We consider the following Problem:

  Strict Trs:
    {  from(X) -> cons(X, n__from(s(X)))
     , first(0(), Z) -> nil()
     , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
     , sel(0(), cons(X, Z)) -> X
     , sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
     , from(X) -> n__from(X)
     , first(X1, X2) -> n__first(X1, X2)
     , activate(n__from(X)) -> from(X)
     , activate(n__first(X1, X2)) -> first(X1, X2)
     , activate(X) -> X}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  from(X) -> cons(X, n__from(s(X)))
       , first(0(), Z) -> nil()
       , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
       , sel(0(), cons(X, Z)) -> X
       , sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
       , from(X) -> n__from(X)
       , first(X1, X2) -> n__first(X1, X2)
       , activate(n__from(X)) -> from(X)
       , activate(n__first(X1, X2)) -> first(X1, X2)
       , activate(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:
      {  from(X) -> cons(X, n__from(s(X)))
       , first(0(), Z) -> nil()
       , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
       , sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(from) = {}, Uargs(cons) = {2}, Uargs(n__from) = {},
        Uargs(s) = {}, Uargs(first) = {}, Uargs(n__first) = {2},
        Uargs(activate) = {}, Uargs(sel) = {2}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       from(x1) = [1 1] x1 + [2]
                  [0 0]      [2]
       cons(x1, x2) = [1 1] x1 + [1 0] x2 + [1]
                      [0 0]      [0 1]      [1]
       n__from(x1) = [0 0] x1 + [0]
                     [1 1]      [0]
       s(x1) = [0 0] x1 + [0]
               [0 0]      [0]
       first(x1, x2) = [0 0] x1 + [1 1] x2 + [1]
                       [0 0]      [0 0]      [1]
       0() = [0]
             [0]
       nil() = [0]
               [0]
       n__first(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                          [0 0]      [0 1]      [0]
       activate(x1) = [1 1] x1 + [0]
                      [0 0]      [0]
       sel(x1, x2) = [0 0] x1 + [1 1] x2 + [1]
                     [0 0]      [0 0]      [1]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  sel(0(), cons(X, Z)) -> X
         , from(X) -> n__from(X)
         , first(X1, X2) -> n__first(X1, X2)
         , activate(n__from(X)) -> from(X)
         , activate(n__first(X1, X2)) -> first(X1, X2)
         , activate(X) -> X}
      Weak Trs:
        {  from(X) -> cons(X, n__from(s(X)))
         , first(0(), Z) -> nil()
         , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
         , sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component:
        {  from(X) -> n__from(X)
         , first(X1, X2) -> n__first(X1, X2)}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(from) = {}, Uargs(cons) = {2}, Uargs(n__from) = {},
          Uargs(s) = {}, Uargs(first) = {}, Uargs(n__first) = {2},
          Uargs(activate) = {}, Uargs(sel) = {2}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         from(x1) = [1 0] x1 + [2]
                    [1 0]      [2]
         cons(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                        [0 0]      [0 0]      [1]
         n__from(x1) = [1 0] x1 + [0]
                       [0 0]      [0]
         s(x1) = [0 0] x1 + [1]
                 [1 1]      [0]
         first(x1, x2) = [0 0] x1 + [1 0] x2 + [1]
                         [0 0]      [0 0]      [1]
         0() = [3]
               [1]
         nil() = [0]
                 [0]
         n__first(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                            [0 0]      [0 0]      [0]
         activate(x1) = [1 0] x1 + [0]
                        [1 0]      [1]
         sel(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                       [1 1]      [1 0]      [3]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  sel(0(), cons(X, Z)) -> X
           , activate(n__from(X)) -> from(X)
           , activate(n__first(X1, X2)) -> first(X1, X2)
           , activate(X) -> X}
        Weak Trs:
          {  from(X) -> n__from(X)
           , first(X1, X2) -> n__first(X1, X2)
           , from(X) -> cons(X, n__from(s(X)))
           , first(0(), Z) -> nil()
           , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
           , sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {activate(X) -> X}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(from) = {}, Uargs(cons) = {2}, Uargs(n__from) = {},
            Uargs(s) = {}, Uargs(first) = {}, Uargs(n__first) = {2},
            Uargs(activate) = {}, Uargs(sel) = {2}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           from(x1) = [1 0] x1 + [2]
                      [1 0]      [2]
           cons(x1, x2) = [1 0] x1 + [1 0] x2 + [2]
                          [0 0]      [0 0]      [0]
           n__from(x1) = [1 0] x1 + [0]
                         [0 0]      [1]
           s(x1) = [0 0] x1 + [0]
                   [0 0]      [0]
           first(x1, x2) = [0 0] x1 + [1 0] x2 + [2]
                           [0 0]      [0 0]      [1]
           0() = [0]
                 [0]
           nil() = [0]
                   [0]
           n__first(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                              [0 0]      [0 0]      [1]
           activate(x1) = [1 0] x1 + [2]
                          [0 1]      [0]
           sel(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                         [0 0]      [1 0]      [3]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs:
            {  sel(0(), cons(X, Z)) -> X
             , activate(n__from(X)) -> from(X)
             , activate(n__first(X1, X2)) -> first(X1, X2)}
          Weak Trs:
            {  activate(X) -> X
             , from(X) -> n__from(X)
             , first(X1, X2) -> n__first(X1, X2)
             , from(X) -> cons(X, n__from(s(X)))
             , first(0(), Z) -> nil()
             , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
             , sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          The weightgap principle applies, where following rules are oriented strictly:
          
