We consider the following Problem:

  Strict Trs:
    {  2nd(cons1(X, cons(Y, Z))) -> Y
     , 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1)))
     , from(X) -> cons(X, n__from(s(X)))
     , from(X) -> n__from(X)
     , activate(n__from(X)) -> from(X)
     , activate(X) -> X}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  2nd(cons1(X, cons(Y, Z))) -> Y
       , 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1)))
       , from(X) -> cons(X, n__from(s(X)))
       , from(X) -> n__from(X)
       , activate(n__from(X)) -> from(X)
       , 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)))
       , from(X) -> n__from(X)}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(2nd) = {1}, Uargs(cons1) = {2}, Uargs(cons) = {},
        Uargs(activate) = {}, Uargs(from) = {}, Uargs(n__from) = {},
        Uargs(s) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       2nd(x1) = [1 1] x1 + [1]
                 [0 0]      [1]
       cons1(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                       [0 0]      [0 0]      [0]
       cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                      [0 0]      [1 0]      [0]
       activate(x1) = [1 0] x1 + [0]
                      [0 0]      [1]
       from(x1) = [1 0] x1 + [2]
                  [0 0]      [0]
       n__from(x1) = [1 0] x1 + [0]
                     [0 0]      [0]
       s(x1) = [0 0] x1 + [0]
               [0 0]      [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  2nd(cons1(X, cons(Y, Z))) -> Y
         , 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1)))
         , activate(n__from(X)) -> from(X)
         , activate(X) -> X}
      Weak Trs:
        {  from(X) -> cons(X, n__from(s(X)))
         , from(X) -> n__from(X)}
      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)
         , activate(X) -> X}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(2nd) = {1}, Uargs(cons1) = {2}, Uargs(cons) = {},
          Uargs(activate) = {}, Uargs(from) = {}, Uargs(n__from) = {},
          Uargs(s) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         2nd(x1) = [1 1] x1 + [1]
                   [0 0]      [1]
         cons1(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                         [0 0]      [0 0]      [0]
         cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                        [0 0]      [1 0]      [0]
         activate(x1) = [1 0] x1 + [2]
                        [0 1]      [1]
         from(x1) = [1 0] x1 + [0]
                    [0 0]      [0]
         n__from(x1) = [1 0] x1 + [0]
                       [0 0]      [0]
         s(x1) = [0 0] x1 + [0]
                 [0 0]      [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  2nd(cons1(X, cons(Y, Z))) -> Y
           , 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1)))}
        Weak Trs:
          {  activate(n__from(X)) -> from(X)
           , activate(X) -> X
           , from(X) -> cons(X, n__from(s(X)))
           , from(X) -> n__from(X)}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1)))}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(2nd) = {1}, Uargs(cons1) = {2}, Uargs(cons) = {},
            Uargs(activate) = {}, Uargs(from) = {}, Uargs(n__from) = {},
            Uargs(s) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           2nd(x1) = [1 1] x1 + [1]
                     [0 0]      [1]
           cons1(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                           [0 0]      [0 0]      [0]
           cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                          [0 0]      [1 0]      [2]
           activate(x1) = [1 0] x1 + [0]
                          [0 1]      [3]
           from(x1) = [1 0] x1 + [0]
                      [0 0]      [2]
           n__from(x1) = [1 0] x1 + [0]
                         [0 0]      [0]
           s(x1) = [0 0] x1 + [0]
                   [0 0]      [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs: {2nd(cons1(X, cons(Y, Z))) -> Y}
          Weak Trs:
            {  2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1)))
             , activate(n__from(X)) -> from(X)
             , activate(X) -> X
             , from(X) -> cons(X, n__from(s(X)))
             , from(X) -> n__from(X)}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          The weightgap principle applies, where following rules are oriented strictly:
          
          TRS Component: {2nd(cons1(X, cons(Y, Z))) -> Y}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(2nd) = {1}, Uargs(cons1) = {2}, Uargs(cons) = {},
              Uargs(activate) = {}, Uargs(from) = {}, Uargs(n__from) = {},
              Uargs(s) = {}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             2nd(x1) = [1 0] x1 + [1]
                       [0 1]      [1]
             cons1(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                             [0 0]      [0 1]      [0]
             cons(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                            [0 1]      [0 1]      [0]
             activate(x1) = [1 0] x1 + [0]
                            [0 1]      [0]
             from(x1) = [1 0] x1 + [0]
                        [0 1]      [0]
             n__from(x1) = [1 0] x1 + [0]
                           [0 1]      [0]
             s(x1) = [0 0] x1 + [0]
                     [0 0]      [0]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Weak Trs:
              {  2nd(cons1(X, cons(Y, Z))) -> Y
               , 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1)))
               , activate(n__from(X)) -> from(X)
               , activate(X) -> X
               , from(X) -> cons(X, n__from(s(X)))
               , from(X) -> n__from(X)}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(O(1),O(1))
          
          Proof:
            We consider the following Problem:
            
              Weak Trs:
                {  2nd(cons1(X, cons(Y, Z))) -> Y
                 , 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1)))
                 , activate(n__from(X)) -> from(X)
                 , activate(X) -> X
                 , from(X) -> cons(X, n__from(s(X)))
                 , from(X) -> n__from(X)}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(O(1),O(1))
            
            Proof:
              Empty rules are trivially bounded

Hurray, we answered YES(?,O(n^1))