We consider the following Problem:

  Strict Trs:
    {  minus(x, x) -> 0()
     , minus(0(), x) -> 0()
     , minus(x, 0()) -> x
     , minus(s(x), s(y)) -> minus(x, y)
     , le(0(), y) -> true()
     , le(s(x), 0()) -> false()
     , le(s(x), s(y)) -> le(x, y)
     , quot(x, y) -> if_quot(minus(x, y), y, le(y, 0()), le(y, x))
     , if_quot(x, y, true(), z) -> divByZeroError()
     , if_quot(x, y, false(), true()) -> s(quot(x, y))
     , if_quot(x, y, false(), false()) -> 0()}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  minus(x, x) -> 0()
       , minus(0(), x) -> 0()
       , minus(x, 0()) -> x
       , minus(s(x), s(y)) -> minus(x, y)
       , le(0(), y) -> true()
       , le(s(x), 0()) -> false()
       , le(s(x), s(y)) -> le(x, y)
       , quot(x, y) -> if_quot(minus(x, y), y, le(y, 0()), le(y, x))
       , if_quot(x, y, true(), z) -> divByZeroError()
       , if_quot(x, y, false(), true()) -> s(quot(x, y))
       , if_quot(x, y, false(), false()) -> 0()}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component:
      {  minus(x, x) -> 0()
       , minus(0(), x) -> 0()
       , le(0(), y) -> true()
       , le(s(x), 0()) -> false()}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(minus) = {}, Uargs(s) = {1}, Uargs(le) = {},
        Uargs(quot) = {}, Uargs(if_quot) = {1, 3, 4}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       minus(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
                       [0 0]      [0 0]      [1]
       0() = [0]
             [0]
       s(x1) = [1 0] x1 + [0]
               [0 0]      [1]
       le(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
                    [0 0]      [0 0]      [1]
       true() = [0]
                [0]
       false() = [0]
                 [0]
       quot(x1, x2) = [1 0] x1 + [1 1] x2 + [0]
                      [0 0]      [0 0]      [0]
       if_quot(x1, x2, x3, x4) = [1 0] x1 + [1 1] x2 + [1 0] x3 + [1 0] x4 + [0]
                                 [0 0]      [0 0]      [0 0]      [0 0]      [1]
       divByZeroError() = [0]
                          [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  minus(x, 0()) -> x
         , minus(s(x), s(y)) -> minus(x, y)
         , le(s(x), s(y)) -> le(x, y)
         , quot(x, y) -> if_quot(minus(x, y), y, le(y, 0()), le(y, x))
         , if_quot(x, y, true(), z) -> divByZeroError()
         , if_quot(x, y, false(), true()) -> s(quot(x, y))
         , if_quot(x, y, false(), false()) -> 0()}
      Weak Trs:
        {  minus(x, x) -> 0()
         , minus(0(), x) -> 0()
         , le(0(), y) -> true()
         , le(s(x), 0()) -> false()}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component:
        {  if_quot(x, y, false(), true()) -> s(quot(x, y))
         , if_quot(x, y, false(), false()) -> 0()}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(minus) = {}, Uargs(s) = {1}, Uargs(le) = {},
          Uargs(quot) = {}, Uargs(if_quot) = {1, 3, 4}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         minus(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
                         [0 0]      [0 0]      [1]
         0() = [0]
               [0]
         s(x1) = [1 0] x1 + [0]
                 [0 0]      [1]
         le(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
                      [0 0]      [0 0]      [1]
         true() = [0]
                  [0]
         false() = [1]
                   [0]
         quot(x1, x2) = [1 0] x1 + [1 1] x2 + [0]
                        [0 0]      [0 0]      [0]
         if_quot(x1, x2, x3, x4) = [1 0] x1 + [1 1] x2 + [1 0] x3 + [1 0] x4 + [0]
                                   [0 0]      [0 0]      [0 0]      [0 0]      [1]
         divByZeroError() = [0]
                            [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  minus(x, 0()) -> x
           , minus(s(x), s(y)) -> minus(x, y)
           , le(s(x), s(y)) -> le(x, y)
           , quot(x, y) -> if_quot(minus(x, y), y, le(y, 0()), le(y, x))
           , if_quot(x, y, true(), z) -> divByZeroError()}
        Weak Trs:
          {  if_quot(x, y, false(), true()) -> s(quot(x, y))
           , if_quot(x, y, false(), false()) -> 0()
           , minus(x, x) -> 0()
           , minus(0(), x) -> 0()
           , le(0(), y) -> true()
           , le(s(x), 0()) -> false()}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {if_quot(x, y, true(), z) -> divByZeroError()}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(minus) = {}, Uargs(s) = {1}, Uargs(le) = {},
            Uargs(quot) = {}, Uargs(if_quot) = {1, 3, 4}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           minus(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
                           [0 0]      [0 0]      [1]
           0() = [0]
                 [0]
           s(x1) = [1 0] x1 + [0]
                   [0 0]      [1]
           le(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
                        [0 0]      [0 0]      [1]
           true() = [1]
                    [0]
           false() = [0]
                     [0]
           quot(x1, x2) = [1 0] x1 + [1 1] x2 + [0]
                          [0 0]      [0 0]      [0]
           if_quot(x1, x2, x3, x4) = [1 0] x1 + [1 1] x2 + [1 0] x3 + [1 0] x4 + [0]
                                     [0 0]      [0 0]      [0 0]      [0 0]      [1]
           divByZeroError() = [0]
                              [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs:
            {  minus(x, 0()) -> x
             , minus(s(x), s(y)) -> minus(x, y)
             , le(s(x), s(y)) -> le(x, y)
             , quot(x, y) -> if_quot(minus(x, y), y, le(y, 0()), le(y, x))}
          Weak Trs:
            {  if_quot(x, y, true(), z) -> divByZeroError()
             , if_quot(x, y, false(), true()) -> s(quot(x, y))
             , if_quot(x, y, false(), false()) -> 0()
             , minus(x, x) -> 0()
             , minus(0(), x) -> 0()
             , le(0(), y) -> true()
             , le(s(x), 0()) -> false()}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          The weightgap principle applies, where following rules are oriented strictly:
          
