We consider the following Problem:

  Strict Trs:
    {  cond1(true(), x, y) -> cond2(gr(x, y), x, y)
     , cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
     , cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
     , cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
     , cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
     , cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
     , cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
     , gr(0(), x) -> false()
     , gr(s(x), 0()) -> true()
     , gr(s(x), s(y)) -> gr(x, y)
     , and(true(), true()) -> true()
     , and(false(), x) -> false()
     , and(x, false()) -> false()
     , 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, y) -> cond2(gr(x, y), x, y)
       , cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
       , cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
       , cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
       , cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
       , cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
       , cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
       , gr(0(), x) -> false()
       , gr(s(x), 0()) -> true()
       , gr(s(x), s(y)) -> gr(x, y)
       , and(true(), true()) -> true()
       , and(false(), x) -> false()
       , and(x, false()) -> false()
       , 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}, Uargs(cond2) = {1}, Uargs(gr) = {},
        Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
        Uargs(and) = {1, 2}, Uargs(s) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       cond1(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                           [0 1]      [0 0]      [0 0]      [1]
       true() = [0]
                [0]
       cond2(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                           [0 0]      [0 0]      [0 0]      [1]
       gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                    [0 0]      [0 1]      [0]
       cond3(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                           [0 0]      [0 0]      [0 0]      [1]
       0() = [0]
             [1]
       false() = [0]
                 [0]
       cond4(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                           [0 0]      [0 0]      [0 0]      [1]
       p(x1) = [0 1] x1 + [0]
               [0 0]      [3]
       and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                     [0 0]      [0 1]      [0]
       s(x1) = [0 0] x1 + [0]
               [1 0]      [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  cond1(true(), x, y) -> cond2(gr(x, y), x, y)
         , cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
         , cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
         , cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
         , cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
         , cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
         , cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
         , gr(0(), x) -> false()
         , gr(s(x), 0()) -> true()
         , gr(s(x), s(y)) -> gr(x, y)
         , and(true(), true()) -> true()
         , and(false(), x) -> false()
         , and(x, false()) -> false()
         , 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:
        {cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(cond1) = {1}, Uargs(cond2) = {1}, Uargs(gr) = {},
          Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
          Uargs(and) = {1, 2}, Uargs(s) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         cond1(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                             [0 0]      [0 0]      [0 0]      [1]
         true() = [0]
                  [0]
         cond2(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                             [0 0]      [0 0]      [0 0]      [1]
         gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                      [0 0]      [0 0]      [0]
         cond3(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                             [0 0]      [0 0]      [0 0]      [1]
         0() = [0]
               [0]
         false() = [0]
                   [2]
         cond4(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [1]
                             [0 0]      [0 0]      [0 0]      [1]
         p(x1) = [0 1] x1 + [0]
                 [0 0]      [0]
         and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                       [0 0]      [0 0]      [1]
         s(x1) = [0 0] x1 + [0]
                 [1 0]      [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  cond1(true(), x, y) -> cond2(gr(x, y), x, y)
           , cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
           , cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
           , cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
           , cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
           , cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
           , gr(0(), x) -> false()
           , gr(s(x), 0()) -> true()
           , gr(s(x), s(y)) -> gr(x, y)
           , and(true(), true()) -> true()
           , and(false(), x) -> false()
           , and(x, false()) -> false()
           , p(s(x)) -> x}
        Weak Trs:
          {  cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
           , 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: {cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(cond1) = {1}, Uargs(cond2) = {1}, Uargs(gr) = {},
            Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
            Uargs(and) = {1, 2}, Uargs(s) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           cond1(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                               [0 0]      [0 0]      [0 0]      [1]
           true() = [0]
                    [2]
           cond2(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                               [0 0]      [0 0]      [0 0]      [1]
           gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                        [0 0]      [0 0]      [0]
           cond3(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                               [0 0]      [0 0]      [0 0]      [1]
           0() = [0]
                 [0]
           false() = [0]
                     [0]
           cond4(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [1]
                               [0 0]      [0 0]      [0 0]      [1]
           p(x1) = [0 1] x1 + [0]
                   [0 0]      [0]
           and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                         [0 0]      [0 0]      [1]
           s(x1) = [0 0] x1 + [0]
                   [1 0]      [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs:
            {  cond1(true(), x, y) -> cond2(gr(x, y), x, y)
             , cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
             , cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
             , cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
             , cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
             , gr(0(), x) -> false()
             , gr(s(x), 0()) -> true()
             , gr(s(x), s(y)) -> gr(x, y)
             , and(true(), true()) -> true()
             , and(false(), x) -> false()
             , and(x, false()) -> false()
             , p(s(x)) -> x}
          Weak Trs:
            {  cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
             , cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
             , 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: {and(true(), true()) -> true()}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(cond1) = {1}, Uargs(cond2) = {1}, Uargs(gr) = {},
              Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
              Uargs(and) = {1, 2}, Uargs(s) = {}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             cond1(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                                 [0 0]      [0 0]      [0 0]      [1]
             true() = [0]
                      [0]
             cond2(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                                 [0 0]      [0 0]      [0 0]      [1]
             gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                          [0 0]      [0 0]      [0]
             cond3(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                                 [0 0]      [0 0]      [0 0]      [1]
             0() = [0]
                   [0]
             false() = [0]
                       [2]
             cond4(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [1]
                                 [0 0]      [0 0]      [0 0]      [1]
             p(x1) = [0 1] x1 + [0]
                     [0 0]      [0]
             and(x1, x2) = [1 0] x1 + [1 0] x2 + [2]
                           [0 0]      [0 0]      [1]
             s(x1) = [0 0] x1 + [0]
                     [1 0]      [0]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Strict Trs:
              {  cond1(true(), x, y) -> cond2(gr(x, y), x, y)
               , cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
               , cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
               , cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
               , cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
               , gr(0(), x) -> false()
               , gr(s(x), 0()) -> true()
               , gr(s(x), s(y)) -> gr(x, y)
               , and(false(), x) -> false()
               , and(x, false()) -> false()
               , p(s(x)) -> x}
            Weak Trs:
              {  and(true(), true()) -> true()
               , cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
               , cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
               , 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}, Uargs(cond2) = {1}, Uargs(gr) = {},
                Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
                Uargs(and) = {1, 2}, Uargs(s) = {}
              We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
              Interpretation Functions:
               cond1(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                                   [0 0]      [0 0]      [0 0]      [1]
               true() = [0]
                        [2]
               cond2(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                                   [0 0]      [0 0]      [0 0]      [1]
               gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                            [0 0]      [0 0]      [0]
               cond3(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                                   [0 0]      [0 0]      [0 0]      [1]
               0() = [0]
                     [2]
               false() = [0]
                         [0]
               cond4(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [1]
                                   [0 0]      [0 0]      [0 0]      [1]
               p(x1) = [1 0] x1 + [0]
                       [0 1]      [2]
               and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                             [0 0]      [0 1]      [1]
               s(x1) = [1 0] x1 + [2]
                       [0 1]      [0]
            
