We consider the following Problem:

  Strict Trs:
    {  zeros() -> cons(0(), n__zeros())
     , tail(cons(X, XS)) -> activate(XS)
     , zeros() -> n__zeros()
     , activate(n__zeros()) -> zeros()
     , activate(X) -> X}
  StartTerms: basic terms
  Strategy: innermost

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

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

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