We consider the following Problem:

  Strict Trs:
    {  cond1(true(), x) -> cond2(even(x), x)
     , cond2(true(), x) -> cond1(neq(x, 0()), div2(x))
     , cond2(false(), x) -> cond1(neq(x, 0()), p(x))
     , neq(0(), 0()) -> false()
     , neq(0(), s(x)) -> true()
     , neq(s(x), 0()) -> true()
     , neq(s(x), s(y())) -> neq(x, y())
     , even(0()) -> true()
     , even(s(0())) -> false()
     , even(s(s(x))) -> even(x)
     , div2(0()) -> 0()
     , div2(s(0())) -> 0()
     , div2(s(s(x))) -> s(div2(x))
     , p(0()) -> 0()
     , p(s(x)) -> x}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  cond1(true(), x) -> cond2(even(x), x)
       , cond2(true(), x) -> cond1(neq(x, 0()), div2(x))
       , cond2(false(), x) -> cond1(neq(x, 0()), p(x))
       , neq(0(), 0()) -> false()
       , neq(0(), s(x)) -> true()
       , neq(s(x), 0()) -> true()
       , neq(s(x), s(y())) -> neq(x, y())
       , even(0()) -> true()
       , even(s(0())) -> false()
       , even(s(s(x))) -> even(x)
       , div2(0()) -> 0()
       , div2(s(0())) -> 0()
       , div2(s(s(x))) -> s(div2(x))
       , p(0()) -> 0()
       , p(s(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: {p(0()) -> 0()}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(cond1) = {1, 2}, Uargs(cond2) = {1}, Uargs(even) = {},
        Uargs(neq) = {}, Uargs(div2) = {}, Uargs(p) = {}, Uargs(s) = {1}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       cond1(x1, x2) = [1 0] x1 + [1 2] x2 + [1]
                       [0 0]      [0 0]      [1]
       true() = [0]
                [0]
       cond2(x1, x2) = [1 0] x1 + [1 1] x2 + [1]
                       [0 0]      [0 0]      [1]
       even(x1) = [0 0] x1 + [0]
                  [0 0]      [1]
       neq(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                     [0 0]      [0 0]      [1]
       0() = [1]
             [0]
       div2(x1) = [0 0] x1 + [0]
                  [0 0]      [0]
       false() = [0]
                 [0]
       p(x1) = [1 0] x1 + [2]
               [0 0]      [3]
       s(x1) = [1 0] x1 + [1]
               [0 0]      [1]
       y() = [0]
             [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  cond1(true(), x) -> cond2(even(x), x)
         , cond2(true(), x) -> cond1(neq(x, 0()), div2(x))
         , cond2(false(), x) -> cond1(neq(x, 0()), p(x))
         , neq(0(), 0()) -> false()
         , neq(0(), s(x)) -> true()
         , neq(s(x), 0()) -> true()
         , neq(s(x), s(y())) -> neq(x, y())
         , even(0()) -> true()
         , even(s(0())) -> false()
         , even(s(s(x))) -> even(x)
         , div2(0()) -> 0()
         , div2(s(0())) -> 0()
         , div2(s(s(x))) -> s(div2(x))
         , p(s(x)) -> x}
      Weak Trs: {p(0()) -> 0()}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {cond2(false(), x) -> cond1(neq(x, 0()), p(x))}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(cond1) = {1, 2}, Uargs(cond2) = {1}, Uargs(even) = {},
          Uargs(neq) = {}, Uargs(div2) = {}, Uargs(p) = {}, Uargs(s) = {1}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         cond1(x1, x2) = [1 0] x1 + [1 1] x2 + [1]
                         [0 0]      [0 0]      [1]
         true() = [0]
                  [0]
         cond2(x1, x2) = [1 2] x1 + [1 1] x2 + [1]
                         [0 0]      [0 0]      [1]
         even(x1) = [0 0] x1 + [0]
                    [0 0]      [0]
         neq(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                       [0 0]      [0 0]      [1]
         0() = [0]
               [0]
         div2(x1) = [0 0] x1 + [0]
                    [0 0]      [0]
         false() = [0]
                   [2]
         p(x1) = [1 0] x1 + [0]
                 [0 0]      [0]
         s(x1) = [1 0] x1 + [1]
                 [0 0]      [1]
         y() = [0]
               [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  cond1(true(), x) -> cond2(even(x), x)
           , cond2(true(), x) -> cond1(neq(x, 0()), div2(x))
           , neq(0(), 0()) -> false()
           , neq(0(), s(x)) -> true()
           , neq(s(x), 0()) -> true()
           , neq(s(x), s(y())) -> neq(x, y())
           , even(0()) -> true()
           , even(s(0())) -> false()
           , even(s(s(x))) -> even(x)
           , div2(0()) -> 0()
           , div2(s(0())) -> 0()
           , div2(s(s(x))) -> s(div2(x))
           , p(s(x)) -> x}
        Weak Trs:
          {  cond2(false(), x) -> cond1(neq(x, 0()), p(x))
           , p(0()) -> 0()}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component:
          {  div2(0()) -> 0()
           , div2(s(0())) -> 0()}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(cond1) = {1, 2}, Uargs(cond2) = {1}, Uargs(even) = {},
            Uargs(neq) = {}, Uargs(div2) = {}, Uargs(p) = {}, Uargs(s) = {1}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           cond1(x1, x2) = [1 0] x1 + [1 2] x2 + [1]
                           [0 0]      [0 0]      [1]
           true() = [0]
                    [0]
           cond2(x1, x2) = [1 0] x1 + [1 1] x2 + [1]
                           [0 0]      [0 0]      [1]
           even(x1) = [0 0] x1 + [0]
                      [0 0]      [1]
           neq(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                         [0 0]      [0 0]      [1]
           0() = [0]
                 [0]
           div2(x1) = [0 0] x1 + [1]
                      [0 0]      [0]
           false() = [3]
                     [0]
           p(x1) = [1 0] x1 + [1]
                   [0 0]      [1]
           s(x1) = [1 1] x1 + [0]
                   [0 0]      [1]
           y() = [0]
                 [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs:
            {  cond1(true(), x) -> cond2(even(x), x)
             , cond2(true(), x) -> cond1(neq(x, 0()), div2(x))
             , neq(0(), 0()) -> false()
             , neq(0(), s(x)) -> true()
             , neq(s(x), 0()) -> true()
             , neq(s(x), s(y())) -> neq(x, y())
             , even(0()) -> true()
             , even(s(0())) -> false()
             , even(s(s(x))) -> even(x)
             , div2(s(s(x))) -> s(div2(x))
             , p(s(x)) -> x}
          Weak Trs:
            {  div2(0()) -> 0()
             , div2(s(0())) -> 0()
             , cond2(false(), x) -> cond1(neq(x, 0()), p(x))
             , p(0()) -> 0()}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          The weightgap principle applies, where following rules are oriented strictly:
          
          TRS Component: {p(s(x)) -> x}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(cond1) = {1, 2}, Uargs(cond2) = {1}, Uargs(even) = {},
              Uargs(neq) = {}, Uargs(div2) = {}, Uargs(p) = {}, Uargs(s) = {1}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             cond1(x1, x2) = [1 0] x1 + [1 1] x2 + [1]
                             [0 0]      [0 0]      [1]
             true() = [0]
                      [0]
             cond2(x1, x2) = [1 0] x1 + [1 1] x2 + [1]
                             [0 0]      [0 0]      [1]
             even(x1) = [0 0] x1 + [0]
                        [0 0]      [1]
             neq(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                           [0 0]      [0 1]      [1]
             0() = [0]
                   [0]
             div2(x1) = [0 0] x1 + [0]
                        [0 0]      [0]
             false() = [0]
                       [0]
             p(x1) = [1 0] x1 + [0]
                     [0 1]      [0]
             s(x1) = [1 0] x1 + [1]
                     [0 1]      [1]
             y() = [0]
                   [0]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Strict Trs:
              {  cond1(true(), x) -> cond2(even(x), x)
               , cond2(true(), x) -> cond1(neq(x, 0()), div2(x))
               , neq(0(), 0()) -> false()
               , neq(0(), s(x)) -> true()
               , neq(s(x), 0()) -> true()
               , neq(s(x), s(y())) -> neq(x, y())
               , even(0()) -> true()
               , even(s(0())) -> false()
               , even(s(s(x))) -> even(x)
               , div2(s(s(x))) -> s(div2(x))}
            Weak Trs:
              {  p(s(x)) -> x
               , div2(0()) -> 0()
               , div2(s(0())) -> 0()
               , cond2(false(), x) -> cond1(neq(x, 0()), p(x))
               , p(0()) -> 0()}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(?,O(n^1))
          