          TRS Component: {le(s(x), s(y)) -> le(x, y)}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(minus) = {}, Uargs(s) = {1}, Uargs(le) = {},
              Uargs(quot) = {}, Uargs(if_quot) = {1, 3, 4}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             minus(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                             [0 0]      [0 0]      [3]
             0() = [0]
                   [2]
             s(x1) = [1 0] x1 + [0]
                     [0 1]      [2]
             le(x1, x2) = [0 0] x1 + [0 2] x2 + [0]
                          [0 0]      [0 0]      [0]
             true() = [0]
                      [0]
             false() = [0]
                       [0]
             quot(x1, x2) = [1 2] x1 + [1 0] x2 + [0]
                            [0 0]      [0 0]      [0]
             if_quot(x1, x2, x3, x4) = [1 3] x1 + [1 0] x2 + [1 0] x3 + [1 0] x4 + [0]
                                       [0 0]      [0 0]      [0 0]      [0 0]      [3]
             divByZeroError() = [0]
                                [0]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Strict Trs:
              {  minus(x, 0()) -> x
               , minus(s(x), s(y)) -> minus(x, y)
               , quot(x, y) -> if_quot(minus(x, y), y, le(y, 0()), le(y, x))}
            Weak Trs:
              {  le(s(x), s(y)) -> le(x, y)
               , if_quot(x, y, true(), z) -> divByZeroError()
               , if_quot(x, y, false(), true()) -> s(quot(x, y))
               , if_quot(x, y, false(), false()) -> 0()
               , minus(x, x) -> 0()
               , minus(0(), x) -> 0()
               , le(0(), y) -> true()
               , le(s(x), 0()) -> false()}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(?,O(n^1))
          
          Proof:
            The weightgap principle applies, where following rules are oriented strictly:
            
            TRS Component: {minus(s(x), s(y)) -> minus(x, y)}
            
            Interpretation of nonconstant growth:
            -------------------------------------
              The following argument positions are usable:
                Uargs(minus) = {}, Uargs(s) = {1}, Uargs(le) = {},
                Uargs(quot) = {}, Uargs(if_quot) = {1, 3, 4}
              We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
              Interpretation Functions:
               minus(x1, x2) = [1 0] x1 + [0 0] x2 + [2]
                               [0 0]      [1 0]      [3]
               0() = [2]
                     [2]
               s(x1) = [1 0] x1 + [2]
                       [0 1]      [2]
               le(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
                            [1 1]      [1 1]      [0]
               true() = [0]
                        [0]
               false() = [1]
                         [0]
               quot(x1, x2) = [1 0] x1 + [1 1] x2 + [0]
                              [0 0]      [0 0]      [0]
               if_quot(x1, x2, x3, x4) = [1 0] x1 + [1 1] x2 + [1 0] x3 + [1 0] x4 + [1]
                                         [0 0]      [0 0]      [0 0]      [0 0]      [3]
               divByZeroError() = [0]
                                  [0]
            
            The strictly oriented rules are moved into the weak component.
            
            We consider the following Problem:
            
              Strict Trs:
                {  minus(x, 0()) -> x
                 , quot(x, y) -> if_quot(minus(x, y), y, le(y, 0()), le(y, x))}
              Weak Trs:
                {  minus(s(x), s(y)) -> minus(x, y)
                 , le(s(x), s(y)) -> le(x, y)
                 , if_quot(x, y, true(), z) -> divByZeroError()
                 , if_quot(x, y, false(), true()) -> s(quot(x, y))
                 , if_quot(x, y, false(), false()) -> 0()
                 , minus(x, x) -> 0()
                 , minus(0(), x) -> 0()
                 , le(0(), y) -> true()
                 , le(s(x), 0()) -> false()}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(?,O(n^1))
            
            Proof:
              The weightgap principle applies, where following rules are oriented strictly:
              