            The strictly oriented rules are moved into the weak component.
            
            We consider the following Problem:
            
              Strict Trs:
                {  cond1(true(), x, y) -> cond2(gr(x, y), x, y)
                 , cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
                 , cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
                 , cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
                 , cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
                 , gr(0(), x) -> false()
                 , gr(s(x), 0()) -> true()
                 , gr(s(x), s(y)) -> gr(x, y)
                 , and(false(), x) -> false()
                 , and(x, false()) -> false()}
              Weak Trs:
                {  p(s(x)) -> x
                 , and(true(), true()) -> true()
                 , cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
                 , cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
                 , 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, y) -> cond4(gr(y, 0()), x, y)}
              
              Interpretation of nonconstant growth:
              -------------------------------------
                The following argument positions are usable:
                  Uargs(cond1) = {1}, Uargs(cond2) = {1}, Uargs(gr) = {},
                  Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
                  Uargs(and) = {1, 2}, Uargs(s) = {}
                We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                Interpretation Functions:
                 cond1(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
                                     [0 0]      [0 1]      [0 1]      [1]
                 true() = [0]
                          [1]
                 cond2(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
                                     [0 0]      [0 1]      [0 1]      [1]
                 gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                              [0 0]      [0 0]      [1]
                 cond3(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
                                     [0 0]      [0 1]      [0 0]      [1]
                 0() = [0]
                       [0]
                 false() = [1]
                           [2]
                 cond4(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
                                     [0 1]      [0 1]      [0 1]      [0]
                 p(x1) = [1 0] x1 + [0]
                         [0 1]      [0]
                 and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                               [0 0]      [0 0]      [1]
                 s(x1) = [1 0] x1 + [0]
                         [0 1]      [0]
              
              The strictly oriented rules are moved into the weak component.
              
              We consider the following Problem:
              
                Strict Trs:
                  {  cond1(true(), x, y) -> cond2(gr(x, y), x, y)
                   , cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
                   , cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
                   , cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
                   , gr(0(), x) -> false()
                   , gr(s(x), 0()) -> true()
                   , gr(s(x), s(y)) -> gr(x, y)
                   , and(false(), x) -> false()
                   , and(x, false()) -> false()}
                Weak Trs:
                  {  cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
                   , p(s(x)) -> x
                   , and(true(), true()) -> true()
                   , cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
                   , cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
                   , 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, y) -> cond2(gr(x, y), x, y)
                   , cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
                   , cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)}
                
                Interpretation of nonconstant growth:
                -------------------------------------
                  The following argument positions are usable:
                    Uargs(cond1) = {1}, Uargs(cond2) = {1}, Uargs(gr) = {},
                    Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
                    Uargs(and) = {1, 2}, Uargs(s) = {}
                  We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                  Interpretation Functions:
                   cond1(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
                                       [1 1]      [1 1]      [1 1]      [1]
                   true() = [2]
                            [2]
                   cond2(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
                                       [0 0]      [1 1]      [1 1]      [1]
                   gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                                [0 0]      [1 0]      [0]
                   cond3(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
                                       [0 0]      [1 1]      [0 0]      [1]
                   0() = [0]
                         [0]
                   false() = [0]
                             [1]
                   cond4(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
                                       [1 1]      [1 1]      [1 1]      [0]
                   p(x1) = [1 0] x1 + [0]
                           [0 1]      [0]
                   and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                                 [0 0]      [1 0]      [0]
                   s(x1) = [1 0] x1 + [0]
                           [0 1]      [1]
                
                The strictly oriented rules are moved into the weak component.
                