          Proof:
            The weightgap principle applies, where following rules are oriented strictly:
            
            TRS Component:
              {  neq(0(), s(x)) -> true()
               , neq(s(x), s(y())) -> neq(x, y())}
            
            Interpretation of nonconstant growth:
            -------------------------------------
              The following argument positions are usable:
                Uargs(cond1) = {1, 2}, Uargs(cond2) = {1}, Uargs(even) = {},
                Uargs(neq) = {}, Uargs(div2) = {}, Uargs(p) = {}, Uargs(s) = {1}
              We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
              Interpretation Functions:
               cond1(x1, x2) = [1 0] x1 + [1 1] x2 + [1]
                               [0 0]      [0 0]      [1]
               true() = [0]
                        [0]
               cond2(x1, x2) = [1 0] x1 + [1 1] x2 + [1]
                               [0 0]      [0 0]      [1]
               even(x1) = [0 0] x1 + [0]
                          [0 0]      [1]
               neq(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                             [0 0]      [0 1]      [1]
               0() = [0]
                     [0]
               div2(x1) = [0 0] x1 + [0]
                          [0 0]      [0]
               false() = [2]
                         [0]
               p(x1) = [1 0] x1 + [1]
                       [0 1]      [0]
               s(x1) = [1 0] x1 + [2]
                       [0 1]      [1]
               y() = [0]
                     [0]
            
            The strictly oriented rules are moved into the weak component.
            
            We consider the following Problem:
            
              Strict Trs:
                {  cond1(true(), x) -> cond2(even(x), x)
                 , cond2(true(), x) -> cond1(neq(x, 0()), div2(x))
                 , neq(0(), 0()) -> false()
                 , neq(s(x), 0()) -> true()
                 , even(0()) -> true()
                 , even(s(0())) -> false()
                 , even(s(s(x))) -> even(x)
                 , div2(s(s(x))) -> s(div2(x))}
              Weak Trs:
                {  neq(0(), s(x)) -> true()
                 , neq(s(x), s(y())) -> neq(x, y())
                 , p(s(x)) -> x
                 , div2(0()) -> 0()
                 , div2(s(0())) -> 0()
                 , cond2(false(), x) -> cond1(neq(x, 0()), p(x))
                 , p(0()) -> 0()}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(?,O(n^1))
            
            Proof:
              The weightgap principle applies, where following rules are oriented strictly:
              
              TRS Component: {neq(s(x), 0()) -> true()}
              
              Interpretation of nonconstant growth:
              -------------------------------------
                The following argument positions are usable:
                  Uargs(cond1) = {1, 2}, Uargs(cond2) = {1}, Uargs(even) = {},
                  Uargs(neq) = {}, Uargs(div2) = {}, Uargs(p) = {}, Uargs(s) = {1}
                We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                Interpretation Functions:
                 cond1(x1, x2) = [1 1] x1 + [1 1] x2 + [1]
                                 [0 0]      [0 0]      [1]
                 true() = [0]
                          [0]
                 cond2(x1, x2) = [1 0] x1 + [1 1] x2 + [1]
                                 [0 0]      [0 0]      [1]
                 even(x1) = [0 0] x1 + [0]
                            [0 1]      [0]
                 neq(x1, x2) = [0 0] x1 + [0 2] x2 + [0]
                               [0 0]      [0 0]      [1]
                 0() = [0]
                       [1]
                 div2(x1) = [0 0] x1 + [2]
                            [0 0]      [2]
                 false() = [3]
                           [0]
                 p(x1) = [1 0] x1 + [0]
                         [0 1]      [0]
                 s(x1) = [1 0] x1 + [1]
                         [0 1]      [0]
                 y() = [0]
                       [0]
              
              The strictly oriented rules are moved into the weak component.
              
              We consider the following Problem:
              
                Strict Trs:
                  {  cond1(true(), x) -> cond2(even(x), x)
                   , cond2(true(), x) -> cond1(neq(x, 0()), div2(x))
                   , neq(0(), 0()) -> false()
                   , even(0()) -> true()
                   , even(s(0())) -> false()
                   , even(s(s(x))) -> even(x)
                   , div2(s(s(x))) -> s(div2(x))}
                Weak Trs:
                  {  neq(s(x), 0()) -> true()
                   , neq(0(), s(x)) -> true()
                   , neq(s(x), s(y())) -> neq(x, y())
                   , p(s(x)) -> x
                   , div2(0()) -> 0()
                   , div2(s(0())) -> 0()
                   , cond2(false(), x) -> cond1(neq(x, 0()), p(x))
                   , p(0()) -> 0()}
                StartTerms: basic terms
                Strategy: innermost
              
              Certificate: YES(?,O(n^1))
              
              Proof:
                The weightgap principle applies, where following rules are oriented strictly:
                
                TRS Component: {neq(0(), 0()) -> false()}
                
                Interpretation of nonconstant growth:
                -------------------------------------
                  The following argument positions are usable:
                    Uargs(cond1) = {1, 2}, Uargs(cond2) = {1}, Uargs(even) = {},
                    Uargs(neq) = {}, Uargs(div2) = {}, Uargs(p) = {}, Uargs(s) = {1}
                  We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                  Interpretation Functions:
                   cond1(x1, x2) = [1 2] x1 + [1 1] x2 + [1]
                                   [0 0]      [0 0]      [1]
                   true() = [0]
                            [0]
                   cond2(x1, x2) = [1 3] x1 + [1 1] x2 + [1]
                                   [0 0]      [0 0]      [1]
                   even(x1) = [0 0] x1 + [0]
                              [0 0]      [0]
                   neq(x1, x2) = [0 0] x1 + [0 2] x2 + [0]
                                 [0 0]      [0 0]      [2]
                   0() = [0]
                         [2]
                   div2(x1) = [0 0] x1 + [1]
                              [0 0]      [3]
                   false() = [2]
                             [2]
                   p(x1) = [1 0] x1 + [0]
                           [0 1]      [0]
                   s(x1) = [1 0] x1 + [0]
                           [0 1]      [2]
                   y() = [0]
                         [2]
                
                The strictly oriented rules are moved into the weak component.
                