              TRS Component: {minus(x, 0()) -> x}
              
              Interpretation of nonconstant growth:
              -------------------------------------
                The following argument positions are usable:
                  Uargs(minus) = {}, Uargs(s) = {1}, Uargs(le) = {},
                  Uargs(quot) = {}, Uargs(if_quot) = {1, 3, 4}
                We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                Interpretation Functions:
                 minus(x1, x2) = [1 0] x1 + [0 0] x2 + [3]
                                 [0 1]      [1 0]      [2]
                 0() = [2]
                       [2]
                 s(x1) = [1 0] x1 + [0]
                         [0 1]      [0]
                 le(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
                              [1 1]      [0 0]      [0]
                 true() = [0]
                          [0]
                 false() = [1]
                           [0]
                 quot(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                                [1 1]      [0 0]      [1]
                 if_quot(x1, x2, x3, x4) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1 0] x4 + [0]
                                           [1 1]      [1 1]      [1 0]      [1 0]      [2]
                 divByZeroError() = [0]
                                    [0]
              
              The strictly oriented rules are moved into the weak component.
              
              We consider the following Problem:
              
                Strict Trs:
                  {quot(x, y) -> if_quot(minus(x, y), y, le(y, 0()), le(y, x))}
                Weak Trs:
                  {  minus(x, 0()) -> x
                   , minus(s(x), s(y)) -> minus(x, y)
                   , le(s(x), s(y)) -> le(x, y)
                   , if_quot(x, y, true(), z) -> divByZeroError()
                   , if_quot(x, y, false(), true()) -> s(quot(x, y))
                   , if_quot(x, y, false(), false()) -> 0()
                   , minus(x, x) -> 0()
                   , minus(0(), x) -> 0()
                   , le(0(), y) -> true()
                   , le(s(x), 0()) -> false()}
                StartTerms: basic terms
                Strategy: innermost
              
              Certificate: YES(?,O(n^1))
              
              Proof:
                We consider the following Problem:
                
                  Strict Trs:
                    {quot(x, y) -> if_quot(minus(x, y), y, le(y, 0()), le(y, x))}
                  Weak Trs:
                    {  minus(x, 0()) -> x
                     , minus(s(x), s(y)) -> minus(x, y)
                     , le(s(x), s(y)) -> le(x, y)
                     , if_quot(x, y, true(), z) -> divByZeroError()
                     , if_quot(x, y, false(), true()) -> s(quot(x, y))
                     , if_quot(x, y, false(), false()) -> 0()
                     , minus(x, x) -> 0()
                     , minus(0(), x) -> 0()
                     , le(0(), y) -> true()
                     , le(s(x), 0()) -> false()}
                  StartTerms: basic terms
                  Strategy: innermost
                
                Certificate: YES(?,O(n^1))
                
                Proof:
                  We have computed the following dependency pairs
                  
                    Strict DPs:
                      {quot^#(x, y) -> if_quot^#(minus(x, y), y, le(y, 0()), le(y, x))}
                    Weak DPs:
                      {  minus^#(x, 0()) -> c_2()
                       , minus^#(s(x), s(y)) -> minus^#(x, y)
                       , le^#(s(x), s(y)) -> le^#(x, y)
                       , if_quot^#(x, y, true(), z) -> c_5()
                       , if_quot^#(x, y, false(), true()) -> quot^#(x, y)
                       , if_quot^#(x, y, false(), false()) -> c_7()
                       , minus^#(x, x) -> c_8()
                       , minus^#(0(), x) -> c_9()
                       , le^#(0(), y) -> c_10()
                       , le^#(s(x), 0()) -> c_11()}
                  
                  We consider the following Problem:
                  
                    Strict DPs:
                      {quot^#(x, y) -> if_quot^#(minus(x, y), y, le(y, 0()), le(y, x))}
                    Strict Trs:
                      {quot(x, y) -> if_quot(minus(x, y), y, le(y, 0()), le(y, x))}
                    Weak DPs:
                      {  minus^#(x, 0()) -> c_2()
                       , minus^#(s(x), s(y)) -> minus^#(x, y)
                       , le^#(s(x), s(y)) -> le^#(x, y)
                       , if_quot^#(x, y, true(), z) -> c_5()
                       , if_quot^#(x, y, false(), true()) -> quot^#(x, y)
                       , if_quot^#(x, y, false(), false()) -> c_7()
                       , minus^#(x, x) -> c_8()
                       , minus^#(0(), x) -> c_9()
                       , le^#(0(), y) -> c_10()
                       , le^#(s(x), 0()) -> c_11()}
                    Weak Trs:
                      {  minus(x, 0()) -> x
                       , minus(s(x), s(y)) -> minus(x, y)
                       , le(s(x), s(y)) -> le(x, y)
                       , if_quot(x, y, true(), z) -> divByZeroError()
                       , if_quot(x, y, false(), true()) -> s(quot(x, y))
                       , if_quot(x, y, false(), false()) -> 0()
                       , minus(x, x) -> 0()
                       , minus(0(), x) -> 0()
                       , le(0(), y) -> true()
                       , le(s(x), 0()) -> false()}
                    StartTerms: basic terms
                    Strategy: innermost
                  