          TRS Component: {activate(n__from(X)) -> from(X)}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(from) = {}, Uargs(cons) = {2}, Uargs(n__from) = {},
              Uargs(s) = {}, Uargs(first) = {}, Uargs(n__first) = {2},
              Uargs(activate) = {}, Uargs(sel) = {2}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             from(x1) = [1 0] x1 + [1]
                        [0 1]      [1]
             cons(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                            [0 1]      [0 0]      [1]
             n__from(x1) = [1 0] x1 + [1]
                           [0 1]      [1]
             s(x1) = [0 0] x1 + [0]
                     [0 1]      [3]
             first(x1, x2) = [0 0] x1 + [1 0] x2 + [1]
                             [0 0]      [0 1]      [1]
             0() = [0]
                   [0]
             nil() = [0]
                     [0]
             n__first(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                                [0 0]      [0 0]      [0]
             activate(x1) = [1 0] x1 + [1]
                            [0 1]      [0]
             sel(x1, x2) = [0 1] x1 + [1 0] x2 + [1]
                           [0 1]      [1 0]      [1]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Strict Trs:
              {  sel(0(), cons(X, Z)) -> X
               , activate(n__first(X1, X2)) -> first(X1, X2)}
            Weak Trs:
              {  activate(n__from(X)) -> from(X)
               , activate(X) -> X
               , from(X) -> n__from(X)
               , first(X1, X2) -> n__first(X1, X2)
               , from(X) -> cons(X, n__from(s(X)))
               , first(0(), Z) -> nil()
               , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
               , sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(?,O(n^1))
          
          Proof:
            The weightgap principle applies, where following rules are oriented strictly:
            
            TRS Component: {sel(0(), cons(X, Z)) -> X}
            
            Interpretation of nonconstant growth:
            -------------------------------------
              The following argument positions are usable:
                Uargs(from) = {}, Uargs(cons) = {2}, Uargs(n__from) = {},
                Uargs(s) = {}, Uargs(first) = {}, Uargs(n__first) = {2},
                Uargs(activate) = {}, Uargs(sel) = {2}
              We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
              Interpretation Functions:
               from(x1) = [1 0] x1 + [0]
                          [0 1]      [1]
               cons(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                              [0 1]      [0 1]      [0]
               n__from(x1) = [1 0] x1 + [0]
                             [0 1]      [1]
               s(x1) = [0 0] x1 + [0]
                       [0 0]      [0]
               first(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                               [0 0]      [0 1]      [1]
               0() = [0]
                     [0]
               nil() = [0]
                       [0]
               n__first(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                                  [0 0]      [0 0]      [0]
               activate(x1) = [1 0] x1 + [0]
                              [0 1]      [0]
               sel(x1, x2) = [0 0] x1 + [1 0] x2 + [1]
                             [0 0]      [0 1]      [1]
            
            The strictly oriented rules are moved into the weak component.
            
            We consider the following Problem:
            
              Strict Trs: {activate(n__first(X1, X2)) -> first(X1, X2)}
              Weak Trs:
                {  sel(0(), cons(X, Z)) -> X
                 , activate(n__from(X)) -> from(X)
                 , activate(X) -> X
                 , from(X) -> n__from(X)
                 , first(X1, X2) -> n__first(X1, X2)
                 , from(X) -> cons(X, n__from(s(X)))
                 , first(0(), Z) -> nil()
                 , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
                 , sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(?,O(n^1))
            
            Proof:
              We consider the following Problem:
              
