We consider the following Problem:

  Strict Trs:
    {  minus(X, 0()) -> X
     , minus(s(X), s(Y)) -> p(minus(X, Y))
     , p(s(X)) -> X
     , div(0(), s(Y)) -> 0()
     , div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  minus(X, 0()) -> X
       , minus(s(X), s(Y)) -> p(minus(X, Y))
       , p(s(X)) -> X
       , div(0(), s(Y)) -> 0()
       , div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {div(0(), s(Y)) -> 0()}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(minus) = {}, Uargs(s) = {1}, Uargs(p) = {1}, Uargs(div) = {1}
      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]
                       [1 0]      [0 0]      [1]
       0() = [0]
             [0]
       s(x1) = [1 0] x1 + [0]
               [0 0]      [1]
       p(x1) = [1 0] x1 + [0]
               [0 0]      [1]
       div(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
                     [0 0]      [0 0]      [1]
    
    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)) -> p(minus(X, Y))
         , p(s(X)) -> X
         , div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))}
      Weak Trs: {div(0(), s(Y)) -> 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(s(X), s(Y)) -> p(minus(X, Y))}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(minus) = {}, Uargs(s) = {1}, Uargs(p) = {1}, Uargs(div) = {1}
        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]
                         [1 0]      [0 0]      [1]
         0() = [0]
               [0]
         s(x1) = [1 0] x1 + [2]
                 [0 0]      [1]
         p(x1) = [1 0] x1 + [0]
                 [0 0]      [1]
         div(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                       [0 0]      [0 0]      [1]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  minus(X, 0()) -> X
           , p(s(X)) -> X
           , div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))}
        Weak Trs:
          {  minus(s(X), s(Y)) -> p(minus(X, Y))
           , div(0(), s(Y)) -> 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, 0()) -> X}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(minus) = {}, Uargs(s) = {1}, Uargs(p) = {1}, Uargs(div) = {1}
          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]
                           [1 1]      [0 0]      [1]
           0() = [0]
                 [0]
           s(x1) = [1 0] x1 + [0]
                   [0 0]      [1]
           p(x1) = [1 0] x1 + [0]
                   [1 0]      [1]
           div(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                         [0 0]      [0 0]      [1]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs:
            {  p(s(X)) -> X
             , div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))}
          Weak Trs:
            {  minus(X, 0()) -> X
             , minus(s(X), s(Y)) -> p(minus(X, Y))
             , div(0(), s(Y)) -> 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(minus) = {}, Uargs(s) = {1}, Uargs(p) = {1}, Uargs(div) = {1}
            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 1]      [0 0]      [1]
             0() = [0]
                   [0]
             s(x1) = [1 0] x1 + [1]
                     [0 1]      [0]
             p(x1) = [1 0] x1 + [1]
                     [0 1]      [0]
             div(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                           [0 0]      [1 0]      [0]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Strict Trs: {div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))}
            Weak Trs:
              {  p(s(X)) -> X
               , minus(X, 0()) -> X
               , minus(s(X), s(Y)) -> p(minus(X, Y))
               , div(0(), s(Y)) -> 0()}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(?,O(n^1))
          
          Proof:
            We consider the following Problem:
            
              Strict Trs: {div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))}
              Weak Trs:
                {  p(s(X)) -> X
                 , minus(X, 0()) -> X
                 , minus(s(X), s(Y)) -> p(minus(X, Y))
                 , div(0(), s(Y)) -> 0()}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(?,O(n^1))
            
            Proof:
              We have computed the following dependency pairs
              
                Strict DPs: {div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))}
                Weak DPs:
                  {  p^#(s(X)) -> c_2()
                   , minus^#(X, 0()) -> c_3()
                   , minus^#(s(X), s(Y)) -> p^#(minus(X, Y))
                   , div^#(0(), s(Y)) -> c_5()}
              
              We consider the following Problem:
              