                  Certificate: YES(?,O(n^1))
                  
                  Proof:
                    We replace strict/weak-rules by the corresponding usable rules:
                    
                      Weak Usable Rules:
                        {  minus(x, 0()) -> x
                         , minus(s(x), s(y)) -> minus(x, y)
                         , le(s(x), s(y)) -> le(x, y)
                         , minus(x, x) -> 0()
                         , minus(0(), x) -> 0()
                         , le(0(), y) -> true()
                         , le(s(x), 0()) -> false()}
                    
                    We consider the following Problem:
                    
                      Strict DPs:
                        {quot^#(x, y) -> if_quot^#(minus(x, y), y, le(y, 0()), le(y, x))}
                      Weak DPs:
                        {  minus^#(x, 0()) -> c_2()
                         , minus^#(s(x), s(y)) -> minus^#(x, y)
                         , le^#(s(x), s(y)) -> le^#(x, y)
                         , if_quot^#(x, y, true(), z) -> c_5()
                         , if_quot^#(x, y, false(), true()) -> quot^#(x, y)
                         , if_quot^#(x, y, false(), false()) -> c_7()
                         , minus^#(x, x) -> c_8()
                         , minus^#(0(), x) -> c_9()
                         , le^#(0(), y) -> c_10()
                         , le^#(s(x), 0()) -> c_11()}
                      Weak Trs:
                        {  minus(x, 0()) -> x
                         , minus(s(x), s(y)) -> minus(x, y)
                         , le(s(x), s(y)) -> le(x, y)
                         , minus(x, x) -> 0()
                         , minus(0(), x) -> 0()
                         , le(0(), y) -> true()
                         , le(s(x), 0()) -> false()}
                      StartTerms: basic terms
                      Strategy: innermost
                    
                    Certificate: YES(?,O(n^1))
                    
                    Proof:
                      We consider the following Problem:
                      
                        Strict DPs:
                          {quot^#(x, y) -> if_quot^#(minus(x, y), y, le(y, 0()), le(y, x))}
                        Weak DPs:
                          {  minus^#(x, 0()) -> c_2()
                           , minus^#(s(x), s(y)) -> minus^#(x, y)
                           , le^#(s(x), s(y)) -> le^#(x, y)
                           , if_quot^#(x, y, true(), z) -> c_5()
                           , if_quot^#(x, y, false(), true()) -> quot^#(x, y)
                           , if_quot^#(x, y, false(), false()) -> c_7()
                           , minus^#(x, x) -> c_8()
                           , minus^#(0(), x) -> c_9()
                           , le^#(0(), y) -> c_10()
                           , le^#(s(x), 0()) -> c_11()}
                        Weak Trs:
                          {  minus(x, 0()) -> x
                           , minus(s(x), s(y)) -> minus(x, y)
                           , le(s(x), s(y)) -> le(x, y)
                           , minus(x, x) -> 0()
                           , minus(0(), x) -> 0()
                           , le(0(), y) -> true()
                           , le(s(x), 0()) -> false()}
                        StartTerms: basic terms
                        Strategy: innermost
                      
                      Certificate: YES(?,O(n^1))
                      
                      Proof:
                        We use following congruence DG for path analysis
                        
                        ->8:{1,6}                                                   [   YES(O(1),O(1))   ]
                           |
                           |->9:{5}                                                 [   YES(O(1),O(1))   ]
                           |
                           `->10:{7}                                                [   YES(O(1),O(1))   ]
                        
                        ->4:{3}                                                     [      subsumed      ]
                           |
                           |->7:{2}                                                 [   YES(O(1),O(1))   ]
                           |
                           |->5:{8}                                                 [   YES(O(1),O(1))   ]
                           |
                           `->6:{9}                                                 [   YES(O(1),O(1))   ]
                        
                        ->1:{4}                                                     [      subsumed      ]
                           |
                           |->2:{10}                                                [   YES(O(1),O(1))   ]
                           |
                           `->3:{11}                                                [   YES(O(1),O(1))   ]
                        
                        
                        Here dependency-pairs are as follows:
                        
                        Strict DPs:
                          {1: quot^#(x, y) ->
                              if_quot^#(minus(x, y), y, le(y, 0()), le(y, x))}
                        WeakDPs DPs:
                          {  2: minus^#(x, 0()) -> c_2()
                           , 3: minus^#(s(x), s(y)) -> minus^#(x, y)
                           , 4: le^#(s(x), s(y)) -> le^#(x, y)
                           , 5: if_quot^#(x, y, true(), z) -> c_5()
                           , 6: if_quot^#(x, y, false(), true()) -> quot^#(x, y)
                           , 7: if_quot^#(x, y, false(), false()) -> c_7()
                           , 8: minus^#(x, x) -> c_8()
                           , 9: minus^#(0(), x) -> c_9()
                           , 10: le^#(0(), y) -> c_10()
                           , 11: le^#(s(x), 0()) -> c_11()}
                        
                        * Path 8:{1,6}: YES(O(1),O(1))
                          ----------------------------
                          