                We consider the following Problem:
                
                  Strict Trs:
                    {  cond1(true(), x) -> cond2(even(x), x)
                     , cond2(true(), x) -> cond1(neq(x, 0()), div2(x))
                     , even(0()) -> true()
                     , even(s(0())) -> false()
                     , even(s(s(x))) -> even(x)
                     , div2(s(s(x))) -> s(div2(x))}
                  Weak Trs:
                    {  neq(0(), 0()) -> false()
                     , neq(s(x), 0()) -> true()
                     , neq(0(), s(x)) -> true()
                     , neq(s(x), s(y())) -> neq(x, y())
                     , p(s(x)) -> x
                     , div2(0()) -> 0()
                     , div2(s(0())) -> 0()
                     , cond2(false(), x) -> cond1(neq(x, 0()), p(x))
                     , p(0()) -> 0()}
                  StartTerms: basic terms
                  Strategy: innermost
                
                Certificate: YES(?,O(n^1))
                
                Proof:
                  The weightgap principle applies, where following rules are oriented strictly:
                  
                  TRS Component:
                    {  cond1(true(), x) -> cond2(even(x), x)
                     , div2(s(s(x))) -> s(div2(x))}
                  
                  Interpretation of nonconstant growth:
                  -------------------------------------
                    The following argument positions are usable:
                      Uargs(cond1) = {1, 2}, Uargs(cond2) = {1}, Uargs(even) = {},
                      Uargs(neq) = {}, Uargs(div2) = {}, Uargs(p) = {}, Uargs(s) = {1}
                    We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                    Interpretation Functions:
                     cond1(x1, x2) = [1 2] x1 + [1 1] x2 + [0]
                                     [0 0]      [0 0]      [0]
                     true() = [0]
                              [3]
                     cond2(x1, x2) = [1 3] x1 + [1 1] x2 + [3]
                                     [0 0]      [0 0]      [0]
                     even(x1) = [0 0] x1 + [0]
                                [0 0]      [0]
                     neq(x1, x2) = [0 0] x1 + [0 1] x2 + [0]
                                   [0 0]      [0 1]      [0]
                     0() = [0]
                           [3]
                     div2(x1) = [1 0] x1 + [2]
                                [0 1]      [2]
                     false() = [2]
                               [2]
                     p(x1) = [1 0] x1 + [0]
                             [0 1]      [0]
                     s(x1) = [1 0] x1 + [2]
                             [0 1]      [3]
                     y() = [0]
                           [0]
                  
                  The strictly oriented rules are moved into the weak component.
                  
                  We consider the following Problem:
                  
                    Strict Trs:
                      {  cond2(true(), x) -> cond1(neq(x, 0()), div2(x))
                       , even(0()) -> true()
                       , even(s(0())) -> false()
                       , even(s(s(x))) -> even(x)}
                    Weak Trs:
                      {  cond1(true(), x) -> cond2(even(x), x)
                       , div2(s(s(x))) -> s(div2(x))
                       , neq(0(), 0()) -> false()
                       , neq(s(x), 0()) -> true()
                       , neq(0(), s(x)) -> true()
                       , neq(s(x), s(y())) -> neq(x, y())
                       , p(s(x)) -> x
                       , div2(0()) -> 0()
                       , div2(s(0())) -> 0()
                       , cond2(false(), x) -> cond1(neq(x, 0()), p(x))
                       , p(0()) -> 0()}
                    StartTerms: basic terms
                    Strategy: innermost
                  
                  Certificate: YES(?,O(n^1))
                  
                  Proof:
                    The weightgap principle applies, where following rules are oriented strictly:
                    
                    TRS Component: {cond2(true(), x) -> cond1(neq(x, 0()), div2(x))}
                    
                    Interpretation of nonconstant growth:
                    -------------------------------------
                      The following argument positions are usable:
                        Uargs(cond1) = {1, 2}, Uargs(cond2) = {1}, Uargs(even) = {},
                        Uargs(neq) = {}, Uargs(div2) = {}, Uargs(p) = {}, Uargs(s) = {1}
                      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                      Interpretation Functions:
                       cond1(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                                       [0 0]      [0 0]      [1]
                       true() = [1]
                                [0]
                       cond2(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                                       [0 1]      [0 0]      [1]
                       even(x1) = [0 0] x1 + [0]
                                  [0 0]      [0]
                       neq(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                                     [0 0]      [0 0]      [1]
                       0() = [1]
                             [0]
                       div2(x1) = [1 0] x1 + [0]
                                  [0 0]      [1]
                       false() = [0]
                                 [0]
                       p(x1) = [1 0] x1 + [0]
                               [0 1]      [1]
                       s(x1) = [1 0] x1 + [2]
                               [0 1]      [0]
                       y() = [0]
                             [0]
                    
                    The strictly oriented rules are moved into the weak component.
                    
                    We consider the following Problem:
                    
                      Strict Trs:
                        {  even(0()) -> true()
                         , even(s(0())) -> false()
                         , even(s(s(x))) -> even(x)}
                      Weak Trs:
                        {  cond2(true(), x) -> cond1(neq(x, 0()), div2(x))
                         , cond1(true(), x) -> cond2(even(x), x)
                         , div2(s(s(x))) -> s(div2(x))
                         , neq(0(), 0()) -> false()
                         , neq(s(x), 0()) -> true()
                         , neq(0(), s(x)) -> true()
                         , neq(s(x), s(y())) -> neq(x, y())
                         , p(s(x)) -> x
                         , div2(0()) -> 0()
                         , div2(s(0())) -> 0()
                         , cond2(false(), x) -> cond1(neq(x, 0()), p(x))
                         , p(0()) -> 0()}
                      StartTerms: basic terms
                      Strategy: innermost
                    
                    Certificate: YES(?,O(n^1))
                    
                    Proof:
                      We consider the following Problem:
                      
                        Strict Trs:
                          {  even(0()) -> true()
                           , even(s(0())) -> false()
                           , even(s(s(x))) -> even(x)}
                        Weak Trs:
                          {  cond2(true(), x) -> cond1(neq(x, 0()), div2(x))
                           , cond1(true(), x) -> cond2(even(x), x)
                           , div2(s(s(x))) -> s(div2(x))
                           , neq(0(), 0()) -> false()
                           , neq(s(x), 0()) -> true()
                           , neq(0(), s(x)) -> true()
                           , neq(s(x), s(y())) -> neq(x, y())
                           , p(s(x)) -> x
                           , div2(0()) -> 0()
                           , div2(s(0())) -> 0()
                           , cond2(false(), x) -> cond1(neq(x, 0()), p(x))
                           , p(0()) -> 0()}
                        StartTerms: basic terms
                        Strategy: innermost
                      
                      Certificate: YES(?,O(n^1))
                      
                      Proof:
                        We have computed the following dependency pairs
                        
                          Strict DPs:
                            {  even^#(0()) -> c_1()
                             , even^#(s(0())) -> c_2()
                             , even^#(s(s(x))) -> even^#(x)}
                          Weak DPs:
                            {  cond2^#(true(), x) -> cond1^#(neq(x, 0()), div2(x))
                             , cond1^#(true(), x) -> cond2^#(even(x), x)
                             , div2^#(s(s(x))) -> div2^#(x)
                             , neq^#(0(), 0()) -> c_7()
                             , neq^#(s(x), 0()) -> c_8()
                             , neq^#(0(), s(x)) -> c_9()
                             , neq^#(s(x), s(y())) -> neq^#(x, y())
                             , p^#(s(x)) -> c_11()
                             , div2^#(0()) -> c_12()
                             , div2^#(s(0())) -> c_13()
                             , cond2^#(false(), x) -> cond1^#(neq(x, 0()), p(x))
                             , p^#(0()) -> c_15()}
                        
                        We consider the following Problem:
                        
                          Strict DPs:
                            {  even^#(0()) -> c_1()
                             , even^#(s(0())) -> c_2()
                             , even^#(s(s(x))) -> even^#(x)}
                          Strict Trs:
                            {  even(0()) -> true()
                             , even(s(0())) -> false()
                             , even(s(s(x))) -> even(x)}
                          Weak DPs:
                            {  cond2^#(true(), x) -> cond1^#(neq(x, 0()), div2(x))
                             , cond1^#(true(), x) -> cond2^#(even(x), x)
                             , div2^#(s(s(x))) -> div2^#(x)
                             , neq^#(0(), 0()) -> c_7()
                             , neq^#(s(x), 0()) -> c_8()
                             , neq^#(0(), s(x)) -> c_9()
                             , neq^#(s(x), s(y())) -> neq^#(x, y())
                             , p^#(s(x)) -> c_11()
                             , div2^#(0()) -> c_12()
                             , div2^#(s(0())) -> c_13()
                             , cond2^#(false(), x) -> cond1^#(neq(x, 0()), p(x))
                             , p^#(0()) -> c_15()}
                          Weak Trs:
                            {  cond2(true(), x) -> cond1(neq(x, 0()), div2(x))
                             , cond1(true(), x) -> cond2(even(x), x)
                             , div2(s(s(x))) -> s(div2(x))
                             , neq(0(), 0()) -> false()
                             , neq(s(x), 0()) -> true()
                             , neq(0(), s(x)) -> true()
                             , neq(s(x), s(y())) -> neq(x, y())
                             , p(s(x)) -> x
                             , div2(0()) -> 0()
                             , div2(s(0())) -> 0()
                             , cond2(false(), x) -> cond1(neq(x, 0()), p(x))
                             , p(0()) -> 0()}
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(?,O(n^1))
                        