                Strict DPs: {div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))}
                Strict Trs: {div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))}
                Weak DPs:
                  {  p^#(s(X)) -> c_2()
                   , minus^#(X, 0()) -> c_3()
                   , minus^#(s(X), s(Y)) -> p^#(minus(X, Y))
                   , div^#(0(), s(Y)) -> c_5()}
                Weak Trs:
                  {  p(s(X)) -> X
                   , minus(X, 0()) -> X
                   , minus(s(X), s(Y)) -> p(minus(X, Y))
                   , div(0(), s(Y)) -> 0()}
                StartTerms: basic terms
                Strategy: innermost
              
              Certificate: YES(?,O(n^1))
              
              Proof:
                We replace strict/weak-rules by the corresponding usable rules:
                
                  Weak Usable Rules:
                    {  p(s(X)) -> X
                     , minus(X, 0()) -> X
                     , minus(s(X), s(Y)) -> p(minus(X, Y))}
                
                We consider the following Problem:
                
                  Strict DPs: {div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))}
                  Weak DPs:
                    {  p^#(s(X)) -> c_2()
                     , minus^#(X, 0()) -> c_3()
                     , minus^#(s(X), s(Y)) -> p^#(minus(X, Y))
                     , div^#(0(), s(Y)) -> c_5()}
                  Weak Trs:
                    {  p(s(X)) -> X
                     , minus(X, 0()) -> X
                     , minus(s(X), s(Y)) -> p(minus(X, Y))}
                  StartTerms: basic terms
                  Strategy: innermost
                
                Certificate: YES(?,O(n^1))
                
                Proof:
                  The weightgap principle applies, where following rules are oriented strictly:
                  
                  Dependency Pairs: {div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))}
                  
                  Interpretation of constant growth:
                  ----------------------------------
                    The following argument positions are usable:
                      Uargs(minus) = {}, Uargs(s) = {}, Uargs(p) = {1}, Uargs(div) = {},
                      Uargs(div^#) = {1}, Uargs(p^#) = {1}, Uargs(minus^#) = {}
                    We have the following constructor-based EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                    Interpretation Functions:
                     minus(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
                                     [0 1]      [1 0]      [1]
                     0() = [0]
                           [0]
                     s(x1) = [1 0] x1 + [2]
                             [0 1]      [0]
                     p(x1) = [1 0] x1 + [1]
                             [0 1]      [1]
                     div(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                                   [0 0]      [0 0]      [0]
                     div^#(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
                                     [0 0]      [0 0]      [1]
                     p^#(x1) = [1 0] x1 + [1]
                               [0 0]      [1]
                     c_2() = [0]
                             [0]
                     minus^#(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                                       [0 0]      [0 0]      [1]
                     c_3() = [0]
                             [0]
                     c_5() = [0]
                             [0]
                  
                  The strictly oriented rules are moved into the weak component.
                  
                  We consider the following Problem:
                  
                    Weak DPs:
                      {  div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))
                       , p^#(s(X)) -> c_2()
                       , minus^#(X, 0()) -> c_3()
                       , minus^#(s(X), s(Y)) -> p^#(minus(X, Y))
                       , div^#(0(), s(Y)) -> c_5()}
                    Weak Trs:
                      {  p(s(X)) -> X
                       , minus(X, 0()) -> X
                       , minus(s(X), s(Y)) -> p(minus(X, Y))}
                    StartTerms: basic terms
                    Strategy: innermost
                  
                  Certificate: YES(?,O(n^1))
                  
                  Proof:
                    We use following congruence DG for path analysis
                    
                    ->4:{1}                                                     [      subsumed      ]
                       |
                       `->5:{5}                                                 [   YES(O(1),O(1))   ]
                    
                    ->2:{3}                                                     [   YES(O(1),O(1))   ]
                    
                    ->1:{4}                                                     [      subsumed      ]
                       |
                       `->3:{2}                                                 [   YES(O(1),O(1))   ]
                    
                    
                    Here dependency-pairs are as follows:
                    
                    WeakDPs DPs:
                      {  1: div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))
                       , 2: p^#(s(X)) -> c_2()
                       , 3: minus^#(X, 0()) -> c_3()
                       , 4: minus^#(s(X), s(Y)) -> p^#(minus(X, Y))
                       , 5: div^#(0(), s(Y)) -> c_5()}
                    
                    * Path 4:{1}: subsumed
                      --------------------
                      
                      This path is subsumed by the proof of paths 4:{1}->5:{5}.
                    