                We consider the following Problem:
                
                  Strict Trs:
                    {  cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
                     , gr(0(), x) -> false()
                     , gr(s(x), 0()) -> true()
                     , gr(s(x), s(y)) -> gr(x, y)
                     , and(false(), x) -> false()
                     , and(x, false()) -> false()}
                  Weak Trs:
                    {  cond1(true(), x, y) -> cond2(gr(x, y), x, y)
                     , cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
                     , cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
                     , cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
                     , p(s(x)) -> x
                     , and(true(), true()) -> true()
                     , cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
                     , cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
                     , 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:
                    {cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)}
                  
                  Interpretation of nonconstant growth:
                  -------------------------------------
                    The following argument positions are usable:
                      Uargs(cond1) = {1}, Uargs(cond2) = {1}, Uargs(gr) = {},
                      Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
                      Uargs(and) = {1, 2}, Uargs(s) = {}
                    We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                    Interpretation Functions:
                     cond1(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [0]
                                         [0 0]      [0 0]      [0 0]      [1]
                     true() = [2]
                              [0]
                     cond2(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [2]
                                         [0 0]      [0 0]      [0 0]      [1]
                     gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                                  [0 0]      [0 0]      [1]
                     cond3(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                                         [0 0]      [0 0]      [0 0]      [1]
                     0() = [0]
                           [0]
                     false() = [0]
                               [0]
                     cond4(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                                         [0 0]      [0 0]      [0 0]      [1]
                     p(x1) = [0 1] x1 + [0]
                             [1 0]      [0]
                     and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                                   [0 0]      [0 0]      [1]
                     s(x1) = [0 1] x1 + [0]
                             [1 0]      [0]
                  
                  The strictly oriented rules are moved into the weak component.
                  
                  We consider the following Problem:
                  
                    Strict Trs:
                      {  gr(0(), x) -> false()
                       , gr(s(x), 0()) -> true()
                       , gr(s(x), s(y)) -> gr(x, y)
                       , and(false(), x) -> false()
                       , and(x, false()) -> false()}
                    Weak Trs:
                      {  cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
                       , cond1(true(), x, y) -> cond2(gr(x, y), x, y)
                       , cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
                       , cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
                       , cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
                       , p(s(x)) -> x
                       , and(true(), true()) -> true()
                       , cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
                       , cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
                       , 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: {and(x, false()) -> false()}
                    
                    Interpretation of nonconstant growth:
                    -------------------------------------
                      The following argument positions are usable:
                        Uargs(cond1) = {1}, Uargs(cond2) = {1}, Uargs(gr) = {},
                        Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
                        Uargs(and) = {1, 2}, Uargs(s) = {}
                      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                      Interpretation Functions:
                       cond1(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
                                           [0 0]      [0 0]      [0 0]      [1]
                       true() = [2]
                                [2]
                       cond2(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [1]
                                           [0 0]      [0 0]      [0 0]      [1]
                       gr(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
                                    [0 0]      [0 0]      [1]
                       cond3(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [1]
                                           [0 0]      [0 0]      [0 0]      [1]
                       0() = [0]
                             [0]
                       false() = [0]
                                 [3]
                       cond4(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [2]
                                           [0 0]      [0 0]      [0 0]      [1]
                       p(x1) = [0 1] x1 + [1]
                               [1 0]      [0]
                       and(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                                     [0 0]      [0 1]      [1]
                       s(x1) = [0 1] x1 + [0]
                               [1 0]      [0]
                    
                    The strictly oriented rules are moved into the weak component.
                    
                    We consider the following Problem:
                    
                      Strict Trs:
                        {  gr(0(), x) -> false()
                         , gr(s(x), 0()) -> true()
                         , gr(s(x), s(y)) -> gr(x, y)
                         , and(false(), x) -> false()}
                      Weak Trs:
                        {  and(x, false()) -> false()
                         , cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
                         , cond1(true(), x, y) -> cond2(gr(x, y), x, y)
                         , cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
                         , cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
                         , cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
                         , p(s(x)) -> x
                         , and(true(), true()) -> true()
                         , cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
                         , cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
                         , 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: {and(false(), x) -> false()}
                      
                      Interpretation of nonconstant growth:
                      -------------------------------------
                        The following argument positions are usable:
                          Uargs(cond1) = {1}, Uargs(cond2) = {1}, Uargs(gr) = {},
                          Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
                          Uargs(and) = {1, 2}, Uargs(s) = {}
                        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                        Interpretation Functions:
                         cond1(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [0]
                                             [0 0]      [0 0]      [0 0]      [1]
                         true() = [3]
                                  [2]
                         cond2(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [3]
                                             [0 0]      [0 0]      [0 0]      [1]
                         gr(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
                                      [0 0]      [0 0]      [1]
                         cond3(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [3]
                                             [0 0]      [0 0]      [0 0]      [1]
                         0() = [0]
                               [0]
                         false() = [0]
                                   [2]
                         cond4(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [3]
                                             [0 0]      [0 0]      [0 0]      [1]
                         p(x1) = [0 1] x1 + [0]
                                 [1 0]      [0]
                         and(x1, x2) = [1 1] x1 + [1 0] x2 + [0]
                                       [0 0]      [0 0]      [2]
                         s(x1) = [0 1] x1 + [0]
                                 [1 0]      [0]
                      
                      The strictly oriented rules are moved into the weak component.
                      
                      We consider the following Problem:
                      
                        Strict Trs:
                          {  gr(0(), x) -> false()
                           , gr(s(x), 0()) -> true()
                           , gr(s(x), s(y)) -> gr(x, y)}
                        Weak Trs:
                          {  and(false(), x) -> false()
                           , and(x, false()) -> false()
                           , cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
                           , cond1(true(), x, y) -> cond2(gr(x, y), x, y)
                           , cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
                           , cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
                           , cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
                           , p(s(x)) -> x
                           , and(true(), true()) -> true()
                           , cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
                           , cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
                           , 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: {gr(0(), x) -> false()}
                        
                        Interpretation of nonconstant growth:
                        -------------------------------------
                          The following argument positions are usable:
                            Uargs(cond1) = {1}, Uargs(cond2) = {1}, Uargs(gr) = {},
                            Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
                            Uargs(and) = {1, 2}, Uargs(s) = {}
                          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                          Interpretation Functions:
                           cond1(x1, x2, x3) = [1 2] x1 + [1 0] x2 + [1 0] x3 + [0]
                                               [0 0]      [0 0]      [0 1]      [0]
                           true() = [3]
                                    [3]
                           cond2(x1, x2, x3) = [1 2] x1 + [1 0] x2 + [1 0] x3 + [3]
                                               [0 0]      [0 0]      [0 1]      [0]
                           gr(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
                                        [0 0]      [0 0]      [0]
                           cond3(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [2]
                                               [0 0]      [0 0]      [0 1]      [0]
                           0() = [0]
                                 [0]
                           false() = [0]
                                     [0]
                           cond4(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [2]
                                               [0 0]      [0 0]      [0 1]      [0]
                           p(x1) = [1 0] x1 + [0]
                                   [0 1]      [0]
                           and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                                         [0 0]      [0 1]      [0]
                           s(x1) = [1 0] x1 + [0]
                                   [0 1]      [0]
                        
                        The strictly oriented rules are moved into the weak component.
                        