                        Proof:
                          We replace strict/weak-rules by the corresponding usable rules:
                          
                            Strict Usable Rules:
                              {  even(0()) -> true()
                               , even(s(0())) -> false()
                               , even(s(s(x))) -> even(x)}
                            Weak Usable Rules:
                              {  div2(s(s(x))) -> s(div2(x))
                               , neq(0(), 0()) -> false()
                               , neq(s(x), 0()) -> true()
                               , neq(0(), s(x)) -> true()
                               , neq(s(x), s(y())) -> neq(x, y())
                               , p(s(x)) -> x
                               , div2(0()) -> 0()
                               , div2(s(0())) -> 0()
                               , p(0()) -> 0()}
                          
                          We consider the following Problem:
                          
                            Strict DPs:
                              {  even^#(0()) -> c_1()
                               , even^#(s(0())) -> c_2()
                               , even^#(s(s(x))) -> even^#(x)}
                            Strict Trs:
                              {  even(0()) -> true()
                               , even(s(0())) -> false()
                               , even(s(s(x))) -> even(x)}
                            Weak DPs:
                              {  cond2^#(true(), x) -> cond1^#(neq(x, 0()), div2(x))
                               , cond1^#(true(), x) -> cond2^#(even(x), x)
                               , div2^#(s(s(x))) -> div2^#(x)
                               , neq^#(0(), 0()) -> c_7()
                               , neq^#(s(x), 0()) -> c_8()
                               , neq^#(0(), s(x)) -> c_9()
                               , neq^#(s(x), s(y())) -> neq^#(x, y())
                               , p^#(s(x)) -> c_11()
                               , div2^#(0()) -> c_12()
                               , div2^#(s(0())) -> c_13()
                               , cond2^#(false(), x) -> cond1^#(neq(x, 0()), p(x))
                               , p^#(0()) -> c_15()}
                            Weak Trs:
                              {  div2(s(s(x))) -> s(div2(x))
                               , neq(0(), 0()) -> false()
                               , neq(s(x), 0()) -> true()
                               , neq(0(), s(x)) -> true()
                               , neq(s(x), s(y())) -> neq(x, y())
                               , p(s(x)) -> x
                               , div2(0()) -> 0()
                               , div2(s(0())) -> 0()
                               , p(0()) -> 0()}
                            StartTerms: basic terms
                            Strategy: innermost
                          
                          Certificate: YES(?,O(n^1))
                          
                          Proof:
                            We consider the following Problem:
                            
                              Strict DPs:
                                {  even^#(0()) -> c_1()
                                 , even^#(s(0())) -> c_2()
                                 , even^#(s(s(x))) -> even^#(x)}
                              Strict Trs:
                                {  even(0()) -> true()
                                 , even(s(0())) -> false()
                                 , even(s(s(x))) -> even(x)}
                              Weak DPs:
                                {  cond2^#(true(), x) -> cond1^#(neq(x, 0()), div2(x))
                                 , cond1^#(true(), x) -> cond2^#(even(x), x)
                                 , div2^#(s(s(x))) -> div2^#(x)
                                 , neq^#(0(), 0()) -> c_7()
                                 , neq^#(s(x), 0()) -> c_8()
                                 , neq^#(0(), s(x)) -> c_9()
                                 , neq^#(s(x), s(y())) -> neq^#(x, y())
                                 , p^#(s(x)) -> c_11()
                                 , div2^#(0()) -> c_12()
                                 , div2^#(s(0())) -> c_13()
                                 , cond2^#(false(), x) -> cond1^#(neq(x, 0()), p(x))
                                 , p^#(0()) -> c_15()}
                              Weak Trs:
                                {  div2(s(s(x))) -> s(div2(x))
                                 , neq(0(), 0()) -> false()
                                 , neq(s(x), 0()) -> true()
                                 , neq(0(), s(x)) -> true()
                                 , neq(s(x), s(y())) -> neq(x, y())
                                 , p(s(x)) -> x
                                 , div2(0()) -> 0()
                                 , div2(s(0())) -> 0()
                                 , p(0()) -> 0()}
                              StartTerms: basic terms
                              Strategy: innermost
                            
                            Certificate: YES(?,O(n^1))
                            
                            Proof:
                              We use following congruence DG for path analysis
                              
                              ->11:{3}                                                    [   YES(?,O(n^1))    ]
                                 |
                                 |->13:{1}                                                [   YES(?,O(n^1))    ]
                                 |
                                 `->12:{2}                                                [   YES(?,O(n^1))    ]
                              
                              ->10:{4,5,14}                                               [   YES(O(1),O(1))   ]
                              
                              ->7:{6}                                                     [      subsumed      ]
                                 |
                                 |->8:{12}                                                [   YES(O(1),O(1))   ]
                                 |
                                 `->9:{13}                                                [   YES(O(1),O(1))   ]
                              
                              ->6:{7}                                                     [   YES(O(1),O(1))   ]
                              
                              ->5:{8}                                                     [   YES(O(1),O(1))   ]
                              
                              ->4:{9}                                                     [   YES(O(1),O(1))   ]
                              
                              ->3:{10}                                                    [   YES(O(1),O(1))   ]
                              
                              ->2:{11}                                                    [   YES(O(1),O(1))   ]
                              
                              ->1:{15}                                                    [   YES(O(1),O(1))   ]
                              
                              
                              Here dependency-pairs are as follows:
                              
                              Strict DPs:
                                {  1: even^#(0()) -> c_1()
                                 , 2: even^#(s(0())) -> c_2()
                                 , 3: even^#(s(s(x))) -> even^#(x)}
                              WeakDPs DPs:
                                {  4: cond2^#(true(), x) -> cond1^#(neq(x, 0()), div2(x))
                                 , 5: cond1^#(true(), x) -> cond2^#(even(x), x)
                                 , 6: div2^#(s(s(x))) -> div2^#(x)
                                 , 7: neq^#(0(), 0()) -> c_7()
                                 , 8: neq^#(s(x), 0()) -> c_8()
                                 , 9: neq^#(0(), s(x)) -> c_9()
                                 , 10: neq^#(s(x), s(y())) -> neq^#(x, y())
                                 , 11: p^#(s(x)) -> c_11()
                                 , 12: div2^#(0()) -> c_12()
                                 , 13: div2^#(s(0())) -> c_13()
                                 , 14: cond2^#(false(), x) -> cond1^#(neq(x, 0()), p(x))
                                 , 15: p^#(0()) -> c_15()}
                              
                              * Path 11:{3}: YES(?,O(n^1))
                                --------------------------
                                
                                We consider the following Problem:
                                
                                  Strict DPs: {even^#(s(s(x))) -> even^#(x)}
                                  Strict Trs:
                                    {  even(0()) -> true()
                                     , even(s(0())) -> false()
                                     , even(s(s(x))) -> even(x)}
                                  Weak Trs:
                                    {  div2(s(s(x))) -> s(div2(x))
                                     , neq(0(), 0()) -> false()
                                     , neq(s(x), 0()) -> true()
                                     , neq(0(), s(x)) -> true()
                                     , neq(s(x), s(y())) -> neq(x, y())
                                     , p(s(x)) -> x
                                     , div2(0()) -> 0()
                                     , div2(s(0())) -> 0()
                                     , p(0()) -> 0()}
                                  StartTerms: basic terms
                                  Strategy: innermost
                                
                                Certificate: YES(?,O(n^1))
                                
                                Proof:
                                  We consider the following Problem:
                                  