                Strict Trs: {activate(n__first(X1, X2)) -> first(X1, X2)}
                Weak Trs:
                  {  sel(0(), cons(X, Z)) -> X
                   , activate(n__from(X)) -> from(X)
                   , activate(X) -> X
                   , from(X) -> n__from(X)
                   , first(X1, X2) -> n__first(X1, X2)
                   , from(X) -> cons(X, n__from(s(X)))
                   , first(0(), Z) -> nil()
                   , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
                   , sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))}
                StartTerms: basic terms
                Strategy: innermost
              
              Certificate: YES(?,O(n^1))
              
              Proof:
                We have computed the following dependency pairs
                
                  Strict DPs: {activate^#(n__first(X1, X2)) -> first^#(X1, X2)}
                  Weak DPs:
                    {  sel^#(0(), cons(X, Z)) -> c_2()
                     , activate^#(n__from(X)) -> from^#(X)
                     , activate^#(X) -> c_4()
                     , from^#(X) -> c_5()
                     , first^#(X1, X2) -> c_6()
                     , from^#(X) -> c_7()
                     , first^#(0(), Z) -> c_8()
                     , first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                     , sel^#(s(X), cons(Y, Z)) -> sel^#(X, activate(Z))}
                
                We consider the following Problem:
                
                  Strict DPs: {activate^#(n__first(X1, X2)) -> first^#(X1, X2)}
                  Strict Trs: {activate(n__first(X1, X2)) -> first(X1, X2)}
                  Weak DPs:
                    {  sel^#(0(), cons(X, Z)) -> c_2()
                     , activate^#(n__from(X)) -> from^#(X)
                     , activate^#(X) -> c_4()
                     , from^#(X) -> c_5()
                     , first^#(X1, X2) -> c_6()
                     , from^#(X) -> c_7()
                     , first^#(0(), Z) -> c_8()
                     , first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                     , sel^#(s(X), cons(Y, Z)) -> sel^#(X, activate(Z))}
                  Weak Trs:
                    {  sel(0(), cons(X, Z)) -> X
                     , activate(n__from(X)) -> from(X)
                     , activate(X) -> X
                     , from(X) -> n__from(X)
                     , first(X1, X2) -> n__first(X1, X2)
                     , from(X) -> cons(X, n__from(s(X)))
                     , first(0(), Z) -> nil()
                     , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
                     , sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))}
                  StartTerms: basic terms
                  Strategy: innermost
                
                Certificate: YES(?,O(n^1))
                
                Proof:
                  We replace strict/weak-rules by the corresponding usable rules:
                  
                    Strict Usable Rules: {activate(n__first(X1, X2)) -> first(X1, X2)}
                    Weak Usable Rules:
                      {  activate(n__from(X)) -> from(X)
                       , activate(X) -> X
                       , from(X) -> n__from(X)
                       , first(X1, X2) -> n__first(X1, X2)
                       , from(X) -> cons(X, n__from(s(X)))
                       , first(0(), Z) -> nil()
                       , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                  
                  We consider the following Problem:
                  
                    Strict DPs: {activate^#(n__first(X1, X2)) -> first^#(X1, X2)}
                    Strict Trs: {activate(n__first(X1, X2)) -> first(X1, X2)}
                    Weak DPs:
                      {  sel^#(0(), cons(X, Z)) -> c_2()
                       , activate^#(n__from(X)) -> from^#(X)
                       , activate^#(X) -> c_4()
                       , from^#(X) -> c_5()
                       , first^#(X1, X2) -> c_6()
                       , from^#(X) -> c_7()
                       , first^#(0(), Z) -> c_8()
                       , first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                       , sel^#(s(X), cons(Y, Z)) -> sel^#(X, activate(Z))}
                    Weak Trs:
                      {  activate(n__from(X)) -> from(X)
                       , activate(X) -> X
                       , from(X) -> n__from(X)
                       , first(X1, X2) -> n__first(X1, X2)
                       , from(X) -> cons(X, n__from(s(X)))
                       , first(0(), Z) -> nil()
                       , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                    StartTerms: basic terms
                    Strategy: innermost
                  
                  Certificate: YES(?,O(n^1))
                  
                  Proof:
                    The weightgap principle applies, where following rules are oriented strictly:
                    
