We consider the following Problem:

  Strict Trs:
    {  2nd(cons(X, n__cons(Y, Z))) -> activate(Y)
     , from(X) -> cons(X, n__from(n__s(X)))
     , cons(X1, X2) -> n__cons(X1, X2)
     , from(X) -> n__from(X)
     , s(X) -> n__s(X)
     , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
     , activate(n__from(X)) -> from(activate(X))
     , activate(n__s(X)) -> s(activate(X))
     , activate(X) -> X}
  StartTerms: basic terms
  Strategy: innermost

Certificate: YES(?,O(n^2))

Proof:
  Arguments of following rules are not normal-forms:
  {2nd(cons(X, n__cons(Y, Z))) -> activate(Y)}
  
  All above mentioned rules can be savely removed.
  
  We consider the following Problem:
  
    Strict Trs:
      {  from(X) -> cons(X, n__from(n__s(X)))
       , cons(X1, X2) -> n__cons(X1, X2)
       , from(X) -> n__from(X)
       , s(X) -> n__s(X)
       , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
       , activate(n__from(X)) -> from(activate(X))
       , activate(n__s(X)) -> s(activate(X))
       , activate(X) -> X}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^2))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {cons(X1, X2) -> n__cons(X1, X2)}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(2nd) = {}, Uargs(cons) = {1}, Uargs(n__cons) = {},
        Uargs(activate) = {}, Uargs(from) = {1}, Uargs(n__from) = {},
        Uargs(n__s) = {}, Uargs(s) = {1}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       2nd(x1) = [0 0] x1 + [0]
                 [0 0]      [0]
       cons(x1, x2) = [1 1] x1 + [0 0] x2 + [1]
                      [0 0]      [0 0]      [1]
       n__cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                         [0 0]      [0 0]      [0]
       activate(x1) = [1 0] x1 + [1]
                      [0 0]      [0]
       from(x1) = [1 1] x1 + [0]
                  [0 0]      [1]
       n__from(x1) = [1 0] x1 + [0]
                     [0 0]      [0]
       n__s(x1) = [1 0] x1 + [0]
                  [0 0]      [0]
       s(x1) = [1 0] x1 + [0]
               [0 0]      [1]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  from(X) -> cons(X, n__from(n__s(X)))
         , from(X) -> n__from(X)
         , s(X) -> n__s(X)
         , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
         , activate(n__from(X)) -> from(activate(X))
         , activate(n__s(X)) -> s(activate(X))
         , activate(X) -> X}
      Weak Trs: {cons(X1, X2) -> n__cons(X1, X2)}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^2))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {s(X) -> n__s(X)}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(2nd) = {}, Uargs(cons) = {1}, Uargs(n__cons) = {},
          Uargs(activate) = {}, Uargs(from) = {1}, Uargs(n__from) = {},
          Uargs(n__s) = {}, Uargs(s) = {1}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         2nd(x1) = [0 0] x1 + [0]
                   [0 0]      [0]
         cons(x1, x2) = [1 1] x1 + [0 0] x2 + [1]
                        [0 0]      [0 0]      [1]
         n__cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                           [0 0]      [0 0]      [0]
         activate(x1) = [1 0] x1 + [1]
                        [0 0]      [0]
         from(x1) = [1 1] x1 + [0]
                    [0 0]      [1]
         n__from(x1) = [1 0] x1 + [0]
                       [0 0]      [0]
         n__s(x1) = [1 0] x1 + [0]
                    [0 0]      [0]
         s(x1) = [1 0] x1 + [2]
                 [0 0]      [1]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  from(X) -> cons(X, n__from(n__s(X)))
           , from(X) -> n__from(X)
           , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
           , activate(n__from(X)) -> from(activate(X))
           , activate(n__s(X)) -> s(activate(X))
           , activate(X) -> X}
        Weak Trs:
          {  s(X) -> n__s(X)
           , cons(X1, X2) -> n__cons(X1, X2)}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^2))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {from(X) -> cons(X, n__from(n__s(X)))}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(2nd) = {}, Uargs(cons) = {1}, Uargs(n__cons) = {},
            Uargs(activate) = {}, Uargs(from) = {1}, Uargs(n__from) = {},
            Uargs(n__s) = {}, Uargs(s) = {1}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           2nd(x1) = [0 0] x1 + [0]
                     [0 0]      [0]
           cons(x1, x2) = [1 1] x1 + [0 0] x2 + [1]
                          [0 0]      [1 0]      [0]
           n__cons(x1, x2) = [1 1] x1 + [0 0] x2 + [0]
                             [0 0]      [1 0]      [0]
           activate(x1) = [1 1] x1 + [0]
                          [0 0]      [0]
           from(x1) = [1 1] x1 + [2]
                      [0 0]      [2]
           n__from(x1) = [0 1] x1 + [0]
                         [1 0]      [0]
           n__s(x1) = [1 1] x1 + [0]
                      [0 0]      [1]
           s(x1) = [1 1] x1 + [0]
                   [0 0]      [1]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs:
            {  from(X) -> n__from(X)
             , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
             , activate(n__from(X)) -> from(activate(X))
             , activate(n__s(X)) -> s(activate(X))
             , activate(X) -> X}
          Weak Trs:
            {  from(X) -> cons(X, n__from(n__s(X)))
             , s(X) -> n__s(X)
             , cons(X1, X2) -> n__cons(X1, X2)}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^2))
        