                                    Strict DPs: {even^#(s(s(x))) -> even^#(x)}
                                    Strict Trs:
                                      {  even(0()) -> true()
                                       , even(s(0())) -> false()
                                       , even(s(s(x))) -> even(x)}
                                    Weak Trs:
                                      {  div2(s(s(x))) -> s(div2(x))
                                       , neq(0(), 0()) -> false()
                                       , neq(s(x), 0()) -> true()
                                       , neq(0(), s(x)) -> true()
                                       , neq(s(x), s(y())) -> neq(x, y())
                                       , p(s(x)) -> x
                                       , div2(0()) -> 0()
                                       , div2(s(0())) -> 0()
                                       , p(0()) -> 0()}
                                    StartTerms: basic terms
                                    Strategy: innermost
                                  
                                  Certificate: YES(?,O(n^1))
                                  
                                  Proof:
                                    We consider the following Problem:
                                    
                                      Strict DPs: {even^#(s(s(x))) -> even^#(x)}
                                      Strict Trs:
                                        {  even(0()) -> true()
                                         , even(s(0())) -> false()
                                         , even(s(s(x))) -> even(x)}
                                      Weak Trs:
                                        {  div2(s(s(x))) -> s(div2(x))
                                         , neq(0(), 0()) -> false()
                                         , neq(s(x), 0()) -> true()
                                         , neq(0(), s(x)) -> true()
                                         , neq(s(x), s(y())) -> neq(x, y())
                                         , p(s(x)) -> x
                                         , div2(0()) -> 0()
                                         , div2(s(0())) -> 0()
                                         , p(0()) -> 0()}
                                      StartTerms: basic terms
                                      Strategy: innermost
                                    
                                    Certificate: YES(?,O(n^1))
                                    
                                    Proof:
                                      No rule is usable.
                                      
                                      We consider the following Problem:
                                      
                                        Strict DPs: {even^#(s(s(x))) -> even^#(x)}
                                        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
                                         , even^#_0(2) -> 1
                                         , even^#_1(2) -> 1}
                              
                              * Path 11:{3}->13:{1}: YES(?,O(n^1))
                                ----------------------------------
                                
                                We consider the following Problem:
                                
                                  Strict DPs: {even^#(0()) -> c_1()}
                                  Strict Trs:
                                    {  even(0()) -> true()
                                     , even(s(0())) -> false()
                                     , even(s(s(x))) -> even(x)}
                                  Weak DPs: {even^#(s(s(x))) -> even^#(x)}
                                  Weak Trs:
                                    {  div2(s(s(x))) -> s(div2(x))
                                     , neq(0(), 0()) -> false()
                                     , neq(s(x), 0()) -> true()
                                     , neq(0(), s(x)) -> true()
                                     , neq(s(x), s(y())) -> neq(x, y())
                                     , p(s(x)) -> x
                                     , div2(0()) -> 0()
                                     , div2(s(0())) -> 0()
                                     , p(0()) -> 0()}
                                  StartTerms: basic terms
                                  Strategy: innermost
                                
                                Certificate: YES(?,O(n^1))
                                
                                Proof:
                                  We consider the following Problem:
                                  
                                    Strict DPs: {even^#(0()) -> c_1()}
                                    Strict Trs:
                                      {  even(0()) -> true()
                                       , even(s(0())) -> false()
                                       , even(s(s(x))) -> even(x)}
                                    Weak DPs: {even^#(s(s(x))) -> even^#(x)}
                                    Weak Trs:
                                      {  div2(s(s(x))) -> s(div2(x))
                                       , neq(0(), 0()) -> false()
                                       , neq(s(x), 0()) -> true()
                                       , neq(0(), s(x)) -> true()
                                       , neq(s(x), s(y())) -> neq(x, y())
                                       , p(s(x)) -> x
                                       , div2(0()) -> 0()
                                       , div2(s(0())) -> 0()
                                       , p(0()) -> 0()}
                                    StartTerms: basic terms
                                    Strategy: innermost
                                  
                                  Certificate: YES(?,O(n^1))
                                  
                                  Proof:
                                    We consider the following Problem:
                                    
                                      Strict DPs: {even^#(0()) -> c_1()}
                                      Strict Trs:
                                        {  even(0()) -> true()
                                         , even(s(0())) -> false()
                                         , even(s(s(x))) -> even(x)}
                                      Weak DPs: {even^#(s(s(x))) -> even^#(x)}
                                      Weak Trs:
                                        {  div2(s(s(x))) -> s(div2(x))
                                         , neq(0(), 0()) -> false()
                                         , neq(s(x), 0()) -> true()
                                         , neq(0(), s(x)) -> true()
                                         , neq(s(x), s(y())) -> neq(x, y())
                                         , p(s(x)) -> x
                                         , div2(0()) -> 0()
                                         , div2(s(0())) -> 0()
                                         , p(0()) -> 0()}
                                      StartTerms: basic terms
                                      Strategy: innermost
                                    
                                    Certificate: YES(?,O(n^1))
                                    
                                    Proof:
                                      No rule is usable.
                                      
                                      We consider the following Problem:
                                      
                                        Strict DPs: {even^#(0()) -> c_1()}
                                        Weak DPs: {even^#(s(s(x))) -> even^#(x)}
                                        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:
                                        {  0_0() -> 2
                                         , s_0(2) -> 2
                                         , even^#_0(2) -> 1
                                         , c_1_1() -> 1}
                              
                              * Path 11:{3}->12:{2}: YES(?,O(n^1))
                                ----------------------------------
                                
                                We consider the following Problem:
                                
                                  Strict DPs: {even^#(s(0())) -> c_2()}
                                  Strict Trs:
                                    {  even(0()) -> true()
                                     , even(s(0())) -> false()
                                     , even(s(s(x))) -> even(x)}
                                  Weak DPs: {even^#(s(s(x))) -> even^#(x)}
                                  Weak Trs:
                                    {  div2(s(s(x))) -> s(div2(x))
                                     , neq(0(), 0()) -> false()
                                     , neq(s(x), 0()) -> true()
                                     , neq(0(), s(x)) -> true()
                                     , neq(s(x), s(y())) -> neq(x, y())
                                     , p(s(x)) -> x
                                     , div2(0()) -> 0()
                                     , div2(s(0())) -> 0()
                                     , p(0()) -> 0()}
                                  StartTerms: basic terms
                                  Strategy: innermost
                                
                                Certificate: YES(?,O(n^1))
                                
                                Proof:
                                  We consider the following Problem:
                                  
                                    Strict DPs: {even^#(s(0())) -> c_2()}
                                    Strict Trs:
                                      {  even(0()) -> true()
                                       , even(s(0())) -> false()
                                       , even(s(s(x))) -> even(x)}
                                    Weak DPs: {even^#(s(s(x))) -> even^#(x)}
                                    Weak Trs:
                                      {  div2(s(s(x))) -> s(div2(x))
                                       , neq(0(), 0()) -> false()
                                       , neq(s(x), 0()) -> true()
                                       , neq(0(), s(x)) -> true()
                                       , neq(s(x), s(y())) -> neq(x, y())
                                       , p(s(x)) -> x
                                       , div2(0()) -> 0()
                                       , div2(s(0())) -> 0()
                                       , p(0()) -> 0()}
                                    StartTerms: basic terms
                                    Strategy: innermost
                                  
                                  Certificate: YES(?,O(n^1))
                                  
                                  Proof:
                                    We consider the following Problem:
                                    
                                      Strict DPs: {even^#(s(0())) -> c_2()}
                                      Strict Trs:
                                        {  even(0()) -> true()
                                         , even(s(0())) -> false()
                                         , even(s(s(x))) -> even(x)}
                                      Weak DPs: {even^#(s(s(x))) -> even^#(x)}
                                      Weak Trs:
                                        {  div2(s(s(x))) -> s(div2(x))
                                         , neq(0(), 0()) -> false()
                                         , neq(s(x), 0()) -> true()
                                         , neq(0(), s(x)) -> true()
                                         , neq(s(x), s(y())) -> neq(x, y())
                                         , p(s(x)) -> x
                                         , div2(0()) -> 0()
                                         , div2(s(0())) -> 0()
                                         , p(0()) -> 0()}
                                      StartTerms: basic terms
                                      Strategy: innermost
                                    
