h(X), h(X), X)
         , u(d(), c(Y), X) -> k(Y)}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component:
        {  h(d()) -> c(a())
         , h(d()) -> c(b())}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(g) = {}, Uargs(u) = {1, 2}, Uargs(h) = {}, Uargs(c) = {},
          Uargs(k) = {}, Uargs(f) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         g(x1) = [1 0] x1 + [3]
                 [1 1]      [2]
         u(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
                         [0 0]      [0 0]      [1 1]      [1]
         h(x1) = [0 0] x1 + [1]
                 [1 1]      [0]
         d() = [2]
               [3]
         c(x1) = [0 0] x1 + [0]
                 [0 0]      [1]
         k(x1) = [0 0] x1 + [0]
                 [0 0]      [0]
         a() = [0]
               [0]
         b() = [0]
               [0]
         f(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [1 0] x3 + [1]
                         [1 1]      [0 0]      [1 1]      [1]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs: {f(k(a()), k(b()), X) -> f(X, X, X)}
        Weak Trs:
          {  h(d()) -> c(a())
           , h(d()) -> c(b())
           , g(X) -> u(h(X), h(X), X)
           , u(d(), c(Y), X) -> k(Y)}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        We consider the following Problem:
        
          Strict Trs: {f(k(a()), k(b()), X) -> f(X, X, X)}
          Weak Trs:
            {  h(d()) -> c(a())
             , h(d()) -> c(b())
             , g(X) -> u(h(X), h(X), X)
             , u(d(), c(Y), X) -> k(Y)}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          We have computed the following dependency pairs
          
            Strict DPs: {f^#(k(a()), k(b()), X) -> f^#(X, X, X)}
            Weak DPs:
              {  h^#(d()) -> c_2()
               , h^#(d()) -> c_3()
               , g^#(X) -> u^#(h(X), h(X), X)
               , u^#(d(), c(Y), X) -> c_5()}
          
          We consider the following Problem:
          
            Strict DPs: {f^#(k(a()), k(b()), X) -> f^#(X, X, X)}
            Strict Trs: {f(k(a()), k(b()), X) -> f(X, X, X)}
            Weak DPs:
              {  h^#(d()) -> c_2()
               , h^#(d()) -> c_3()
               , g^#(X) -> u^#(h(X), h(X), X)
               , u^#(d(), c(Y), X) -> c_5()}
            Weak Trs:
              {  h(d()) -> c(a())
               , h(d()) -> c(b())
               , g(X) -> u(h(X), h(X), X)
               , u(d(), c(Y), X) -> k(Y)}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(?,O(n^1))
          
          Proof:
            We replace strict/weak-rules by the corresponding usable rules:
            
              Weak Usable Rules:
                {  h(d()) -> c(a())
                 , h(d()) -> c(b())}
            
            We consider the following Problem:
            
              Strict DPs: {f^#(k(a()), k(b()), X) -> f^#(X, X, X)}
              Weak DPs:
                {  h^#(d()) -> c_2()
                 , h^#(d()) -> c_3()
                 , g^#(X) -> u^#(h(X), h(X), X)
                 , u^#(d(), c(Y), X) -> c_5()}
              Weak Trs:
                {  h(d()) -> c(a())
                 , h(d()) -> c(b())}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(?,O(n^1))
            
            Proof:
              We consider the following Problem:
              
                Strict DPs: {f^#(k(a()), k(b()), X) -> f^#(X, X, X)}
                Weak DPs:
                  {  h^#(d()) -> c_2()
                   , h^#(d()) -> c_3()
                   , g^#(X) -> u^#(h(X), h(X), X)
                   , u^#(d(), c(Y), X) -> c_5()}
                Weak Trs:
                  {  h(d()) -> c(a())
                   , h(d()) -> c(b())}
                StartTerms: basic terms
                Strategy: innermost
              
              Certificate: YES(?,O(n^1))
              
              Proof:
                We use following congruence DG for path analysis
                
                ->5:{1}                                                     [   YES(O(1),O(1))   ]
                
                ->4:{2}                                                     [   YES(O(1),O(1))   ]
                
                ->3:{3}                                                     [   YES(O(1),O(1))   ]
                
                ->2:{4}                                                     [   YES(O(1),O(1))   ]
                
                ->1:{5}                                                     [   YES(O(1),O(1))   ]
                
                
                Here dependency-pairs are as follows:
                
                Strict DPs:
                  {1: f^#(k(a()), k(b()), X) -> f^#(X, X, X)}
                WeakDPs DPs:
                  {  2: h^#(d()) -> c_2()
                   , 3: h^#(d()) -> c_3()
                   , 4: g^#(X) -> u^#(h(X), h(X), X)
                   , 5: u^#(d(), c(Y), X) -> c_5()}
                
                * Path 5:{1}: YES(O(1),O(1))
                  --------------------------
                  
                  We consider the following Problem:
                  
                    Strict DPs: {f^#(k(a()), k(b()), X) -> f^#(X, X, X)}
                    Weak Trs:
                      {  h(d()) -> c(a())
                       , h(d()) -> c(b())}
                    StartTerms: basic terms
                    Strategy: innermost
                  
                  Certificate: YES(O(1),O(1))
                  
                  Proof:
                    We consider the the dependency-graph
                    
                      1: f^#(k(a()), k(b()), X) -> f^#(X, X, X)
                      
                    
                    together with the congruence-graph
                    
                      ->1:{1}                                                     Noncyclic, trivial, SCC
                      