                    TRS Component: {activate(n__first(X1, X2)) -> first(X1, X2)}
                    
                    Interpretation of constant growth:
                    ----------------------------------
                      The following argument positions are usable:
                        Uargs(from) = {}, Uargs(cons) = {2}, Uargs(n__from) = {},
                        Uargs(s) = {}, Uargs(first) = {}, Uargs(n__first) = {2},
                        Uargs(activate) = {}, Uargs(sel) = {}, Uargs(activate^#) = {},
                        Uargs(first^#) = {}, Uargs(sel^#) = {2}, Uargs(from^#) = {}
                      We have the following constructor-based EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                      Interpretation Functions:
                       from(x1) = [0 0] x1 + [0]
                                  [0 0]      [0]
                       cons(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                                      [0 0]      [0 0]      [0]
                       n__from(x1) = [0 0] x1 + [0]
                                     [0 0]      [0]
                       s(x1) = [1 0] x1 + [2]
                               [0 0]      [0]
                       first(x1, x2) = [1 0] x1 + [1 0] x2 + [2]
                                       [0 0]      [0 0]      [0]
                       0() = [0]
                             [0]
                       nil() = [0]
                               [0]
                       n__first(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                                          [0 0]      [0 0]      [0]
                       activate(x1) = [1 0] x1 + [3]
                                      [0 2]      [0]
                       sel(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                                     [0 0]      [0 0]      [0]
                       activate^#(x1) = [0 0] x1 + [0]
                                        [0 0]      [1]
                       first^#(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
                                         [0 0]      [0 0]      [1]
                       sel^#(x1, x2) = [3 0] x1 + [1 0] x2 + [2]
                                       [0 0]      [0 0]      [1]
                       c_2() = [0]
                               [0]
                       from^#(x1) = [0 0] x1 + [0]
                                    [0 0]      [0]
                       c_4() = [0]
                               [0]
                       c_5() = [0]
                               [0]
                       c_6() = [0]
                               [0]
                       c_7() = [0]
                               [0]
                       c_8() = [0]
                               [0]
                    
                    The strictly oriented rules are moved into the weak component.
                    
                    We consider the following Problem:
                    
                      Strict DPs: {activate^#(n__first(X1, X2)) -> first^#(X1, X2)}
                      Weak DPs:
                        {  sel^#(0(), cons(X, Z)) -> c_2()
                         , activate^#(n__from(X)) -> from^#(X)
                         , activate^#(X) -> c_4()
                         , from^#(X) -> c_5()
                         , first^#(X1, X2) -> c_6()
                         , from^#(X) -> c_7()
                         , first^#(0(), Z) -> c_8()
                         , first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                         , sel^#(s(X), cons(Y, Z)) -> sel^#(X, activate(Z))}
                      Weak Trs:
                        {  activate(n__first(X1, X2)) -> first(X1, X2)
                         , activate(n__from(X)) -> from(X)
                         , activate(X) -> X
                         , from(X) -> n__from(X)
                         , first(X1, X2) -> n__first(X1, X2)
                         , from(X) -> cons(X, n__from(s(X)))
                         , first(0(), Z) -> nil()
                         , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                      StartTerms: basic terms
                      Strategy: innermost
                    
                    Certificate: YES(?,O(n^1))
                    
                    Proof:
                      We use following congruence DG for path analysis
                      
                      ->3:{1,9}                                                   [   YES(O(1),O(1))   ]
                         |
                         |->6:{3}                                                 [      subsumed      ]
                         |   |
                         |   |->7:{5}                                             [   YES(O(1),O(1))   ]
                         |   |
                         |   `->8:{7}                                             [   YES(O(1),O(1))   ]
                         |
                         |->9:{4}                                                 [   YES(O(1),O(1))   ]
                         |
                         |->4:{6}                                                 [   YES(O(1),O(1))   ]
                         |
                         `->5:{8}                                                 [   YES(O(1),O(1))   ]
                      
                      ->1:{10}                                                    [      subsumed      ]
                         |
                         `->2:{2}                                                 [   YES(O(1),O(1))   ]
                      
                      
                      Here dependency-pairs are as follows:
                      
                      Strict DPs:
                        {1: activate^#(n__first(X1, X2)) -> first^#(X1, X2)}
                      WeakDPs DPs:
                        {  2: sel^#(0(), cons(X, Z)) -> c_2()
                         , 3: activate^#(n__from(X)) -> from^#(X)
                         , 4: activate^#(X) -> c_4()
                         , 5: from^#(X) -> c_5()
                         , 6: first^#(X1, X2) -> c_6()
                         , 7: from^#(X) -> c_7()
                         , 8: first^#(0(), Z) -> c_8()
                         , 9: first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                         , 10: sel^#(s(X), cons(Y, Z)) -> sel^#(X, activate(Z))}
                      
                      * Path 3:{1,9}: YES(O(1),O(1))
                        ----------------------------
                        
                        We consider the following Problem:
                        