                                    Certificate: YES(?,O(n^1))
                                    
                                    Proof:
                                      No rule is usable.
                                      
                                      We consider the following Problem:
                                      
                                        Strict DPs: {even^#(s(0())) -> c_2()}
                                        Weak DPs: {even^#(s(s(x))) -> even^#(x)}
                                        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:
                                        {  0_0() -> 2
                                         , s_0(2) -> 2
                                         , even^#_0(2) -> 1
                                         , c_2_1() -> 1}
                              
                              * Path 10:{4,5,14}: YES(O(1),O(1))
                                --------------------------------
                                
                                We consider the following Problem:
                                
                                  Strict Trs:
                                    {  even(0()) -> true()
                                     , even(s(0())) -> false()
                                     , even(s(s(x))) -> even(x)}
                                  Weak Trs:
                                    {  div2(s(s(x))) -> s(div2(x))
                                     , neq(0(), 0()) -> false()
                                     , neq(s(x), 0()) -> true()
                                     , neq(0(), s(x)) -> true()
                                     , neq(s(x), s(y())) -> neq(x, y())
                                     , p(s(x)) -> x
                                     , div2(0()) -> 0()
                                     , div2(s(0())) -> 0()
                                     , p(0()) -> 0()}
                                  StartTerms: basic terms
                                  Strategy: innermost
                                
                                Certificate: YES(O(1),O(1))
                                
                                Proof:
                                  We consider the following Problem:
                                  
                                    Strict Trs:
                                      {  even(0()) -> true()
                                       , even(s(0())) -> false()
                                       , even(s(s(x))) -> even(x)}
                                    Weak Trs:
                                      {  div2(s(s(x))) -> s(div2(x))
                                       , neq(0(), 0()) -> false()
                                       , neq(s(x), 0()) -> true()
                                       , neq(0(), s(x)) -> true()
                                       , neq(s(x), s(y())) -> neq(x, y())
                                       , p(s(x)) -> x
                                       , div2(0()) -> 0()
                                       , div2(s(0())) -> 0()
                                       , p(0()) -> 0()}
                                    StartTerms: basic terms
                                    Strategy: innermost
                                  
                                  Certificate: YES(O(1),O(1))
                                  
                                  Proof:
                                    We consider the following Problem:
                                    
                                      Strict Trs:
                                        {  even(0()) -> true()
                                         , even(s(0())) -> false()
                                         , even(s(s(x))) -> even(x)}
                                      Weak Trs:
                                        {  div2(s(s(x))) -> s(div2(x))
                                         , neq(0(), 0()) -> false()
                                         , neq(s(x), 0()) -> true()
                                         , neq(0(), s(x)) -> true()
                                         , neq(s(x), s(y())) -> neq(x, y())
                                         , p(s(x)) -> x
                                         , div2(0()) -> 0()
                                         , div2(s(0())) -> 0()
                                         , p(0()) -> 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
                              
                              * Path 7:{6}: subsumed
                                --------------------
                                
                                This path is subsumed by the proof of paths 7:{6}->9:{13},
                                                                            7:{6}->8:{12}.
                              
                              * Path 7:{6}->8:{12}: YES(O(1),O(1))
                                ----------------------------------
                                
                                We consider the following Problem:
                                
                                  Strict Trs:
                                    {  even(0()) -> true()
                                     , even(s(0())) -> false()
                                     , even(s(s(x))) -> even(x)}
                                  Weak DPs: {div2^#(s(s(x))) -> div2^#(x)}
                                  Weak Trs:
                                    {  div2(s(s(x))) -> s(div2(x))
                                     , neq(0(), 0()) -> false()
                                     , neq(s(x), 0()) -> true()
                                     , neq(0(), s(x)) -> true()
                                     , neq(s(x), s(y())) -> neq(x, y())
                                     , p(s(x)) -> x
                                     , div2(0()) -> 0()
                                     , div2(s(0())) -> 0()
                                     , p(0()) -> 0()}
                                  StartTerms: basic terms
                                  Strategy: innermost
                                
                                Certificate: YES(O(1),O(1))
                                
                                Proof:
                                  We consider the following Problem:
                                  
                                    Strict Trs:
                                      {  even(0()) -> true()
                                       , even(s(0())) -> false()
                                       , even(s(s(x))) -> even(x)}
                                    Weak DPs: {div2^#(s(s(x))) -> div2^#(x)}
                                    Weak Trs:
                                      {  div2(s(s(x))) -> s(div2(x))
                                       , neq(0(), 0()) -> false()
                                       , neq(s(x), 0()) -> true()
                                       , neq(0(), s(x)) -> true()
                                       , neq(s(x), s(y())) -> neq(x, y())
                                       , p(s(x)) -> x
                                       , div2(0()) -> 0()
                                       , div2(s(0())) -> 0()
                                       , p(0()) -> 0()}
                                    StartTerms: basic terms
                                    Strategy: innermost
                                  
                                  Certificate: YES(O(1),O(1))
                                  
                                  Proof:
                                    We consider the following Problem:
                                    
                                      Strict Trs:
                                        {  even(0()) -> true()
                                         , even(s(0())) -> false()
                                         , even(s(s(x))) -> even(x)}
                                      Weak DPs: {div2^#(s(s(x))) -> div2^#(x)}
                                      Weak Trs:
                                        {  div2(s(s(x))) -> s(div2(x))
                                         , neq(0(), 0()) -> false()
                                         , neq(s(x), 0()) -> true()
                                         , neq(0(), s(x)) -> true()
                                         , neq(s(x), s(y())) -> neq(x, y())
                                         , p(s(x)) -> x
                                         , div2(0()) -> 0()
                                         , div2(s(0())) -> 0()
                                         , p(0()) -> 0()}
                                      StartTerms: basic terms
                                      Strategy: innermost
                                    
                                    Certificate: YES(O(1),O(1))
                                    
                                    Proof:
                                      No rule is usable.
                                      
                                      We consider the following Problem:
                                      
                                        Weak DPs: {div2^#(s(s(x))) -> div2^#(x)}
                                        StartTerms: basic terms
                                        Strategy: innermost
                                      
                                      Certificate: YES(O(1),O(1))
                                      
                                      Proof:
                                        Empty rules are trivially bounded
                              
                              * Path 7:{6}->9:{13}: YES(O(1),O(1))
                                ----------------------------------
                                
                                We consider the following Problem:
                                
                                  Strict Trs:
                                    {  even(0()) -> true()
                                     , even(s(0())) -> false()
                                     , even(s(s(x))) -> even(x)}
                                  Weak DPs: {div2^#(s(s(x))) -> div2^#(x)}
                                  Weak Trs:
                                    {  div2(s(s(x))) -> s(div2(x))
                                     , neq(0(), 0()) -> false()
                                     , neq(s(x), 0()) -> true()
                                     , neq(0(), s(x)) -> true()
                                     , neq(s(x), s(y())) -> neq(x, y())
                                     , p(s(x)) -> x
                                     , div2(0()) -> 0()
                                     , div2(s(0())) -> 0()
                                     , p(0()) -> 0()}
                                  StartTerms: basic terms
                                  Strategy: innermost
                                
                                Certificate: YES(O(1),O(1))
                                
                                Proof:
                                  We consider the following Problem:
                                  
                                    Strict Trs:
                                      {  even(0()) -> true()
                                       , even(s(0())) -> false()
                                       , even(s(s(x))) -> even(x)}
                                    Weak DPs: {div2^#(s(s(x))) -> div2^#(x)}
                                    Weak Trs:
                                      {  div2(s(s(x))) -> s(div2(x))
                                       , neq(0(), 0()) -> false()
                                       , neq(s(x), 0()) -> true()
                                       , neq(0(), s(x)) -> true()
                                       , neq(s(x), s(y())) -> neq(x, y())
                                       , p(s(x)) -> x
                                       , div2(0()) -> 0()
                                       , div2(s(0())) -> 0()
                                       , p(0()) -> 0()}
                                    StartTerms: basic terms
                                    Strategy: innermost
                                  
