We consider the following Problem:

  Strict Trs:
    {  f(x, 0()) -> s(0())
     , f(s(x), s(y)) -> s(f(x, y))
     , g(0(), x) -> g(f(x, x), x)}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  f(x, 0()) -> s(0())
       , f(s(x), s(y)) -> s(f(x, y))
       , g(0(), x) -> g(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(0(), x) -> g(f(x, x), x)}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(f) = {}, Uargs(s) = {1}, Uargs(g) = {1}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       f(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
                   [0 0]      [0 0]      [1]
       0() = [0]
             [2]
       s(x1) = [1 0] x1 + [1]
               [0 0]      [1]
       g(x1, x2) = [1 2] x1 + [1 1] x2 + [0]
                   [0 0]      [0 0]      [1]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  f(x, 0()) -> s(0())
         , f(s(x), s(y)) -> s(f(x, y))}
      Weak Trs: {g(0(), x) -> g(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: {f(x, 0()) -> s(0())}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(f) = {}, Uargs(s) = {1}, Uargs(g) = {1}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         f(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
                     [0 0]      [0 0]      [0]
         0() = [0]
               [2]
         s(x1) = [1 0] x1 + [0]
                 [0 0]      [0]
         g(x1, x2) = [1 2] x1 + [1 0] x2 + [0]
                     [0 0]      [0 0]      [1]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs: {f(s(x), s(y)) -> s(f(x, y))}
        Weak Trs:
          {  f(x, 0()) -> s(0())
           , g(0(), x) -> g(f(x, x), x)}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        We consider the following Problem:
        
          Strict Trs: {f(s(x), s(y)) -> s(f(x, y))}
          Weak Trs:
            {  f(x, 0()) -> s(0())
             , g(0(), x) -> g(f(x, x), x)}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          We have computed the following dependency pairs
          
            Strict DPs: {f^#(s(x), s(y)) -> f^#(x, y)}
            Weak DPs:
              {  f^#(x, 0()) -> c_2()
               , g^#(0(), x) -> g^#(f(x, x), x)}
          
          We consider the following Problem:
          
            Strict DPs: {f^#(s(x), s(y)) -> f^#(x, y)}
            Strict Trs: {f(s(x), s(y)) -> s(f(x, y))}
            Weak DPs:
              {  f^#(x, 0()) -> c_2()
               , g^#(0(), x) -> g^#(f(x, x), x)}
            Weak Trs:
              {  f(x, 0()) -> s(0())
               , g(0(), x) -> g(f(x, x), x)}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(?,O(n^1))
          
          Proof:
            We replace strict/weak-rules by the corresponding usable rules:
            
              Strict Usable Rules: {f(s(x), s(y)) -> s(f(x, y))}
              Weak Usable Rules: {f(x, 0()) -> s(0())}
            
            We consider the following Problem:
            
              Strict DPs: {f^#(s(x), s(y)) -> f^#(x, y)}
              Strict Trs: {f(s(x), s(y)) -> s(f(x, y))}
              Weak DPs:
                {  f^#(x, 0()) -> c_2()
                 , g^#(0(), x) -> g^#(f(x, x), x)}
              Weak Trs: {f(x, 0()) -> s(0())}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(?,O(n^1))
            
            Proof:
              We consider the following Problem:
              
                Strict DPs: {f^#(s(x), s(y)) -> f^#(x, y)}
                Strict Trs: {f(s(x), s(y)) -> s(f(x, y))}
                Weak DPs:
                  {  f^#(x, 0()) -> c_2()
                   , g^#(0(), x) -> g^#(f(x, x), x)}
                Weak Trs: {f(x, 0()) -> s(0())}
                StartTerms: basic terms
                Strategy: innermost
              
              Certificate: YES(?,O(n^1))
              
              Proof:
                We use following congruence DG for path analysis
                
                ->2:{1}                                                     [   YES(?,O(n^1))    ]
                   |
                   `->3:{2}                                                 [   YES(O(1),O(1))   ]
                
                ->1:{3}                                                     [   YES(O(1),O(1))   ]
                
                
                Here dependency-pairs are as follows:
                
                Strict DPs:
                  {1: f^#(s(x), s(y)) -> f^#(x, y)}
                WeakDPs DPs:
                  {  2: f^#(x, 0()) -> c_2()
                   , 3: g^#(0(), x) -> g^#(f(x, x), x)}
                