                        We consider the following Problem:
                        
                          Strict Trs:
                            {  gr(s(x), 0()) -> true()
                             , gr(s(x), s(y)) -> gr(x, y)}
                          Weak Trs:
                            {  gr(0(), x) -> false()
                             , and(false(), x) -> false()
                             , and(x, false()) -> false()
                             , cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
                             , cond1(true(), x, y) -> cond2(gr(x, y), x, y)
                             , cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
                             , cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
                             , cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
                             , p(s(x)) -> x
                             , and(true(), true()) -> true()
                             , cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
                             , cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
                             , p(0()) -> 0()}
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(?,O(n^1))
                        
                        Proof:
                          We consider the following Problem:
                          
                            Strict Trs:
                              {  gr(s(x), 0()) -> true()
                               , gr(s(x), s(y)) -> gr(x, y)}
                            Weak Trs:
                              {  gr(0(), x) -> false()
                               , and(false(), x) -> false()
                               , and(x, false()) -> false()
                               , cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
                               , cond1(true(), x, y) -> cond2(gr(x, y), x, y)
                               , cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
                               , cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
                               , cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
                               , p(s(x)) -> x
                               , and(true(), true()) -> true()
                               , cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
                               , cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
                               , p(0()) -> 0()}
                            StartTerms: basic terms
                            Strategy: innermost
                          
                          Certificate: YES(?,O(n^1))
                          
                          Proof:
                            We have computed the following dependency pairs
                            
                              Strict DPs:
                                {  gr^#(s(x), 0()) -> c_1()
                                 , gr^#(s(x), s(y)) -> gr^#(x, y)}
                              Weak DPs:
                                {  gr^#(0(), x) -> c_3()
                                 , and^#(false(), x) -> c_4()
                                 , and^#(x, false()) -> c_5()
                                 , cond3^#(false(), x, y) ->
                                   cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
                                 , cond1^#(true(), x, y) -> cond2^#(gr(x, y), x, y)
                                 , cond2^#(true(), x, y) -> cond3^#(gr(x, 0()), x, y)
                                 , cond3^#(true(), x, y) -> cond3^#(gr(x, 0()), p(x), y)
                                 , cond2^#(false(), x, y) -> cond4^#(gr(y, 0()), x, y)
                                 , p^#(s(x)) -> c_11()
                                 , and^#(true(), true()) -> c_12()
                                 , cond4^#(true(), x, y) -> cond4^#(gr(y, 0()), x, p(y))
                                 , cond4^#(false(), x, y) ->
                                   cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
                                 , p^#(0()) -> c_15()}
                            
                            We consider the following Problem:
                            
                              Strict DPs:
                                {  gr^#(s(x), 0()) -> c_1()
                                 , gr^#(s(x), s(y)) -> gr^#(x, y)}
                              Strict Trs:
                                {  gr(s(x), 0()) -> true()
                                 , gr(s(x), s(y)) -> gr(x, y)}
                              Weak DPs:
                                {  gr^#(0(), x) -> c_3()
                                 , and^#(false(), x) -> c_4()
                                 , and^#(x, false()) -> c_5()
                                 , cond3^#(false(), x, y) ->
                                   cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
                                 , cond1^#(true(), x, y) -> cond2^#(gr(x, y), x, y)
                                 , cond2^#(true(), x, y) -> cond3^#(gr(x, 0()), x, y)
                                 , cond3^#(true(), x, y) -> cond3^#(gr(x, 0()), p(x), y)
                                 , cond2^#(false(), x, y) -> cond4^#(gr(y, 0()), x, y)
                                 , p^#(s(x)) -> c_11()
                                 , and^#(true(), true()) -> c_12()
                                 , cond4^#(true(), x, y) -> cond4^#(gr(y, 0()), x, p(y))
                                 , cond4^#(false(), x, y) ->
                                   cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
                                 , p^#(0()) -> c_15()}
                              Weak Trs:
                                {  gr(0(), x) -> false()
                                 , and(false(), x) -> false()
                                 , and(x, false()) -> false()
                                 , cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
                                 , cond1(true(), x, y) -> cond2(gr(x, y), x, y)
                                 , cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
                                 , cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
                                 , cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
                                 , p(s(x)) -> x
                                 , and(true(), true()) -> true()
                                 , cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
                                 , cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
                                 , 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:
                                  {  gr(s(x), 0()) -> true()
                                   , gr(s(x), s(y)) -> gr(x, y)}
                                Weak Usable Rules:
                                  {  gr(0(), x) -> false()
                                   , and(false(), x) -> false()
                                   , and(x, false()) -> false()
                                   , p(s(x)) -> x
                                   , and(true(), true()) -> true()
                                   , p(0()) -> 0()}
                              
                              We consider the following Problem:
                              