                                  Certificate: YES(O(1),O(1))
                                  
                                  Proof:
                                    We consider the following Problem:
                                    
                                      Strict Trs:
                                        {  even(0()) -> true()
                                         , even(s(0())) -> false()
                                         , even(s(s(x))) -> even(x)}
                                      Weak DPs: {div2^#(s(s(x))) -> div2^#(x)}
                                      Weak Trs:
                                        {  div2(s(s(x))) -> s(div2(x))
                                         , neq(0(), 0()) -> false()
                                         , neq(s(x), 0()) -> true()
                                         , neq(0(), s(x)) -> true()
                                         , neq(s(x), s(y())) -> neq(x, y())
                                         , p(s(x)) -> x
                                         , div2(0()) -> 0()
                                         , div2(s(0())) -> 0()
                                         , p(0()) -> 0()}
                                      StartTerms: basic terms
                                      Strategy: innermost
                                    
                                    Certificate: YES(O(1),O(1))
                                    
                                    Proof:
                                      No rule is usable.
                                      
                                      We consider the following Problem:
                                      
                                        Weak DPs: {div2^#(s(s(x))) -> div2^#(x)}
                                        StartTerms: basic terms
                                        Strategy: innermost
                                      
                                      Certificate: YES(O(1),O(1))
                                      
                                      Proof:
                                        Empty rules are trivially bounded
                              
                              * Path 6:{7}: YES(O(1),O(1))
                                --------------------------
                                
                                We consider the following Problem:
                                
                                  Strict Trs:
                                    {  even(0()) -> true()
                                     , even(s(0())) -> false()
                                     , even(s(s(x))) -> even(x)}
                                  Weak Trs:
                                    {  div2(s(s(x))) -> s(div2(x))
                                     , neq(0(), 0()) -> false()
                                     , neq(s(x), 0()) -> true()
                                     , neq(0(), s(x)) -> true()
                                     , neq(s(x), s(y())) -> neq(x, y())
                                     , p(s(x)) -> x
                                     , div2(0()) -> 0()
                                     , div2(s(0())) -> 0()
                                     , p(0()) -> 0()}
                                  StartTerms: basic terms
                                  Strategy: innermost
                                
                                Certificate: YES(O(1),O(1))
                                
                                Proof:
                                  We consider the following Problem:
                                  
                                    Strict Trs:
                                      {  even(0()) -> true()
                                       , even(s(0())) -> false()
                                       , even(s(s(x))) -> even(x)}
                                    Weak Trs:
                                      {  div2(s(s(x))) -> s(div2(x))
                                       , neq(0(), 0()) -> false()
                                       , neq(s(x), 0()) -> true()
                                       , neq(0(), s(x)) -> true()
                                       , neq(s(x), s(y())) -> neq(x, y())
                                       , p(s(x)) -> x
                                       , div2(0()) -> 0()
                                       , div2(s(0())) -> 0()
                                       , p(0()) -> 0()}
                                    StartTerms: basic terms
                                    Strategy: innermost
                                  
                                  Certificate: YES(O(1),O(1))
                                  
                                  Proof:
                                    We consider the following Problem:
                                    
                                      Strict Trs:
                                        {  even(0()) -> true()
                                         , even(s(0())) -> false()
                                         , even(s(s(x))) -> even(x)}
                                      Weak Trs:
                                        {  div2(s(s(x))) -> s(div2(x))
                                         , neq(0(), 0()) -> false()
                                         , neq(s(x), 0()) -> true()
                                         , neq(0(), s(x)) -> true()
                                         , neq(s(x), s(y())) -> neq(x, y())
                                         , p(s(x)) -> x
                                         , div2(0()) -> 0()
                                         , div2(s(0())) -> 0()
                                         , p(0()) -> 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
                              
                              * Path 5:{8}: YES(O(1),O(1))
                                --------------------------
                                
                                We consider the following Problem:
                                
                                  Strict Trs:
                                    {  even(0()) -> true()
                                     , even(s(0())) -> false()
                                     , even(s(s(x))) -> even(x)}
                                  Weak Trs:
                                    {  div2(s(s(x))) -> s(div2(x))
                                     , neq(0(), 0()) -> false()
                                     , neq(s(x), 0()) -> true()
                                     , neq(0(), s(x)) -> true()
                                     , neq(s(x), s(y())) -> neq(x, y())
                                     , p(s(x)) -> x
                                     , div2(0()) -> 0()
                                     , div2(s(0())) -> 0()
                                     , p(0()) -> 0()}
                                  StartTerms: basic terms
                                  Strategy: innermost
                                
                                Certificate: YES(O(1),O(1))
                                
                                Proof:
                                  We consider the following Problem:
                                  
                                    Strict Trs:
                                      {  even(0()) -> true()
                                       , even(s(0())) -> false()
                                       , even(s(s(x))) -> even(x)}
                                    Weak Trs:
                                      {  div2(s(s(x))) -> s(div2(x))
                                       , neq(0(), 0()) -> false()
                                       , neq(s(x), 0()) -> true()
                                       , neq(0(), s(x)) -> true()
                                       , neq(s(x), s(y())) -> neq(x, y())
                                       , p(s(x)) -> x
                                       , div2(0()) -> 0()
                                       , div2(s(0())) -> 0()
                                       , p(0()) -> 0()}
                                    StartTerms: basic terms
                                    Strategy: innermost
                                  
                                  Certificate: YES(O(1),O(1))
                                  
                                  Proof:
                                    We consider the following Problem:
                                    
                                      Strict Trs:
                                        {  even(0()) -> true()
                                         , even(s(0())) -> false()
                                         , even(s(s(x))) -> even(x)}
                                      Weak Trs:
                                        {  div2(s(s(x))) -> s(div2(x))
                                         , neq(0(), 0()) -> false()
                                         , neq(s(x), 0()) -> true()
                                         , neq(0(), s(x)) -> true()
                                         , neq(s(x), s(y())) -> neq(x, y())
                                         , p(s(x)) -> x
                                         , div2(0()) -> 0()
                                         , div2(s(0())) -> 0()
                                         , p(0()) -> 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
                              
                              * Path 4:{9}: YES(O(1),O(1))
                                --------------------------
                                
                                We consider the following Problem:
                                
                                  Strict Trs:
                                    {  even(0()) -> true()
                                     , even(s(0())) -> false()
                                     , even(s(s(x))) -> even(x)}
                                  Weak Trs:
                                    {  div2(s(s(x))) -> s(div2(x))
                                     , neq(0(), 0()) -> false()
                                     , neq(s(x), 0()) -> true()
                                     , neq(0(), s(x)) -> true()
                                     , neq(s(x), s(y())) -> neq(x, y())
                                     , p(s(x)) -> x
                                     , div2(0()) -> 0()
                                     , div2(s(0())) -> 0()
                                     , p(0()) -> 0()}
                                  StartTerms: basic terms
                                  Strategy: innermost
                                
                                Certificate: YES(O(1),O(1))
                                
                                Proof:
                                  We consider the following Problem:
                                  
                                    Strict Trs:
                                      {  even(0()) -> true()
                                       , even(s(0())) -> false()
                                       , even(s(s(x))) -> even(x)}
                                    Weak Trs:
                                      {  div2(s(s(x))) -> s(div2(x))
                                       , neq(0(), 0()) -> false()
                                       , neq(s(x), 0()) -> true()
                                       , neq(0(), s(x)) -> true()
                                       , neq(s(x), s(y())) -> neq(x, y())
                                       , p(s(x)) -> x
                                       , div2(0()) -> 0()
                                       , div2(s(0())) -> 0()
                                       , p(0()) -> 0()}
                                    StartTerms: basic terms
                                    Strategy: innermost
                                  