                      
                      Here dependency-pairs are as follows:
                      
                      Strict DPs:
                        {1: f^#(k(a()), k(b()), X) -> f^#(X, X, X)}
                    
                    The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
                    
                      {1: f^#(k(a()), k(b()), X) -> f^#(X, X, X)}
                    
                    We consider the following Problem:
                    
                      Weak Trs:
                        {  h(d()) -> c(a())
                         , h(d()) -> c(b())}
                      StartTerms: basic terms
                      Strategy: innermost
                    
                    Certificate: YES(O(1),O(1))
                    
                    Proof:
                      We consider the following Problem:
                      
                        Weak Trs:
                          {  h(d()) -> c(a())
                           , h(d()) -> c(b())}
                        StartTerms: basic terms
                        Strategy: innermost
                      
                      Certificate: YES(O(1),O(1))
                      
                      Proof:
                        No rule is usable.
                        
                        We consider the following Problem:
                        
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(O(1),O(1))
                        
                        Proof:
                          Empty rules are trivially bounded
                
                * Path 4:{2}: YES(O(1),O(1))
                  --------------------------
                  
                  We consider the following Problem:
                  
                    Weak Trs:
                      {  h(d()) -> c(a())
                       , h(d()) -> c(b())}
                    StartTerms: basic terms
                    Strategy: innermost
                  
                  Certificate: YES(O(1),O(1))
                  
                  Proof:
                    We consider the following Problem:
                    
                      Weak Trs:
                        {  h(d()) -> c(a())
                         , h(d()) -> c(b())}
                      StartTerms: basic terms
                      Strategy: innermost
                    
                    Certificate: YES(O(1),O(1))
                    
                    Proof:
                      We consider the following Problem:
                      
                        Weak Trs:
                          {  h(d()) -> c(a())
                           , h(d()) -> c(b())}
                        StartTerms: basic terms
                        Strategy: innermost
                      
                      Certificate: YES(O(1),O(1))
                      
                      Proof:
                        No rule is usable.
                        
                        We consider the following Problem:
                        
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(O(1),O(1))
                        
                        Proof:
                          Empty rules are trivially bounded
                
                * Path 3:{3}: YES(O(1),O(1))
                  --------------------------
                  
                  We consider the following Problem:
                  
                    Weak Trs:
                      {  h(d()) -> c(a())
                       , h(d()) -> c(b())}
                    StartTerms: basic terms
                    Strategy: innermost
                  
                  Certificate: YES(O(1),O(1))
                  
                  Proof:
                    We consider the following Problem:
                    
                      Weak Trs:
                        {  h(d()) -> c(a())
                         , h(d()) -> c(b())}
                      StartTerms: basic terms
                      Strategy: innermost
                    
                    Certificate: YES(O(1),O(1))
                    
                    Proof:
                      We consider the following Problem:
                      
                        Weak Trs:
                          {  h(d()) -> c(a())
                           , h(d()) -> c(b())}
                        StartTerms: basic terms
                        Strategy: innermost
                      
                      Certificate: YES(O(1),O(1))
                      
                      Proof:
                        No rule is usable.
                        
                        We consider the following Problem:
                        
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(O(1),O(1))
                        
                        Proof:
                          Empty rules are trivially bounded
                
                * Path 2:{4}: YES(O(1),O(1))
                  --------------------------
                  
                  We consider the following Problem:
                  
                    Weak Trs:
                      {  h(d()) -> c(a())
                       , h(d()) -> c(b())}
                    StartTerms: basic terms
                    Strategy: innermost
                  
                  Certificate: YES(O(1),O(1))
                  
                  Proof:
                    We consider the following Problem:
                    
                      Weak Trs:
                        {  h(d()) -> c(a())
                         , h(d()) -> c(b())}
                      StartTerms: basic terms
                      Strategy: innermost
                    
                    Certificate: YES(O(1),O(1))
                    
                    Proof:
                      We consider the following Problem:
                      
                        Weak Trs:
                          {  h(d()) -> c(a())
                           , h(d()) -> c(b())}
                        StartTerms: basic terms
                        Strategy: innermost
                      
                      Certificate: YES(O(1),O(1))
                      
                      Proof:
                        No rule is usable.
                        
                        We consider the following Problem:
                        
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(O(1),O(1))
                        
                        Proof:
                          Empty rules are trivially bounded
                
                * Path 1:{5}: YES(O(1),O(1))
                  --------------------------
                  
                  We consider the following Problem:
                  
                    Weak Trs:
                      {  h(d()) -> c(a())
                       , h(d()) -> c(b())}
                    StartTerms: basic terms
                    Strategy: innermost
                  
                  Certificate: YES(O(1),O(1))
                  
                  Proof:
                    We consider the following Problem:
                    
                      Weak Trs:
                        {  h(d()) -> c(a())
                         , h(d()) -> c(b())}
                      StartTerms: basic terms
                      Strategy: innermost
                    
                    Certificate: YES(O(1),O(1))
                    
                    Proof:
                      We consider the following Problem:
                      
                        Weak Trs:
                          {  h(d()) -> c(a())
                           , h(d()) -> c(b())}
                        StartTerms: basic terms
                        Strategy: innermost
                      
                      Certificate: YES(O(1),O(1))
                      
                      Proof:
                        No rule is usable.
                        
                        We consider the following Problem:
                        
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(O(1),O(1))
                        
                        Proof:
                          Empty rules are trivially bounded

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