                                Strict DPs:
                                  {  gr^#(s(x), 0()) -> c_1()
                                   , gr^#(s(x), s(y)) -> gr^#(x, y)}
                                Strict Trs:
                                  {  gr(s(x), 0()) -> true()
                                   , gr(s(x), s(y)) -> gr(x, y)}
                                Weak DPs:
                                  {  gr^#(0(), x) -> c_3()
                                   , and^#(false(), x) -> c_4()
                                   , and^#(x, false()) -> c_5()
                                   , cond3^#(false(), x, y) ->
                                     cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
                                   , cond1^#(true(), x, y) -> cond2^#(gr(x, y), x, y)
                                   , cond2^#(true(), x, y) -> cond3^#(gr(x, 0()), x, y)
                                   , cond3^#(true(), x, y) -> cond3^#(gr(x, 0()), p(x), y)
                                   , cond2^#(false(), x, y) -> cond4^#(gr(y, 0()), x, y)
                                   , p^#(s(x)) -> c_11()
                                   , and^#(true(), true()) -> c_12()
                                   , cond4^#(true(), x, y) -> cond4^#(gr(y, 0()), x, p(y))
                                   , cond4^#(false(), x, y) ->
                                     cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
                                   , p^#(0()) -> c_15()}
                                Weak Trs:
                                  {  gr(0(), x) -> false()
                                   , and(false(), x) -> false()
                                   , and(x, false()) -> false()
                                   , p(s(x)) -> x
                                   , and(true(), true()) -> true()
                                   , p(0()) -> 0()}
                                StartTerms: basic terms
                                Strategy: innermost
                              
                              Certificate: YES(?,O(n^1))
                              
                              Proof:
                                We consider the following Problem:
                                
                                  Strict DPs:
                                    {  gr^#(s(x), 0()) -> c_1()
                                     , gr^#(s(x), s(y)) -> gr^#(x, y)}
                                  Strict Trs:
                                    {  gr(s(x), 0()) -> true()
                                     , gr(s(x), s(y)) -> gr(x, y)}
                                  Weak DPs:
                                    {  gr^#(0(), x) -> c_3()
                                     , and^#(false(), x) -> c_4()
                                     , and^#(x, false()) -> c_5()
                                     , cond3^#(false(), x, y) ->
                                       cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
                                     , cond1^#(true(), x, y) -> cond2^#(gr(x, y), x, y)
                                     , cond2^#(true(), x, y) -> cond3^#(gr(x, 0()), x, y)
                                     , cond3^#(true(), x, y) -> cond3^#(gr(x, 0()), p(x), y)
                                     , cond2^#(false(), x, y) -> cond4^#(gr(y, 0()), x, y)
                                     , p^#(s(x)) -> c_11()
                                     , and^#(true(), true()) -> c_12()
                                     , cond4^#(true(), x, y) -> cond4^#(gr(y, 0()), x, p(y))
                                     , cond4^#(false(), x, y) ->
                                       cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
                                     , p^#(0()) -> c_15()}
                                  Weak Trs:
                                    {  gr(0(), x) -> false()
                                     , and(false(), x) -> false()
                                     , and(x, false()) -> false()
                                     , p(s(x)) -> x
                                     , and(true(), true()) -> true()
                                     , p(0()) -> 0()}
                                  StartTerms: basic terms
                                  Strategy: innermost
                                
                                Certificate: YES(?,O(n^1))
                                
                                Proof:
                                  We use following congruence DG for path analysis
                                  
                                  ->7:{2}                                                     [   YES(?,O(n^1))    ]
                                     |
                                     |->9:{1}                                                 [   YES(?,O(n^1))    ]
                                     |
                                     `->8:{3}                                                 [   YES(O(1),O(1))   ]
                                  
                                  ->6:{4}                                                     [   YES(O(1),O(1))   ]
                                  
                                  ->5:{5}                                                     [   YES(O(1),O(1))   ]
                                  
                                  ->4:{6,9,8,7,14,13,10}                                      [   YES(O(1),O(1))   ]
                                  
                                  ->3:{11}                                                    [   YES(O(1),O(1))   ]
                                  
                                  ->2:{12}                                                    [   YES(O(1),O(1))   ]
                                  
                                  ->1:{15}                                                    [   YES(O(1),O(1))   ]
                                  
                                  
                                  Here dependency-pairs are as follows:
                                  
                                  Strict DPs:
                                    {  1: gr^#(s(x), 0()) -> c_1()
                                     , 2: gr^#(s(x), s(y)) -> gr^#(x, y)}
                                  WeakDPs DPs:
                                    {  3: gr^#(0(), x) -> c_3()
                                     , 4: and^#(false(), x) -> c_4()
                                     , 5: and^#(x, false()) -> c_5()
                                     , 6: cond3^#(false(), x, y) ->
                                          cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
                                     , 7: cond1^#(true(), x, y) -> cond2^#(gr(x, y), x, y)
                                     , 8: cond2^#(true(), x, y) -> cond3^#(gr(x, 0()), x, y)
                                     , 9: cond3^#(true(), x, y) -> cond3^#(gr(x, 0()), p(x), y)
                                     , 10: cond2^#(false(), x, y) -> cond4^#(gr(y, 0()), x, y)
                                     , 11: p^#(s(x)) -> c_11()
                                     , 12: and^#(true(), true()) -> c_12()
                                     , 13: cond4^#(true(), x, y) -> cond4^#(gr(y, 0()), x, p(y))
                                     , 14: cond4^#(false(), x, y) ->
                                           cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
                                     , 15: p^#(0()) -> c_15()}
                                  
                                  * Path 7:{2}: YES(?,O(n^1))
                                    -------------------------
                                    
                                    We consider the following Problem:
                                    