                                  Certificate: YES(O(1),O(1))
                                  
                                  Proof:
                                    We consider the following Problem:
                                    
                                      Strict Trs:
                                        {  even(0()) -> true()
                                         , even(s(0())) -> false()
                                         , even(s(s(x))) -> even(x)}
                                      Weak Trs:
                                        {  div2(s(s(x))) -> s(div2(x))
                                         , neq(0(), 0()) -> false()
                                         , neq(s(x), 0()) -> true()
                                         , neq(0(), s(x)) -> true()
                                         , neq(s(x), s(y())) -> neq(x, y())
                                         , p(s(x)) -> x
                                         , div2(0()) -> 0()
                                         , div2(s(0())) -> 0()
                                         , p(0()) -> 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
                              
                              * Path 3:{10}: YES(O(1),O(1))
                                ---------------------------
                                
                                We consider the following Problem:
                                
                                  Strict Trs:
                                    {  even(0()) -> true()
                                     , even(s(0())) -> false()
                                     , even(s(s(x))) -> even(x)}
                                  Weak Trs:
                                    {  div2(s(s(x))) -> s(div2(x))
                                     , neq(0(), 0()) -> false()
                                     , neq(s(x), 0()) -> true()
                                     , neq(0(), s(x)) -> true()
                                     , neq(s(x), s(y())) -> neq(x, y())
                                     , p(s(x)) -> x
                                     , div2(0()) -> 0()
                                     , div2(s(0())) -> 0()
                                     , p(0()) -> 0()}
                                  StartTerms: basic terms
                                  Strategy: innermost
                                
                                Certificate: YES(O(1),O(1))
                                
                                Proof:
                                  We consider the following Problem:
                                  
                                    Strict Trs:
                                      {  even(0()) -> true()
                                       , even(s(0())) -> false()
                                       , even(s(s(x))) -> even(x)}
                                    Weak Trs:
                                      {  div2(s(s(x))) -> s(div2(x))
                                       , neq(0(), 0()) -> false()
                                       , neq(s(x), 0()) -> true()
                                       , neq(0(), s(x)) -> true()
                                       , neq(s(x), s(y())) -> neq(x, y())
                                       , p(s(x)) -> x
                                       , div2(0()) -> 0()
                                       , div2(s(0())) -> 0()
                                       , p(0()) -> 0()}
                                    StartTerms: basic terms
                                    Strategy: innermost
                                  
                                  Certificate: YES(O(1),O(1))
                                  
                                  Proof:
                                    We consider the following Problem:
                                    
                                      Strict Trs:
                                        {  even(0()) -> true()
                                         , even(s(0())) -> false()
                                         , even(s(s(x))) -> even(x)}
                                      Weak Trs:
                                        {  div2(s(s(x))) -> s(div2(x))
                                         , neq(0(), 0()) -> false()
                                         , neq(s(x), 0()) -> true()
                                         , neq(0(), s(x)) -> true()
                                         , neq(s(x), s(y())) -> neq(x, y())
                                         , p(s(x)) -> x
                                         , div2(0()) -> 0()
                                         , div2(s(0())) -> 0()
                                         , p(0()) -> 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
                              
                              * Path 2:{11}: YES(O(1),O(1))
                                ---------------------------
                                
                                We consider the following Problem:
                                
                                  Strict Trs:
                                    {  even(0()) -> true()
                                     , even(s(0())) -> false()
                                     , even(s(s(x))) -> even(x)}
                                  Weak Trs:
                                    {  div2(s(s(x))) -> s(div2(x))
                                     , neq(0(), 0()) -> false()
                                     , neq(s(x), 0()) -> true()
                                     , neq(0(), s(x)) -> true()
                                     , neq(s(x), s(y())) -> neq(x, y())
                                     , p(s(x)) -> x
                                     , div2(0()) -> 0()
                                     , div2(s(0())) -> 0()
                                     , p(0()) -> 0()}
                                  StartTerms: basic terms
                                  Strategy: innermost
                                
                                Certificate: YES(O(1),O(1))
                                
                                Proof:
                                  We consider the following Problem:
                                  
                                    Strict Trs:
                                      {  even(0()) -> true()
                                       , even(s(0())) -> false()
                                       , even(s(s(x))) -> even(x)}
                                    Weak Trs:
                                      {  div2(s(s(x))) -> s(div2(x))
                                       , neq(0(), 0()) -> false()
                                       , neq(s(x), 0()) -> true()
                                       , neq(0(), s(x)) -> true()
                                       , neq(s(x), s(y())) -> neq(x, y())
                                       , p(s(x)) -> x
                                       , div2(0()) -> 0()
                                       , div2(s(0())) -> 0()
                                       , p(0()) -> 0()}
                                    StartTerms: basic terms
                                    Strategy: innermost
                                  
                                  Certificate: YES(O(1),O(1))
                                  
                                  Proof:
                                    We consider the following Problem:
                                    
                                      Strict Trs:
                                        {  even(0()) -> true()
                                         , even(s(0())) -> false()
                                         , even(s(s(x))) -> even(x)}
                                      Weak Trs:
                                        {  div2(s(s(x))) -> s(div2(x))
                                         , neq(0(), 0()) -> false()
                                         , neq(s(x), 0()) -> true()
                                         , neq(0(), s(x)) -> true()
                                         , neq(s(x), s(y())) -> neq(x, y())
                                         , p(s(x)) -> x
                                         , div2(0()) -> 0()
                                         , div2(s(0())) -> 0()
                                         , p(0()) -> 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
                              
                              * Path 1:{15}: YES(O(1),O(1))
                                ---------------------------
                                
                                We consider the following Problem:
                                
                                  Strict Trs:
                                    {  even(0()) -> true()
                                     , even(s(0())) -> false()
                                     , even(s(s(x))) -> even(x)}
                                  Weak Trs:
                                    {  div2(s(s(x))) -> s(div2(x))
                                     , neq(0(), 0()) -> false()
                                     , neq(s(x), 0()) -> true()
                                     , neq(0(), s(x)) -> true()
                                     , neq(s(x), s(y())) -> neq(x, y())
                                     , p(s(x)) -> x
                                     , div2(0()) -> 0()
                                     , div2(s(0())) -> 0()
                                     , p(0()) -> 0()}
                                  StartTerms: basic terms
                                  Strategy: innermost
                                
                                Certificate: YES(O(1),O(1))
                                
                                Proof:
                                  We consider the following Problem:
                                  
                                    Strict Trs:
                                      {  even(0()) -> true()
                                       , even(s(0())) -> false()
                                       , even(s(s(x))) -> even(x)}
                                    Weak Trs:
                                      {  div2(s(s(x))) -> s(div2(x))
                                       , neq(0(), 0()) -> false()
                                       , neq(s(x), 0()) -> true()
                                       , neq(0(), s(x)) -> true()
                                       , neq(s(x), s(y())) -> neq(x, y())
                                       , p(s(x)) -> x
                                       , div2(0()) -> 0()
                                       , div2(s(0())) -> 0()
                                       , p(0()) -> 0()}
                                    StartTerms: basic terms
                                    Strategy: innermost
                                  
                                  Certificate: YES(O(1),O(1))
                                  
                                  Proof:
                                    We consider the following Problem:
                                    
                                      Strict Trs:
                                        {  even(0()) -> true()
                                         , even(s(0())) -> false()
                                         , even(s(s(x))) -> even(x)}
                                      Weak Trs:
                                        {  div2(s(s(x))) -> s(div2(x))
                                         , neq(0(), 0()) -> false()
                                         , neq(s(x), 0()) -> true()
                                         , neq(0(), s(x)) -> true()
                                         , neq(s(x), s(y())) -> neq(x, y())
                                         , p(s(x)) -> x
                                         , div2(0()) -> 0()
                                         , div2(s(0())) -> 0()
                                         , p(0()) -> 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))