                          We consider the following Problem:
                          
                            Strict DPs:
                              {quot^#(x, y) -> if_quot^#(minus(x, y), y, le(y, 0()), le(y, x))}
                            Weak Trs:
                              {  minus(x, 0()) -> x
                               , minus(s(x), s(y)) -> minus(x, y)
                               , le(s(x), s(y)) -> le(x, y)
                               , minus(x, x) -> 0()
                               , minus(0(), x) -> 0()
                               , le(0(), y) -> true()
                               , le(s(x), 0()) -> false()}
                            StartTerms: basic terms
                            Strategy: innermost
                          
                          Certificate: YES(O(1),O(1))
                          
                          Proof:
                            We consider the the dependency-graph
                            
                              1: quot^#(x, y) -> if_quot^#(minus(x, y), y, le(y, 0()), le(y, x))
                              
                            
                            together with the congruence-graph
                            
                              ->1:{1}                                                     Noncyclic, trivial, SCC
                              
                              
                              Here dependency-pairs are as follows:
                              
                              Strict DPs:
                                {1: quot^#(x, y) ->
                                    if_quot^#(minus(x, y), y, le(y, 0()), le(y, x))}
                            
                            The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
                            
                              {1: quot^#(x, y) ->
                                  if_quot^#(minus(x, y), y, le(y, 0()), le(y, x))}
                            
                            We consider the following Problem:
                            
                              Weak Trs:
                                {  minus(x, 0()) -> x
                                 , minus(s(x), s(y)) -> minus(x, y)
                                 , le(s(x), s(y)) -> le(x, y)
                                 , minus(x, x) -> 0()
                                 , minus(0(), x) -> 0()
                                 , le(0(), y) -> true()
                                 , le(s(x), 0()) -> false()}
                              StartTerms: basic terms
                              Strategy: innermost
                            
                            Certificate: YES(O(1),O(1))
                            
                            Proof:
                              We consider the following Problem:
                              
                                Weak Trs:
                                  {  minus(x, 0()) -> x
                                   , minus(s(x), s(y)) -> minus(x, y)
                                   , le(s(x), s(y)) -> le(x, y)
                                   , minus(x, x) -> 0()
                                   , minus(0(), x) -> 0()
                                   , le(0(), y) -> true()
                                   , le(s(x), 0()) -> false()}
                                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 8:{1,6}->9:{5}: YES(O(1),O(1))
                          -----------------------------------
                          
                          We consider the following Problem:
                          
                            Weak DPs:
                              {  if_quot^#(x, y, false(), true()) -> quot^#(x, y)
                               , quot^#(x, y) -> if_quot^#(minus(x, y), y, le(y, 0()), le(y, x))}
                            Weak Trs:
                              {  minus(x, 0()) -> x
                               , minus(s(x), s(y)) -> minus(x, y)
                               , le(s(x), s(y)) -> le(x, y)
                               , minus(x, x) -> 0()
                               , minus(0(), x) -> 0()
                               , le(0(), y) -> true()
                               , le(s(x), 0()) -> false()}
                            StartTerms: basic terms
                            Strategy: innermost
                          
                          Certificate: YES(O(1),O(1))
                          
                          Proof:
                            We consider the the dependency-graph
                            
                              1: if_quot^#(x, y, false(), true()) -> quot^#(x, y)
                                 -->_1 quot^#(x, y) ->
                                       if_quot^#(minus(x, y), y, le(y, 0()), le(y, x)) :2
                              
                              2: quot^#(x, y) -> if_quot^#(minus(x, y), y, le(y, 0()), le(y, x))
                                 -->_1 if_quot^#(x, y, false(), true()) -> quot^#(x, y) :1
                              
                            
                            together with the congruence-graph
                            
                              ->1:{1,2}                                                   Weak SCC
                              
                              
                              Here dependency-pairs are as follows:
                              
                              WeakDPs DPs:
                                {  1: if_quot^#(x, y, false(), true()) -> quot^#(x, y)
                                 , 2: quot^#(x, y) ->
                                      if_quot^#(minus(x, y), y, le(y, 0()), le(y, x))}
                            
                            The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
                            
                              {  1: if_quot^#(x, y, false(), true()) -> quot^#(x, y)
                               , 2: quot^#(x, y) ->
                                    if_quot^#(minus(x, y), y, le(y, 0()), le(y, x))}
                            
                            We consider the following Problem:
                            
                              Weak Trs:
                                {  minus(x, 0()) -> x
                                 , minus(s(x), s(y)) -> minus(x, y)
                                 , le(s(x), s(y)) -> le(x, y)
                                 , minus(x, x) -> 0()
                                 , minus(0(), x) -> 0()
                                 , le(0(), y) -> true()
                                 , le(s(x), 0()) -> false()}
                              StartTerms: basic terms
                              Strategy: innermost
                            
                            Certificate: YES(O(1),O(1))
                            
                            Proof:
                              We consider the following Problem:
                              