                                      Strict DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
                                      Strict Trs:
                                        {  gr(s(x), 0()) -> true()
                                         , gr(s(x), s(y)) -> gr(x, y)}
                                      Weak Trs:
                                        {  gr(0(), x) -> false()
                                         , and(false(), x) -> false()
                                         , and(x, false()) -> false()
                                         , p(s(x)) -> x
                                         , and(true(), true()) -> true()
                                         , p(0()) -> 0()}
                                      StartTerms: basic terms
                                      Strategy: innermost
                                    
                                    Certificate: YES(?,O(n^1))
                                    
                                    Proof:
                                      We consider the following Problem:
                                      
                                        Strict DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
                                        Strict Trs:
                                          {  gr(s(x), 0()) -> true()
                                           , gr(s(x), s(y)) -> gr(x, y)}
                                        Weak Trs:
                                          {  gr(0(), x) -> false()
                                           , and(false(), x) -> false()
                                           , and(x, false()) -> false()
                                           , p(s(x)) -> x
                                           , and(true(), true()) -> true()
                                           , p(0()) -> 0()}
                                        StartTerms: basic terms
                                        Strategy: innermost
                                      
                                      Certificate: YES(?,O(n^1))
                                      
                                      Proof:
                                        We consider the following Problem:
                                        
                                          Strict DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
                                          Strict Trs:
                                            {  gr(s(x), 0()) -> true()
                                             , gr(s(x), s(y)) -> gr(x, y)}
                                          Weak Trs:
                                            {  gr(0(), x) -> false()
                                             , and(false(), x) -> false()
                                             , and(x, false()) -> false()
                                             , p(s(x)) -> x
                                             , and(true(), true()) -> true()
                                             , 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: {gr^#(s(x), s(y)) -> gr^#(x, y)}
                                            StartTerms: basic terms
                                            Strategy: innermost
                                          
                                          Certificate: YES(?,O(n^1))
                                          
                                          Proof:
                                            The problem is match-bounded by 1.
                                            The enriched problem is compatible with the following automaton:
                                            {  s_0(2) -> 2
                                             , gr^#_0(2, 2) -> 1
                                             , gr^#_1(2, 2) -> 1}
                                  
                                  * Path 7:{2}->9:{1}: YES(?,O(n^1))
                                    --------------------------------
                                    
                                    We consider the following Problem:
                                    
                                      Strict DPs: {gr^#(s(x), 0()) -> c_1()}
                                      Strict Trs:
                                        {  gr(s(x), 0()) -> true()
                                         , gr(s(x), s(y)) -> gr(x, y)}
                                      Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
                                      Weak Trs:
                                        {  gr(0(), x) -> false()
                                         , and(false(), x) -> false()
                                         , and(x, false()) -> false()
                                         , p(s(x)) -> x
                                         , and(true(), true()) -> true()
                                         , p(0()) -> 0()}
                                      StartTerms: basic terms
                                      Strategy: innermost
                                    
                                    Certificate: YES(?,O(n^1))
                                    
                                    Proof:
                                      We consider the following Problem:
                                      
                                        Strict DPs: {gr^#(s(x), 0()) -> c_1()}
                                        Strict Trs:
                                          {  gr(s(x), 0()) -> true()
                                           , gr(s(x), s(y)) -> gr(x, y)}
                                        Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
                                        Weak Trs:
                                          {  gr(0(), x) -> false()
                                           , and(false(), x) -> false()
                                           , and(x, false()) -> false()
                                           , p(s(x)) -> x
                                           , and(true(), true()) -> true()
                                           , p(0()) -> 0()}
                                        StartTerms: basic terms
                                        Strategy: innermost
                                      
                                      Certificate: YES(?,O(n^1))
                                      
                                      Proof:
                                        We consider the following Problem:
                                        
                                          Strict DPs: {gr^#(s(x), 0()) -> c_1()}
                                          Strict Trs:
                                            {  gr(s(x), 0()) -> true()
                                             , gr(s(x), s(y)) -> gr(x, y)}
                                          Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
                                          Weak Trs:
                                            {  gr(0(), x) -> false()
                                             , and(false(), x) -> false()
                                             , and(x, false()) -> false()
                                             , p(s(x)) -> x
                                             , and(true(), true()) -> true()
                                             , 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: {gr^#(s(x), 0()) -> c_1()}
                                            Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
                                            StartTerms: basic terms
                                            Strategy: innermost
                                          
                                          Certificate: YES(?,O(n^1))
                                          
                                          Proof:
                                            The problem is match-bounded by 1.
                                            The enriched problem is compatible with the following automaton:
                                            {  0_0() -> 2
                                             , s_0(2) -> 2
                                             , gr^#_0(2, 2) -> 1
                                             , c_1_1() -> 1}
                                  
                                  * Path 7:{2}->8:{3}: YES(O(1),O(1))
                                    ---------------------------------
                                    
                                    We consider the following Problem:
                                    
                                      Strict Trs:
                                        {  gr(s(x), 0()) -> true()
                                         , gr(s(x), s(y)) -> gr(x, y)}
                                      Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
                                      Weak Trs:
                                        {  gr(0(), x) -> false()
                                         , and(false(), x) -> false()
                                         , and(x, false()) -> false()
                                         , p(s(x)) -> x
                                         , and(true(), true()) -> true()
                                         , p(0()) -> 0()}
                                      StartTerms: basic terms
                                      Strategy: innermost
                                    
                                    Certificate: YES(O(1),O(1))
                                    
                                    Proof:
                                      We consider the following Problem:
                                      
                                        Strict Trs:
                                          {  gr(s(x), 0()) -> true()
                                           , gr(s(x), s(y)) -> gr(x, y)}
                                        Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
                                        Weak Trs:
                                          {  gr(0(), x) -> false()
                                           , and(false(), x) -> false()
                                           , and(x, false()) -> false()
                                           , p(s(x)) -> x
                                           , and(true(), true()) -> true()
                                           , p(0()) -> 0()}
                                        StartTerms: basic terms
                                        Strategy: innermost
                                      