                          Strict DPs: {activate^#(n__first(X1, X2)) -> first^#(X1, X2)}
                          Weak Trs:
                            {  activate(n__first(X1, X2)) -> first(X1, X2)
                             , activate(n__from(X)) -> from(X)
                             , activate(X) -> X
                             , from(X) -> n__from(X)
                             , first(X1, X2) -> n__first(X1, X2)
                             , from(X) -> cons(X, n__from(s(X)))
                             , first(0(), Z) -> nil()
                             , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(O(1),O(1))
                        
                        Proof:
                          We consider the the dependency-graph
                          
                            1: activate^#(n__first(X1, X2)) -> first^#(X1, X2)
                            
                          
                          together with the congruence-graph
                          
                            ->1:{1}                                                     Noncyclic, trivial, SCC
                            
                            
                            Here dependency-pairs are as follows:
                            
                            Strict DPs:
                              {1: activate^#(n__first(X1, X2)) -> first^#(X1, X2)}
                          
                          The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
                          
                            {1: activate^#(n__first(X1, X2)) -> first^#(X1, X2)}
                          
                          We consider the following Problem:
                          
                            Weak Trs:
                              {  activate(n__first(X1, X2)) -> first(X1, X2)
                               , activate(n__from(X)) -> from(X)
                               , activate(X) -> X
                               , from(X) -> n__from(X)
                               , first(X1, X2) -> n__first(X1, X2)
                               , from(X) -> cons(X, n__from(s(X)))
                               , first(0(), Z) -> nil()
                               , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                            StartTerms: basic terms
                            Strategy: innermost
                          
                          Certificate: YES(O(1),O(1))
                          
                          Proof:
                            We consider the following Problem:
                            
                              Weak Trs:
                                {  activate(n__first(X1, X2)) -> first(X1, X2)
                                 , activate(n__from(X)) -> from(X)
                                 , activate(X) -> X
                                 , from(X) -> n__from(X)
                                 , first(X1, X2) -> n__first(X1, X2)
                                 , from(X) -> cons(X, n__from(s(X)))
                                 , first(0(), Z) -> nil()
                                 , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                              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:{1,9}->6:{3}: subsumed
                        -----------------------------
                        
                        This path is subsumed by the proof of paths 3:{1,9}->6:{3}->8:{7},
                                                                    3:{1,9}->6:{3}->7:{5}.
                      
                      * Path 3:{1,9}->6:{3}->7:{5}: YES(O(1),O(1))
                        ------------------------------------------
                        
                        We consider the following Problem:
                        
                          Weak DPs:
                            {  first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                             , activate^#(n__first(X1, X2)) -> first^#(X1, X2)
                             , activate^#(n__from(X)) -> from^#(X)}
                          Weak Trs:
                            {  activate(n__first(X1, X2)) -> first(X1, X2)
                             , activate(n__from(X)) -> from(X)
                             , activate(X) -> X
                             , from(X) -> n__from(X)
                             , first(X1, X2) -> n__first(X1, X2)
                             , from(X) -> cons(X, n__from(s(X)))
                             , first(0(), Z) -> nil()
                             , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(O(1),O(1))
                        
                        Proof:
                          We consider the the dependency-graph
                          
                            1: first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                               -->_1 activate^#(n__first(X1, X2)) -> first^#(X1, X2) :2
                               -->_1 activate^#(n__from(X)) -> from^#(X) :3
                            
                            2: activate^#(n__first(X1, X2)) -> first^#(X1, X2)
                               -->_1 first^#(s(X), cons(Y, Z)) -> activate^#(Z) :1
                            
                            3: activate^#(n__from(X)) -> from^#(X)
                            
                          
                          together with the congruence-graph
                          
                            ->1:{1,2}                                                   Weak SCC
                               |
                               `->2:{3}                                                 Weak SCC
                            
                            
                            Here dependency-pairs are as follows:
                            
                            WeakDPs DPs:
                              {  1: first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                               , 2: activate^#(n__first(X1, X2)) -> first^#(X1, X2)
                               , 3: activate^#(n__from(X)) -> from^#(X)}
                          
                          The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
                          
                            {  1: first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                             , 2: activate^#(n__first(X1, X2)) -> first^#(X1, X2)
                             , 3: activate^#(n__from(X)) -> from^#(X)}
                          
                          We consider the following Problem:
                          
                            Weak Trs:
                              {  activate(n__first(X1, X2)) -> first(X1, X2)
                               , activate(n__from(X)) -> from(X)
                               , activate(X) -> X
                               , from(X) -> n__from(X)
                               , first(X1, X2) -> n__first(X1, X2)
                               , from(X) -> cons(X, n__from(s(X)))
                               , first(0(), Z) -> nil()
                               , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                            StartTerms: basic terms
                            Strategy: innermost
                          