                                Weak Trs:
                                  {  minus(x, 0()) -> x
                                   , minus(s(x), s(y)) -> minus(x, y)
                                   , le(s(x), s(y)) -> le(x, y)
                                   , minus(x, x) -> 0()
                                   , minus(0(), x) -> 0()
                                   , le(0(), y) -> true()
                                   , le(s(x), 0()) -> false()}
                                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 8:{1,6}->10:{7}: YES(O(1),O(1))
                          ------------------------------------
                          
                          We consider the following Problem:
                          
                            Weak DPs:
                              {  if_quot^#(x, y, false(), true()) -> quot^#(x, y)
                               , quot^#(x, y) -> if_quot^#(minus(x, y), y, le(y, 0()), le(y, x))}
                            Weak Trs:
                              {  minus(x, 0()) -> x
                               , minus(s(x), s(y)) -> minus(x, y)
                               , le(s(x), s(y)) -> le(x, y)
                               , minus(x, x) -> 0()
                               , minus(0(), x) -> 0()
                               , le(0(), y) -> true()
                               , le(s(x), 0()) -> false()}
                            StartTerms: basic terms
                            Strategy: innermost
                          
                          Certificate: YES(O(1),O(1))
                          
                          Proof:
                            We consider the the dependency-graph
                            
                              1: if_quot^#(x, y, false(), true()) -> quot^#(x, y)
                                 -->_1 quot^#(x, y) ->
                                       if_quot^#(minus(x, y), y, le(y, 0()), le(y, x)) :2
                              
                              2: quot^#(x, y) -> if_quot^#(minus(x, y), y, le(y, 0()), le(y, x))
                                 -->_1 if_quot^#(x, y, false(), true()) -> quot^#(x, y) :1
                              
                            
                            together with the congruence-graph
                            
                              ->1:{1,2}                                                   Weak SCC
                              
                              
                              Here dependency-pairs are as follows:
                              
                              WeakDPs DPs:
                                {  1: if_quot^#(x, y, false(), true()) -> quot^#(x, y)
                                 , 2: quot^#(x, y) ->
                                      if_quot^#(minus(x, y), y, le(y, 0()), le(y, x))}
                            
                            The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
                            
                              {  1: if_quot^#(x, y, false(), true()) -> quot^#(x, y)
                               , 2: quot^#(x, y) ->
                                    if_quot^#(minus(x, y), y, le(y, 0()), le(y, x))}
                            
                            We consider the following Problem:
                            
                              Weak Trs:
                                {  minus(x, 0()) -> x
                                 , minus(s(x), s(y)) -> minus(x, y)
                                 , le(s(x), s(y)) -> le(x, y)
                                 , minus(x, x) -> 0()
                                 , minus(0(), x) -> 0()
                                 , le(0(), y) -> true()
                                 , le(s(x), 0()) -> false()}
                              StartTerms: basic terms
                              Strategy: innermost
                            
                            Certificate: YES(O(1),O(1))
                            
                            Proof:
                              We consider the following Problem:
                              
                                Weak Trs:
                                  {  minus(x, 0()) -> x
                                   , minus(s(x), s(y)) -> minus(x, y)
                                   , le(s(x), s(y)) -> le(x, y)
                                   , minus(x, x) -> 0()
                                   , minus(0(), x) -> 0()
                                   , le(0(), y) -> true()
                                   , le(s(x), 0()) -> false()}
                                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:{3}: subsumed
                          --------------------
                          
                          This path is subsumed by the proof of paths 4:{3}->7:{2},
                                                                      4:{3}->6:{9},
                                                                      4:{3}->5:{8}.
                        
                        * Path 4:{3}->7:{2}: YES(O(1),O(1))
                          ---------------------------------
                          
                          We consider the following Problem:
                          
                            Weak DPs: {minus^#(s(x), s(y)) -> minus^#(x, y)}
                            Weak Trs:
                              {  minus(x, 0()) -> x
                               , minus(s(x), s(y)) -> minus(x, y)
                               , le(s(x), s(y)) -> le(x, y)
                               , minus(x, x) -> 0()
                               , minus(0(), x) -> 0()
                               , le(0(), y) -> true()
                               , le(s(x), 0()) -> false()}
                            StartTerms: basic terms
                            Strategy: innermost
                          
                          Certificate: YES(O(1),O(1))
                          
                          Proof:
                            We consider the the dependency-graph
                            
                              1: minus^#(s(x), s(y)) -> minus^#(x, y)
                                 -->_1 minus^#(s(x), s(y)) -> minus^#(x, y) :1
                              
                            
                            together with the congruence-graph
                            
                              ->1:{1}                                                     Weak SCC
                              
                              
                              Here dependency-pairs are as follows:
                              
                              WeakDPs DPs:
                                {1: minus^#(s(x), s(y)) -> minus^#(x, y)}
                            
                            The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
                            
                              {1: minus^#(s(x), s(y)) -> minus^#(x, y)}
                            
                            We consider the following Problem:
                            