                                      Certificate: YES(O(1),O(1))
                                      
                                      Proof:
                                        We consider the following Problem:
                                        
                                          Strict Trs:
                                            {  gr(s(x), 0()) -> true()
                                             , gr(s(x), s(y)) -> gr(x, y)}
                                          Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
                                          Weak Trs:
                                            {  gr(0(), x) -> false()
                                             , and(false(), x) -> false()
                                             , and(x, false()) -> false()
                                             , p(s(x)) -> x
                                             , and(true(), true()) -> true()
                                             , 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: {gr^#(s(x), s(y)) -> gr^#(x, y)}
                                            StartTerms: basic terms
                                            Strategy: innermost
                                          
                                          Certificate: YES(O(1),O(1))
                                          
                                          Proof:
                                            Empty rules are trivially bounded
                                  
                                  * Path 6:{4}: YES(O(1),O(1))
                                    --------------------------
                                    
                                    We consider the following Problem:
                                    
                                      Strict Trs:
                                        {  gr(s(x), 0()) -> true()
                                         , gr(s(x), s(y)) -> gr(x, y)}
                                      Weak Trs:
                                        {  gr(0(), x) -> false()
                                         , and(false(), x) -> false()
                                         , and(x, false()) -> false()
                                         , p(s(x)) -> x
                                         , and(true(), true()) -> true()
                                         , p(0()) -> 0()}
                                      StartTerms: basic terms
                                      Strategy: innermost
                                    
                                    Certificate: YES(O(1),O(1))
                                    
                                    Proof:
                                      We consider the following Problem:
                                      
                                        Strict Trs:
                                          {  gr(s(x), 0()) -> true()
                                           , gr(s(x), s(y)) -> gr(x, y)}
                                        Weak Trs:
                                          {  gr(0(), x) -> false()
                                           , and(false(), x) -> false()
                                           , and(x, false()) -> false()
                                           , p(s(x)) -> x
                                           , and(true(), true()) -> true()
                                           , p(0()) -> 0()}
                                        StartTerms: basic terms
                                        Strategy: innermost
                                      
                                      Certificate: YES(O(1),O(1))
                                      
                                      Proof:
                                        We consider the following Problem:
                                        
                                          Strict Trs:
                                            {  gr(s(x), 0()) -> true()
                                             , gr(s(x), s(y)) -> gr(x, y)}
                                          Weak Trs:
                                            {  gr(0(), x) -> false()
                                             , and(false(), x) -> false()
                                             , and(x, false()) -> false()
                                             , p(s(x)) -> x
                                             , and(true(), true()) -> true()
                                             , 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:{5}: YES(O(1),O(1))
                                    --------------------------
                                    
                                    We consider the following Problem:
                                    
                                      Strict Trs:
                                        {  gr(s(x), 0()) -> true()
                                         , gr(s(x), s(y)) -> gr(x, y)}
                                      Weak Trs:
                                        {  gr(0(), x) -> false()
                                         , and(false(), x) -> false()
                                         , and(x, false()) -> false()
                                         , p(s(x)) -> x
                                         , and(true(), true()) -> true()
                                         , p(0()) -> 0()}
                                      StartTerms: basic terms
                                      Strategy: innermost
                                    
                                    Certificate: YES(O(1),O(1))
                                    
                                    Proof:
                                      We consider the following Problem:
                                      
                                        Strict Trs:
                                          {  gr(s(x), 0()) -> true()
                                           , gr(s(x), s(y)) -> gr(x, y)}
                                        Weak Trs:
                                          {  gr(0(), x) -> false()
                                           , and(false(), x) -> false()
                                           , and(x, false()) -> false()
                                           , p(s(x)) -> x
                                           , and(true(), true()) -> true()
                                           , p(0()) -> 0()}
                                        StartTerms: basic terms
                                        Strategy: innermost
                                      
                                      Certificate: YES(O(1),O(1))
                                      
                                      Proof:
                                        We consider the following Problem:
                                        
                                          Strict Trs:
                                            {  gr(s(x), 0()) -> true()
                                             , gr(s(x), s(y)) -> gr(x, y)}
                                          Weak Trs:
                                            {  gr(0(), x) -> false()
                                             , and(false(), x) -> false()
                                             , and(x, false()) -> false()
                                             , p(s(x)) -> x
                                             , and(true(), true()) -> true()
                                             , 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:{6,9,8,7,14,13,10}: YES(O(1),O(1))
                                    -----------------------------------------
                                    
                                    We consider the following Problem:
                                    
                                      Strict Trs:
                                        {  gr(s(x), 0()) -> true()
                                         , gr(s(x), s(y)) -> gr(x, y)}
                                      Weak Trs:
                                        {  gr(0(), x) -> false()
                                         , and(false(), x) -> false()
                                         , and(x, false()) -> false()
                                         , p(s(x)) -> x
                                         , and(true(), true()) -> true()
                                         , p(0()) -> 0()}
                                      StartTerms: basic terms
                                      Strategy: innermost
                                    
                                    Certificate: YES(O(1),O(1))
                                    
                                    Proof:
                                      We consider the following Problem:
                                      
                                        Strict Trs:
                                          {  gr(s(x), 0()) -> true()
                                           , gr(s(x), s(y)) -> gr(x, y)}
                                        Weak Trs:
                                          {  gr(0(), x) -> false()
                                           , and(false(), x) -> false()
                                           , and(x, false()) -> false()
                                           , p(s(x)) -> x
                                           , and(true(), true()) -> true()
                                           , p(0()) -> 0()}
                                        StartTerms: basic terms
                                        Strategy: innermost
                                      