                          Certificate: YES(O(1),O(1))
                          
                          Proof:
                            We consider the following Problem:
                            
                              Weak Trs:
                                {  activate(n__first(X1, X2)) -> first(X1, X2)
                                 , activate(n__from(X)) -> from(X)
                                 , activate(X) -> X
                                 , from(X) -> n__from(X)
                                 , first(X1, X2) -> n__first(X1, X2)
                                 , from(X) -> cons(X, n__from(s(X)))
                                 , first(0(), Z) -> nil()
                                 , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                              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:{1,9}->6:{3}->8:{7}: YES(O(1),O(1))
                        ------------------------------------------
                        
                        We consider the following Problem:
                        
                          Weak DPs:
                            {  first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                             , activate^#(n__first(X1, X2)) -> first^#(X1, X2)
                             , activate^#(n__from(X)) -> from^#(X)}
                          Weak Trs:
                            {  activate(n__first(X1, X2)) -> first(X1, X2)
                             , activate(n__from(X)) -> from(X)
                             , activate(X) -> X
                             , from(X) -> n__from(X)
                             , first(X1, X2) -> n__first(X1, X2)
                             , from(X) -> cons(X, n__from(s(X)))
                             , first(0(), Z) -> nil()
                             , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(O(1),O(1))
                        
                        Proof:
                          We consider the the dependency-graph
                          
                            1: first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                               -->_1 activate^#(n__first(X1, X2)) -> first^#(X1, X2) :2
                               -->_1 activate^#(n__from(X)) -> from^#(X) :3
                            
                            2: activate^#(n__first(X1, X2)) -> first^#(X1, X2)
                               -->_1 first^#(s(X), cons(Y, Z)) -> activate^#(Z) :1
                            
                            3: activate^#(n__from(X)) -> from^#(X)
                            
                          
                          together with the congruence-graph
                          
                            ->1:{1,2}                                                   Weak SCC
                               |
                               `->2:{3}                                                 Weak SCC
                            
                            
                            Here dependency-pairs are as follows:
                            
                            WeakDPs DPs:
                              {  1: first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                               , 2: activate^#(n__first(X1, X2)) -> first^#(X1, X2)
                               , 3: activate^#(n__from(X)) -> from^#(X)}
                          
                          The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
                          
                            {  1: first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                             , 2: activate^#(n__first(X1, X2)) -> first^#(X1, X2)
                             , 3: activate^#(n__from(X)) -> from^#(X)}
                          
                          We consider the following Problem:
                          
                            Weak Trs:
                              {  activate(n__first(X1, X2)) -> first(X1, X2)
                               , activate(n__from(X)) -> from(X)
                               , activate(X) -> X
                               , from(X) -> n__from(X)
                               , first(X1, X2) -> n__first(X1, X2)
                               , from(X) -> cons(X, n__from(s(X)))
                               , first(0(), Z) -> nil()
                               , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                            StartTerms: basic terms
                            Strategy: innermost
                          
                          Certificate: YES(O(1),O(1))
                          
                          Proof:
                            We consider the following Problem:
                            
                              Weak Trs:
                                {  activate(n__first(X1, X2)) -> first(X1, X2)
                                 , activate(n__from(X)) -> from(X)
                                 , activate(X) -> X
                                 , from(X) -> n__from(X)
                                 , first(X1, X2) -> n__first(X1, X2)
                                 , from(X) -> cons(X, n__from(s(X)))
                                 , first(0(), Z) -> nil()
                                 , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                              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:{1,9}->9:{4}: YES(O(1),O(1))
                        -----------------------------------
                        
                        We consider the following Problem:
                        
                          Weak DPs:
                            {  first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                             , activate^#(n__first(X1, X2)) -> first^#(X1, X2)}
                          Weak Trs:
                            {  activate(n__first(X1, X2)) -> first(X1, X2)
                             , activate(n__from(X)) -> from(X)
                             , activate(X) -> X
                             , from(X) -> n__from(X)
                             , first(X1, X2) -> n__first(X1, X2)
                             , from(X) -> cons(X, n__from(s(X)))
                             , first(0(), Z) -> nil()
                             , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(O(1),O(1))
                        
                        Proof:
                          We consider the the dependency-graph
                          
                            1: first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                               -->_1 activate^#(n__first(X1, X2)) -> first^#(X1, X2) :2
                            
                            2: activate^#(n__first(X1, X2)) -> first^#(X1, X2)
                               -->_1 first^#(s(X), cons(Y, Z)) -> activate^#(Z) :1
                            
