We consider the following Problem:

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

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  g(X) -> h(X)
       , c() -> d()
       , h(d()) -> g(c())}
    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) -> h(X)}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(g) = {1}, Uargs(h) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       g(x1) = [1 0] x1 + [3]
               [0 0]      [1]
       h(x1) = [0 0] x1 + [1]
               [0 0]      [1]
       c() = [0]
             [0]
       d() = [0]
             [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  c() -> d()
         , h(d()) -> g(c())}
      Weak Trs: {g(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: {c() -> d()}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(g) = {1}, Uargs(h) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         g(x1) = [1 3] x1 + [1]
                 [0 0]      [1]
         h(x1) = [0 0] x1 + [1]
                 [0 0]      [1]
         c() = [3]
               [3]
         d() = [0]
               [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs: {h(d()) -> g(c())}
        Weak Trs:
          {  c() -> d()
           , g(X) -> h(X)}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        We consider the following Problem:
        
          Strict Trs: {h(d()) -> g(c())}
          Weak Trs:
            {  c() -> d()
             , g(X) -> h(X)}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          We have computed the following dependency pairs
          
            Strict DPs: {h^#(d()) -> g^#(c())}
            Weak DPs:
              {  c^#() -> c_2()
               , g^#(X) -> h^#(X)}
          
          We consider the following Problem:
          
            Strict DPs: {h^#(d()) -> g^#(c())}
            Strict Trs: {h(d()) -> g(c())}
            Weak DPs:
              {  c^#() -> c_2()
               , g^#(X) -> h^#(X)}
            Weak Trs:
              {  c() -> d()
               , g(X) -> h(X)}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(?,O(n^1))
          
          Proof:
            We replace strict/weak-rules by the corresponding usable rules:
            
              Weak Usable Rules: {c() -> d()}
            
            We consider the following Problem:
            
              Strict DPs: {h^#(d()) -> g^#(c())}
              Weak DPs:
                {  c^#() -> c_2()
                 , g^#(X) -> h^#(X)}
              Weak Trs: {c() -> d()}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(?,O(n^1))
            
            Proof:
              We consider the following Problem:
              
                Strict DPs: {h^#(d()) -> g^#(c())}
                Weak DPs:
                  {  c^#() -> c_2()
                   , g^#(X) -> h^#(X)}
                Weak Trs: {c() -> d()}
                StartTerms: basic terms
                Strategy: innermost
              
              Certificate: YES(?,O(n^1))
              
              Proof:
                We use following congruence DG for path analysis
                
                ->2:{1,3}                                                   [   YES(O(1),O(1))   ]
                
                ->1:{2}                                                     [   YES(O(1),O(1))   ]
                
                
                Here dependency-pairs are as follows:
                
                Strict DPs:
                  {1: h^#(d()) -> g^#(c())}
                WeakDPs DPs:
                  {  2: c^#() -> c_2()
                   , 3: g^#(X) -> h^#(X)}
                
                * Path 2:{1,3}: YES(O(1),O(1))
                  ----------------------------
                  
                  We consider the following Problem:
                  
                    Strict DPs: {h^#(d()) -> g^#(c())}
                    Weak Trs: {c() -> d()}
                    StartTerms: basic terms
                    Strategy: innermost
                  
                  Certificate: YES(O(1),O(1))
                  
                  Proof:
                    We consider the the dependency-graph
                    
                      1: h^#(d()) -> g^#(c())
                      
                    
                    together with the congruence-graph
                    
                      ->1:{1}                                                     Noncyclic, trivial, SCC
                      
                      
                      Here dependency-pairs are as follows:
                      
                      Strict DPs:
                        {1: h^#(d()) -> g^#(c())}
                    
                    The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
                    
                      {1: h^#(d()) -> g^#(c())}
                    
                    We consider the following Problem:
                    
                      Weak Trs: {c() -> d()}
                      StartTerms: basic terms
                      Strategy: innermost
                    
                    Certificate: YES(O(1),O(1))
                    
                    Proof:
                      We consider the following Problem:
                      
                        Weak Trs: {c() -> d()}
                        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:{2}: YES(O(1),O(1))
                  --------------------------
                  
                  We consider the following Problem:
                  
                    Weak Trs: {c() -> d()}
                    StartTerms: basic terms
                    Strategy: innermost
                  
                  Certificate: YES(O(1),O(1))
                  
                  Proof:
                    We consider the following Problem:
                    
                      Weak Trs: {c() -> d()}
                      StartTerms: basic terms
                      Strategy: innermost
                    
                    Certificate: YES(O(1),O(1))
                    
                    Proof:
                      We consider the following Problem:
                      
                        Weak Trs: {c() -> d()}
                        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))