                    * Path 4:{1}->5:{5}: YES(O(1),O(1))
                      ---------------------------------
                      
                      We consider the following Problem:
                      
                        Weak DPs: {div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))}
                        Weak Trs:
                          {  p(s(X)) -> X
                           , minus(X, 0()) -> X
                           , minus(s(X), s(Y)) -> p(minus(X, Y))}
                        StartTerms: basic terms
                        Strategy: innermost
                      
                      Certificate: YES(O(1),O(1))
                      
                      Proof:
                        We consider the the dependency-graph
                        
                          1: div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))
                             -->_1 div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y)) :1
                          
                        
                        together with the congruence-graph
                        
                          ->1:{1}                                                     Weak SCC
                          
                          
                          Here dependency-pairs are as follows:
                          
                          WeakDPs DPs:
                            {1: div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))}
                        
                        The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
                        
                          {1: div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))}
                        
                        We consider the following Problem:
                        
                          Weak Trs:
                            {  p(s(X)) -> X
                             , minus(X, 0()) -> X
                             , minus(s(X), s(Y)) -> p(minus(X, Y))}
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(O(1),O(1))
                        
                        Proof:
                          We consider the following Problem:
                          
                            Weak Trs:
                              {  p(s(X)) -> X
                               , minus(X, 0()) -> X
                               , minus(s(X), s(Y)) -> p(minus(X, Y))}
                            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:{3}: YES(O(1),O(1))
                      --------------------------
                      
                      We consider the following Problem:
                      
                        Weak Trs:
                          {  p(s(X)) -> X
                           , minus(X, 0()) -> X
                           , minus(s(X), s(Y)) -> p(minus(X, Y))}
                        StartTerms: basic terms
                        Strategy: innermost
                      
                      Certificate: YES(O(1),O(1))
                      
                      Proof:
                        We consider the following Problem:
                        
                          Weak Trs:
                            {  p(s(X)) -> X
                             , minus(X, 0()) -> X
                             , minus(s(X), s(Y)) -> p(minus(X, Y))}
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(O(1),O(1))
                        
                        Proof:
                          We consider the following Problem:
                          
                            Weak Trs:
                              {  p(s(X)) -> X
                               , minus(X, 0()) -> X
                               , minus(s(X), s(Y)) -> p(minus(X, Y))}
                            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:{2}.
                    
                    * Path 1:{4}->3:{2}: YES(O(1),O(1))
                      ---------------------------------
                      
                      We consider the following Problem:
                      
                        Weak DPs: {minus^#(s(X), s(Y)) -> p^#(minus(X, Y))}
                        Weak Trs:
                          {  p(s(X)) -> X
                           , minus(X, 0()) -> X
                           , minus(s(X), s(Y)) -> p(minus(X, Y))}
                        StartTerms: basic terms
                        Strategy: innermost
                      
                      Certificate: YES(O(1),O(1))
                      
                      Proof:
                        We consider the the dependency-graph
                        
                          1: minus^#(s(X), s(Y)) -> p^#(minus(X, Y))
                          
                        
                        together with the congruence-graph
                        
                          ->1:{1}                                                     Weak SCC
                          
                          
                          Here dependency-pairs are as follows:
                          
                          WeakDPs DPs:
                            {1: minus^#(s(X), s(Y)) -> p^#(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)) -> p^#(minus(X, Y))}
                        
                        We consider the following Problem:
                        
                          Weak Trs:
                            {  p(s(X)) -> X
                             , minus(X, 0()) -> X
                             , minus(s(X), s(Y)) -> p(minus(X, Y))}
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(O(1),O(1))
                        
                        Proof:
                          We consider the following Problem:
                          
                            Weak Trs:
                              {  p(s(X)) -> X
                               , minus(X, 0()) -> X
                               , minus(s(X), s(Y)) -> p(minus(X, Y))}
                            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))