                          
                          together with the congruence-graph
                          
                            ->1:{1,2}                                                   Weak SCC
                            
                            
                            Here dependency-pairs are as follows:
                            
                            WeakDPs DPs:
                              {  1: first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                               , 2: activate^#(n__first(X1, X2)) -> first^#(X1, X2)}
                          
                          The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
                          
                            {  1: first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                             , 2: activate^#(n__first(X1, X2)) -> first^#(X1, X2)}
                          
                          We consider the following Problem:
                          
                            Weak Trs:
                              {  activate(n__first(X1, X2)) -> first(X1, X2)
                               , activate(n__from(X)) -> from(X)
                               , activate(X) -> X
                               , from(X) -> n__from(X)
                               , first(X1, X2) -> n__first(X1, X2)
                               , from(X) -> cons(X, n__from(s(X)))
                               , first(0(), Z) -> nil()
                               , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                            StartTerms: basic terms
                            Strategy: innermost
                          
                          Certificate: YES(O(1),O(1))
                          
                          Proof:
                            We consider the following Problem:
                            
                              Weak Trs:
                                {  activate(n__first(X1, X2)) -> first(X1, X2)
                                 , activate(n__from(X)) -> from(X)
                                 , activate(X) -> X
                                 , from(X) -> n__from(X)
                                 , first(X1, X2) -> n__first(X1, X2)
                                 , from(X) -> cons(X, n__from(s(X)))
                                 , first(0(), Z) -> nil()
                                 , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                              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:{1,9}->4:{6}: YES(O(1),O(1))
                        -----------------------------------
                        
                        We consider the following Problem:
                        
                          Weak DPs:
                            {  first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                             , activate^#(n__first(X1, X2)) -> first^#(X1, X2)}
                          Weak Trs:
                            {  activate(n__first(X1, X2)) -> first(X1, X2)
                             , activate(n__from(X)) -> from(X)
                             , activate(X) -> X
                             , from(X) -> n__from(X)
                             , first(X1, X2) -> n__first(X1, X2)
                             , from(X) -> cons(X, n__from(s(X)))
                             , first(0(), Z) -> nil()
                             , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(O(1),O(1))
                        
                        Proof:
                          We consider the the dependency-graph
                          
                            1: first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                               -->_1 activate^#(n__first(X1, X2)) -> first^#(X1, X2) :2
                            
                            2: activate^#(n__first(X1, X2)) -> first^#(X1, X2)
                               -->_1 first^#(s(X), cons(Y, Z)) -> activate^#(Z) :1
                            
                          
                          together with the congruence-graph
                          
                            ->1:{1,2}                                                   Weak SCC
                            
                            
                            Here dependency-pairs are as follows:
                            
                            WeakDPs DPs:
                              {  1: first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                               , 2: activate^#(n__first(X1, X2)) -> first^#(X1, X2)}
                          
                          The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
                          
                            {  1: first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                             , 2: activate^#(n__first(X1, X2)) -> first^#(X1, X2)}
                          
                          We consider the following Problem:
                          
                            Weak Trs:
                              {  activate(n__first(X1, X2)) -> first(X1, X2)
                               , activate(n__from(X)) -> from(X)
                               , activate(X) -> X
                               , from(X) -> n__from(X)
                               , first(X1, X2) -> n__first(X1, X2)
                               , from(X) -> cons(X, n__from(s(X)))
                               , first(0(), Z) -> nil()
                               , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                            StartTerms: basic terms
                            Strategy: innermost
                          
                          Certificate: YES(O(1),O(1))
                          
                          Proof:
                            We consider the following Problem:
                            
                              Weak Trs:
                                {  activate(n__first(X1, X2)) -> first(X1, X2)
                                 , activate(n__from(X)) -> from(X)
                                 , activate(X) -> X
                                 , from(X) -> n__from(X)
                                 , first(X1, X2) -> n__first(X1, X2)
                                 , from(X) -> cons(X, n__from(s(X)))
                                 , first(0(), Z) -> nil()
                                 , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                              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:{1,9}->5:{8}: YES(O(1),O(1))
                        -----------------------------------
                        
                        We consider the following Problem:
                        
                          Weak DPs:
                            {  first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                             , activate^#(n__first(X1, X2)) -> first^#(X1, X2)}
                          Weak Trs:
                            {  activate(n__first(X1, X2)) -> first(X1, X2)
                             , activate(n__from(X)) -> from(X)
                             , activate(X) -> X
                             , from(X) -> n__from(X)
                             , first(X1, X2) -> n__first(X1, X2)
                             , from(X) -> cons(X, n__from(s(X)))
                             , first(0(), Z) -> nil()
                             , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(O(1),O(1))
                        