                                      Certificate: YES(O(1),O(1))
                                      
                                      Proof:
                                        We consider the following Problem:
                                        
                                          Strict Trs:
                                            {  gr(s(x), 0()) -> true()
                                             , gr(s(x), s(y)) -> gr(x, y)}
                                          Weak Trs:
                                            {  gr(0(), x) -> false()
                                             , and(false(), x) -> false()
                                             , and(x, false()) -> false()
                                             , p(s(x)) -> x
                                             , and(true(), true()) -> true()
                                             , 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:{11}: YES(O(1),O(1))
                                    ---------------------------
                                    
                                    We consider the following Problem:
                                    
                                      Strict Trs:
                                        {  gr(s(x), 0()) -> true()
                                         , gr(s(x), s(y)) -> gr(x, y)}
                                      Weak Trs:
                                        {  gr(0(), x) -> false()
                                         , and(false(), x) -> false()
                                         , and(x, false()) -> false()
                                         , p(s(x)) -> x
                                         , and(true(), true()) -> true()
                                         , p(0()) -> 0()}
                                      StartTerms: basic terms
                                      Strategy: innermost
                                    
                                    Certificate: YES(O(1),O(1))
                                    
                                    Proof:
                                      We consider the following Problem:
                                      
                                        Strict Trs:
                                          {  gr(s(x), 0()) -> true()
                                           , gr(s(x), s(y)) -> gr(x, y)}
                                        Weak Trs:
                                          {  gr(0(), x) -> false()
                                           , and(false(), x) -> false()
                                           , and(x, false()) -> false()
                                           , p(s(x)) -> x
                                           , and(true(), true()) -> true()
                                           , p(0()) -> 0()}
                                        StartTerms: basic terms
                                        Strategy: innermost
                                      
                                      Certificate: YES(O(1),O(1))
                                      
                                      Proof:
                                        We consider the following Problem:
                                        
                                          Strict Trs:
                                            {  gr(s(x), 0()) -> true()
                                             , gr(s(x), s(y)) -> gr(x, y)}
                                          Weak Trs:
                                            {  gr(0(), x) -> false()
                                             , and(false(), x) -> false()
                                             , and(x, false()) -> false()
                                             , p(s(x)) -> x
                                             , and(true(), true()) -> true()
                                             , 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:{12}: YES(O(1),O(1))
                                    ---------------------------
                                    
                                    We consider the following Problem:
                                    
                                      Strict Trs:
                                        {  gr(s(x), 0()) -> true()
                                         , gr(s(x), s(y)) -> gr(x, y)}
                                      Weak Trs:
                                        {  gr(0(), x) -> false()
                                         , and(false(), x) -> false()
                                         , and(x, false()) -> false()
                                         , p(s(x)) -> x
                                         , and(true(), true()) -> true()
                                         , p(0()) -> 0()}
                                      StartTerms: basic terms
                                      Strategy: innermost
                                    
                                    Certificate: YES(O(1),O(1))
                                    
                                    Proof:
                                      We consider the following Problem:
                                      
                                        Strict Trs:
                                          {  gr(s(x), 0()) -> true()
                                           , gr(s(x), s(y)) -> gr(x, y)}
                                        Weak Trs:
                                          {  gr(0(), x) -> false()
                                           , and(false(), x) -> false()
                                           , and(x, false()) -> false()
                                           , p(s(x)) -> x
                                           , and(true(), true()) -> true()
                                           , p(0()) -> 0()}
                                        StartTerms: basic terms
                                        Strategy: innermost
                                      
                                      Certificate: YES(O(1),O(1))
                                      
                                      Proof:
                                        We consider the following Problem:
                                        
                                          Strict Trs:
                                            {  gr(s(x), 0()) -> true()
                                             , gr(s(x), s(y)) -> gr(x, y)}
                                          Weak Trs:
                                            {  gr(0(), x) -> false()
                                             , and(false(), x) -> false()
                                             , and(x, false()) -> false()
                                             , p(s(x)) -> x
                                             , and(true(), true()) -> true()
                                             , 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:
                                        {  gr(s(x), 0()) -> true()
                                         , gr(s(x), s(y)) -> gr(x, y)}
                                      Weak Trs:
                                        {  gr(0(), x) -> false()
                                         , and(false(), x) -> false()
                                         , and(x, false()) -> false()
                                         , p(s(x)) -> x
                                         , and(true(), true()) -> true()
                                         , p(0()) -> 0()}
                                      StartTerms: basic terms
                                      Strategy: innermost
                                    
                                    Certificate: YES(O(1),O(1))
                                    
                                    Proof:
                                      We consider the following Problem:
                                      
                                        Strict Trs:
                                          {  gr(s(x), 0()) -> true()
                                           , gr(s(x), s(y)) -> gr(x, y)}
                                        Weak Trs:
                                          {  gr(0(), x) -> false()
                                           , and(false(), x) -> false()
                                           , and(x, false()) -> false()
                                           , p(s(x)) -> x
                                           , and(true(), true()) -> true()
                                           , p(0()) -> 0()}
                                        StartTerms: basic terms
                                        Strategy: innermost
                                      
                                      Certificate: YES(O(1),O(1))
                                      
                                      Proof:
                                        We consider the following Problem:
                                        
                                          Strict Trs:
                                            {  gr(s(x), 0()) -> true()
                                             , gr(s(x), s(y)) -> gr(x, y)}
                                          Weak Trs:
                                            {  gr(0(), x) -> false()
                                             , and(false(), x) -> false()
                                             , and(x, false()) -> false()
                                             , p(s(x)) -> x
                                             , and(true(), true()) -> true()
                                             , 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))