        Proof:
          The weightgap principle applies, where following rules are oriented strictly:
          
          TRS Component: {from(X) -> n__from(X)}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(2nd) = {}, Uargs(cons) = {1}, Uargs(n__cons) = {},
              Uargs(activate) = {}, Uargs(from) = {1}, Uargs(n__from) = {},
              Uargs(n__s) = {}, Uargs(s) = {1}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             2nd(x1) = [1 0] x1 + [0]
                       [0 1]      [0]
             cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                            [0 1]      [0 0]      [1]
             n__cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                               [0 0]      [0 0]      [0]
             activate(x1) = [1 0] x1 + [1]
                            [0 0]      [1]
             from(x1) = [1 0] x1 + [2]
                        [0 1]      [1]
             n__from(x1) = [1 0] x1 + [0]
                           [0 0]      [0]
             n__s(x1) = [1 0] x1 + [0]
                        [0 0]      [0]
             s(x1) = [1 0] x1 + [0]
                     [0 0]      [1]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Strict Trs:
              {  activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
               , activate(n__from(X)) -> from(activate(X))
               , activate(n__s(X)) -> s(activate(X))
               , activate(X) -> X}
            Weak Trs:
              {  from(X) -> n__from(X)
               , from(X) -> cons(X, n__from(n__s(X)))
               , s(X) -> n__s(X)
               , cons(X1, X2) -> n__cons(X1, X2)}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(?,O(n^2))
          
          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(2nd) = {}, Uargs(cons) = {1}, Uargs(n__cons) = {},
                Uargs(activate) = {}, Uargs(from) = {1}, Uargs(n__from) = {},
                Uargs(n__s) = {}, Uargs(s) = {1}
              We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
              Interpretation Functions:
               2nd(x1) = [1 0] x1 + [0]
                         [0 1]      [0]
               cons(x1, x2) = [1 3] x1 + [0 0] x2 + [0]
                              [0 0]      [0 1]      [0]
               n__cons(x1, x2) = [1 3] x1 + [0 0] x2 + [0]
                                 [0 0]      [0 1]      [0]
               activate(x1) = [1 0] x1 + [2]
                              [0 1]      [2]
               from(x1) = [1 3] x1 + [0]
                          [0 0]      [0]
               n__from(x1) = [1 3] x1 + [0]
                             [0 0]      [0]
               n__s(x1) = [1 3] x1 + [1]
                          [0 0]      [0]
               s(x1) = [1 3] x1 + [1]
                       [0 0]      [0]
            
            The strictly oriented rules are moved into the weak component.
            
            We consider the following Problem:
            
              Strict Trs:
                {  activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
                 , activate(n__from(X)) -> from(activate(X))
                 , activate(n__s(X)) -> s(activate(X))}
              Weak Trs:
                {  activate(X) -> X
                 , from(X) -> n__from(X)
                 , from(X) -> cons(X, n__from(n__s(X)))
                 , s(X) -> n__s(X)
                 , cons(X1, X2) -> n__cons(X1, X2)}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(?,O(n^2))
            
            Proof:
              We consider the following Problem:
              
                Strict Trs:
                  {  activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
                   , activate(n__from(X)) -> from(activate(X))
                   , activate(n__s(X)) -> s(activate(X))}
                Weak Trs:
                  {  activate(X) -> X
                   , from(X) -> n__from(X)
                   , from(X) -> cons(X, n__from(n__s(X)))
                   , s(X) -> n__s(X)
                   , cons(X1, X2) -> n__cons(X1, X2)}
                StartTerms: basic terms
                Strategy: innermost
              
              Certificate: YES(?,O(n^2))
              
              Proof:
                The following argument positions are usable:
                  Uargs(cons) = {1}, Uargs(n__cons) = {}, Uargs(activate) = {},
                  Uargs(from) = {1}, Uargs(n__from) = {}, Uargs(n__s) = {},
                  Uargs(s) = {1}
                We have the following restricted  polynomial interpretation:
                Interpretation Functions:
                 [cons](x1, x2) = 3 + x1
                 [n__cons](x1, x2) = 3 + x1
                 [activate](x1) = 1 + x1 + 2*x1^2
                 [from](x1) = 3 + x1
                 [n__from](x1) = 2 + x1
                 [n__s](x1) = 2 + x1
                 [s](x1) = 2 + x1

Hurray, we answered YES(?,O(n^2))