                        Proof:
                          We consider the the dependency-graph
                          
                            1: first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                               -->_1 activate^#(n__first(X1, X2)) -> first^#(X1, X2) :2
                            
                            2: activate^#(n__first(X1, X2)) -> first^#(X1, X2)
                               -->_1 first^#(s(X), cons(Y, Z)) -> activate^#(Z) :1
                            
                          
                          together with the congruence-graph
                          
                            ->1:{1,2}                                                   Weak SCC
                            
                            
                            Here dependency-pairs are as follows:
                            
                            WeakDPs DPs:
                              {  1: first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                               , 2: activate^#(n__first(X1, X2)) -> first^#(X1, X2)}
                          
                          The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
                          
                            {  1: first^#(s(X), cons(Y, Z)) -> activate^#(Z)
                             , 2: activate^#(n__first(X1, X2)) -> first^#(X1, X2)}
                          
                          We consider the following Problem:
                          
                            Weak Trs:
                              {  activate(n__first(X1, X2)) -> first(X1, X2)
                               , activate(n__from(X)) -> from(X)
                               , activate(X) -> X
                               , from(X) -> n__from(X)
                               , first(X1, X2) -> n__first(X1, X2)
                               , from(X) -> cons(X, n__from(s(X)))
                               , first(0(), Z) -> nil()
                               , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                            StartTerms: basic terms
                            Strategy: innermost
                          
                          Certificate: YES(O(1),O(1))
                          
                          Proof:
                            We consider the following Problem:
                            
                              Weak Trs:
                                {  activate(n__first(X1, X2)) -> first(X1, X2)
                                 , activate(n__from(X)) -> from(X)
                                 , activate(X) -> X
                                 , from(X) -> n__from(X)
                                 , first(X1, X2) -> n__first(X1, X2)
                                 , from(X) -> cons(X, n__from(s(X)))
                                 , first(0(), Z) -> nil()
                                 , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                              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:{10}: subsumed
                        ---------------------
                        
                        This path is subsumed by the proof of paths 1:{10}->2:{2}.
                      
                      * Path 1:{10}->2:{2}: YES(O(1),O(1))
                        ----------------------------------
                        
                        We consider the following Problem:
                        
                          Weak DPs: {sel^#(s(X), cons(Y, Z)) -> sel^#(X, activate(Z))}
                          Weak Trs:
                            {  activate(n__first(X1, X2)) -> first(X1, X2)
                             , activate(n__from(X)) -> from(X)
                             , activate(X) -> X
                             , from(X) -> n__from(X)
                             , first(X1, X2) -> n__first(X1, X2)
                             , from(X) -> cons(X, n__from(s(X)))
                             , first(0(), Z) -> nil()
                             , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                          StartTerms: basic terms
                          Strategy: innermost
                        
                        Certificate: YES(O(1),O(1))
                        
                        Proof:
                          We consider the the dependency-graph
                          
                            1: sel^#(s(X), cons(Y, Z)) -> sel^#(X, activate(Z))
                               -->_1 sel^#(s(X), cons(Y, Z)) -> sel^#(X, activate(Z)) :1
                            
                          
                          together with the congruence-graph
                          
                            ->1:{1}                                                     Weak SCC
                            
                            
                            Here dependency-pairs are as follows:
                            
                            WeakDPs DPs:
                              {1: sel^#(s(X), cons(Y, Z)) -> sel^#(X, activate(Z))}
                          
                          The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
                          
                            {1: sel^#(s(X), cons(Y, Z)) -> sel^#(X, activate(Z))}
                          
                          We consider the following Problem:
                          
                            Weak Trs:
                              {  activate(n__first(X1, X2)) -> first(X1, X2)
                               , activate(n__from(X)) -> from(X)
                               , activate(X) -> X
                               , from(X) -> n__from(X)
                               , first(X1, X2) -> n__first(X1, X2)
                               , from(X) -> cons(X, n__from(s(X)))
                               , first(0(), Z) -> nil()
                               , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                            StartTerms: basic terms
                            Strategy: innermost
                          
                          Certificate: YES(O(1),O(1))
                          
                          Proof:
                            We consider the following Problem:
                            
                              Weak Trs:
                                {  activate(n__first(X1, X2)) -> first(X1, X2)
                                 , activate(n__from(X)) -> from(X)
                                 , activate(X) -> X
                                 , from(X) -> n__from(X)
                                 , first(X1, X2) -> n__first(X1, X2)
                                 , from(X) -> cons(X, n__from(s(X)))
                                 , first(0(), Z) -> nil()
                                 , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))}
                              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))