                * Path 2:{1}: YES(?,O(n^1))
                  -------------------------
                  
                  We consider the following Problem:
                  
                    Strict DPs: {f^#(s(x), s(y)) -> f^#(x, y)}
                    Strict Trs: {f(s(x), s(y)) -> s(f(x, y))}
                    Weak Trs: {f(x, 0()) -> s(0())}
                    StartTerms: basic terms
                    Strategy: innermost
                  
                  Certificate: YES(?,O(n^1))
                  
                  Proof:
                    We consider the following Problem:
                    
                      Strict DPs: {f^#(s(x), s(y)) -> f^#(x, y)}
                      Strict Trs: {f(s(x), s(y)) -> s(f(x, y))}
                      Weak Trs: {f(x, 0()) -> s(0())}
                      StartTerms: basic terms
                      Strategy: innermost
                    
                    Certificate: YES(?,O(n^1))
                    
                    Proof:
                      We consider the following Problem:
                      
                        Strict DPs: {f^#(s(x), s(y)) -> f^#(x, y)}
                        Strict Trs: {f(s(x), s(y)) -> s(f(x, y))}
                        Weak Trs: {f(x, 0()) -> s(0())}
                        StartTerms: basic terms
                        Strategy: innermost
                      
                      Certificate: YES(?,O(n^1))
                      
                      Proof:
                        No rule is usable.
                        
                        We consider the following Problem:
                        
                          Strict DPs: {f^#(s(x), s(y)) -> f^#(x, y)}
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(?,O(n^1))
                        
                        Proof:
                          The problem is match-bounded by 1.
                          The enriched problem is compatible with the following automaton:
                          {  s_0(2) -> 2
                           , f^#_0(2, 2) -> 1
                           , f^#_1(2, 2) -> 1}
                
                * Path 2:{1}->3:{2}: YES(O(1),O(1))
                  ---------------------------------
                  
                  We consider the following Problem:
                  
                    Strict Trs: {f(s(x), s(y)) -> s(f(x, y))}
                    Weak DPs: {f^#(s(x), s(y)) -> f^#(x, y)}
                    Weak Trs: {f(x, 0()) -> s(0())}
                    StartTerms: basic terms
                    Strategy: innermost
                  
                  Certificate: YES(O(1),O(1))
                  
                  Proof:
                    We consider the following Problem:
                    
                      Strict Trs: {f(s(x), s(y)) -> s(f(x, y))}
                      Weak DPs: {f^#(s(x), s(y)) -> f^#(x, y)}
                      Weak Trs: {f(x, 0()) -> s(0())}
                      StartTerms: basic terms
                      Strategy: innermost
                    
                    Certificate: YES(O(1),O(1))
                    
                    Proof:
                      We consider the following Problem:
                      
                        Strict Trs: {f(s(x), s(y)) -> s(f(x, y))}
                        Weak DPs: {f^#(s(x), s(y)) -> f^#(x, y)}
                        Weak Trs: {f(x, 0()) -> s(0())}
                        StartTerms: basic terms
                        Strategy: innermost
                      
                      Certificate: YES(O(1),O(1))
                      
                      Proof:
                        No rule is usable.
                        
                        We consider the following Problem:
                        
                          Weak DPs: {f^#(s(x), s(y)) -> f^#(x, y)}
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(O(1),O(1))
                        
                        Proof:
                          Empty rules are trivially bounded
                
                * Path 1:{3}: YES(O(1),O(1))
                  --------------------------
                  
                  We consider the following Problem:
                  
                    Strict Trs: {f(s(x), s(y)) -> s(f(x, y))}
                    Weak Trs: {f(x, 0()) -> s(0())}
                    StartTerms: basic terms
                    Strategy: innermost
                  
                  Certificate: YES(O(1),O(1))
                  
                  Proof:
                    We consider the following Problem:
                    
                      Strict Trs: {f(s(x), s(y)) -> s(f(x, y))}
                      Weak Trs: {f(x, 0()) -> s(0())}
                      StartTerms: basic terms
                      Strategy: innermost
                    
                    Certificate: YES(O(1),O(1))
                    
                    Proof:
                      We consider the following Problem:
                      
                        Strict Trs: {f(s(x), s(y)) -> s(f(x, y))}
                        Weak Trs: {f(x, 0()) -> s(0())}
                        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))