                              Weak Trs:
                                {  minus(x, 0()) -> x
                                 , minus(s(x), s(y)) -> minus(x, y)
                                 , le(s(x), s(y)) -> le(x, y)
                                 , minus(x, x) -> 0()
                                 , minus(0(), x) -> 0()
                                 , le(0(), y) -> true()
                                 , le(s(x), 0()) -> false()}
                              StartTerms: basic terms
                              Strategy: innermost
                            
                            Certificate: YES(O(1),O(1))
                            
                            Proof:
                              We consider the following Problem:
                              
                                Weak Trs:
                                  {  minus(x, 0()) -> x
                                   , minus(s(x), s(y)) -> minus(x, y)
                                   , le(s(x), s(y)) -> le(x, y)
                                   , minus(x, x) -> 0()
                                   , minus(0(), x) -> 0()
                                   , le(0(), y) -> true()
                                   , le(s(x), 0()) -> false()}
                                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:{3}->5:{8}: YES(O(1),O(1))
                          ---------------------------------
                          
                          We consider the following Problem:
                          
                            Weak DPs: {minus^#(s(x), s(y)) -> minus^#(x, y)}
                            Weak Trs:
                              {  minus(x, 0()) -> x
                               , minus(s(x), s(y)) -> minus(x, y)
                               , le(s(x), s(y)) -> le(x, y)
                               , minus(x, x) -> 0()
                               , minus(0(), x) -> 0()
                               , le(0(), y) -> true()
                               , le(s(x), 0()) -> false()}
                            StartTerms: basic terms
                            Strategy: innermost
                          
                          Certificate: YES(O(1),O(1))
                          
                          Proof:
                            We consider the the dependency-graph
                            
                              1: minus^#(s(x), s(y)) -> minus^#(x, y)
                                 -->_1 minus^#(s(x), s(y)) -> minus^#(x, y) :1
                              
                            
                            together with the congruence-graph
                            
                              ->1:{1}                                                     Weak SCC
                              
                              
                              Here dependency-pairs are as follows:
                              
                              WeakDPs DPs:
                                {1: minus^#(s(x), s(y)) -> minus^#(x, y)}
                            
                            The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
                            
                              {1: minus^#(s(x), s(y)) -> minus^#(x, y)}
                            
                            We consider the following Problem:
                            
                              Weak Trs:
                                {  minus(x, 0()) -> x
                                 , minus(s(x), s(y)) -> minus(x, y)
                                 , le(s(x), s(y)) -> le(x, y)
                                 , minus(x, x) -> 0()
                                 , minus(0(), x) -> 0()
                                 , le(0(), y) -> true()
                                 , le(s(x), 0()) -> false()}
                              StartTerms: basic terms
                              Strategy: innermost
                            
                            Certificate: YES(O(1),O(1))
                            
                            Proof:
                              We consider the following Problem:
                              
                                Weak Trs:
                                  {  minus(x, 0()) -> x
                                   , minus(s(x), s(y)) -> minus(x, y)
                                   , le(s(x), s(y)) -> le(x, y)
                                   , minus(x, x) -> 0()
                                   , minus(0(), x) -> 0()
                                   , le(0(), y) -> true()
                                   , le(s(x), 0()) -> false()}
                                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:{3}->6:{9}: YES(O(1),O(1))
                          ---------------------------------
                          
                          We consider the following Problem:
                          
                            Weak DPs: {minus^#(s(x), s(y)) -> minus^#(x, y)}
                            Weak Trs:
                              {  minus(x, 0()) -> x
                               , minus(s(x), s(y)) -> minus(x, y)
                               , le(s(x), s(y)) -> le(x, y)
                               , minus(x, x) -> 0()
                               , minus(0(), x) -> 0()
                               , le(0(), y) -> true()
                               , le(s(x), 0()) -> false()}
                            StartTerms: basic terms
                            Strategy: innermost
                          
                          Certificate: YES(O(1),O(1))
                          
                          Proof:
                            We consider the the dependency-graph
                            
                              1: minus^#(s(x), s(y)) -> minus^#(x, y)
                                 -->_1 minus^#(s(x), s(y)) -> minus^#(x, y) :1
                              
                            
                            together with the congruence-graph
                            
                              ->1:{1}                                                     Weak SCC
                              
                              
                              Here dependency-pairs are as follows:
                              
                              WeakDPs DPs:
                                {1: minus^#(s(x), s(y)) -> minus^#(x, y)}
                            
                            The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
                            
                              {1: minus^#(s(x), s(y)) -> minus^#(x, y)}
                            
                            We consider the following Problem:
                            
                              Weak Trs:
                                {  minus(x, 0()) -> x
                                 , minus(s(x), s(y)) -> minus(x, y)
                                 , le(s(x), s(y)) -> le(x, y)
                                 , minus(x, x) -> 0()
                                 , minus(0(), x) -> 0()
                                 , le(0(), y) -> true()
                                 , le(s(x), 0()) -> false()}
                              StartTerms: basic terms
                              Strategy: innermost
                            
                            Certificate: YES(O(1),O(1))
                            
                            Proof:
                              We consider the following Problem:
                              
                                Weak Trs:
                                  {  minus(x, 0()) -> x
                                   , minus(s(x), s(y)) -> minus(x, y)
                                   , le(s(x), s(y)) -> le(x, y)
                                   , minus(x, x) -> 0()
                                   , minus(0(), x) -> 0()
                                   , le(0(), y) -> true()
                                   , le(s(x), 0()) -> false()}
                                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:{4}: subsumed
                          --------------------
                          
                          This path is subsumed by the proof of paths 1:{4}->3:{11},
                                                                      1:{4}->2:{10}.
                        
                        * Path 1:{4}->2:{10}: YES(O(1),O(1))
                          ----------------------------------
                          
                          We consider the following Problem:
                          
                            Weak DPs: {le^#(s(x), s(y)) -> le^#(x, y)}
                            Weak Trs:
                              {  minus(x, 0()) -> x
                               , minus(s(x), s(y)) -> minus(x, y)
                               , le(s(x), s(y)) -> le(x, y)
                               , minus(x, x) -> 0()
                               , minus(0(), x) -> 0()
                               , le(0(), y) -> true()
                               , le(s(x), 0()) -> false()}
                            StartTerms: basic terms
                            Strategy: innermost
                          
                          Certificate: YES(O(1),O(1))
                          
                          Proof:
                            We consider the the dependency-graph
                            
                              1: le^#(s(x), s(y)) -> le^#(x, y)
                                 -->_1 le^#(s(x), s(y)) -> le^#(x, y) :1
                              
                            
                            together with the congruence-graph
                            
                              ->1:{1}                                                     Weak SCC
                              
                              
                              Here dependency-pairs are as follows:
                              
                              WeakDPs DPs:
                                {1: le^#(s(x), s(y)) -> le^#(x, y)}
                            
                            The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
                            
                              {1: le^#(s(x), s(y)) -> le^#(x, y)}
                            
                            We consider the following Problem:
                            
                              Weak Trs:
                                {  minus(x, 0()) -> x
                                 , minus(s(x), s(y)) -> minus(x, y)
                                 , le(s(x), s(y)) -> le(x, y)
                                 , minus(x, x) -> 0()
                                 , minus(0(), x) -> 0()
                                 , le(0(), y) -> true()
                                 , le(s(x), 0()) -> false()}
                              StartTerms: basic terms
                              Strategy: innermost
                            
                            Certificate: YES(O(1),O(1))
                            
                            Proof:
                              We consider the following Problem:
                              
                                Weak Trs:
                                  {  minus(x, 0()) -> x
                                   , minus(s(x), s(y)) -> minus(x, y)
                                   , le(s(x), s(y)) -> le(x, y)
                                   , minus(x, x) -> 0()
                                   , minus(0(), x) -> 0()
                                   , le(0(), y) -> true()
                                   , le(s(x), 0()) -> false()}
                                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:{4}->3:{11}: YES(O(1),O(1))
                          ----------------------------------
                          
                          We consider the following Problem:
                          
                            Weak DPs: {le^#(s(x), s(y)) -> le^#(x, y)}
                            Weak Trs:
                              {  minus(x, 0()) -> x
                               , minus(s(x), s(y)) -> minus(x, y)
                               , le(s(x), s(y)) -> le(x, y)
                               , minus(x, x) -> 0()
                               , minus(0(), x) -> 0()
                               , le(0(), y) -> true()
                               , le(s(x), 0()) -> false()}
                            StartTerms: basic terms
                            Strategy: innermost
                          
                          Certificate: YES(O(1),O(1))
                          
                          Proof:
                            We consider the the dependency-graph
                            
                              1: le^#(s(x), s(y)) -> le^#(x, y)
                                 -->_1 le^#(s(x), s(y)) -> le^#(x, y) :1
                              
                            
                            together with the congruence-graph
                            
                              ->1:{1}                                                     Weak SCC
                              
                              
                              Here dependency-pairs are as follows:
                              
                              WeakDPs DPs:
                                {1: le^#(s(x), s(y)) -> le^#(x, y)}
                            
                            The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
                            
                              {1: le^#(s(x), s(y)) -> le^#(x, y)}
                            
                            We consider the following Problem:
                            
                              Weak Trs:
                                {  minus(x, 0()) -> x
                                 , minus(s(x), s(y)) -> minus(x, y)
                                 , le(s(x), s(y)) -> le(x, y)
                                 , minus(x, x) -> 0()
                                 , minus(0(), x) -> 0()
                                 , le(0(), y) -> true()
                                 , le(s(x), 0()) -> false()}
                              StartTerms: basic terms
                              Strategy: innermost
                            
                            Certificate: YES(O(1),O(1))
                            
                            Proof:
                              We consider the following Problem:
                              
                                Weak Trs:
                                  {  minus(x, 0()) -> x
                                   , minus(s(x), s(y)) -> minus(x, y)
                                   , le(s(x), s(y)) -> le(x, y)
                                   , minus(x, x) -> 0()
                                   , minus(0(), x) -> 0()
                                   , le(0(), y) -> true()
                                   , le(s(x), 0()) -> false()}
                                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))