We consider the following Problem:

  Strict Trs:
    {  f(0()) -> cons(0())
     , f(s(0())) -> f(p(s(0())))
     , p(s(X)) -> X}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  f(0()) -> cons(0())
       , f(s(0())) -> f(p(s(0())))
       , p(s(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: {p(s(X)) -> X}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(f) = {1}, Uargs(cons) = {}, Uargs(s) = {}, Uargs(p) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       f(x1) = [1 0] x1 + [1]
               [0 0]      [1]
       0() = [0]
             [0]
       cons(x1) = [1 0] x1 + [1]
                  [0 1]      [1]
       s(x1) = [1 0] x1 + [0]
               [0 1]      [0]
       p(x1) = [1 0] x1 + [3]
               [0 1]      [1]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  f(0()) -> cons(0())
         , f(s(0())) -> f(p(s(0())))}
      Weak Trs: {p(s(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: {f(0()) -> cons(0())}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(f) = {1}, Uargs(cons) = {}, Uargs(s) = {}, Uargs(p) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         f(x1) = [1 1] x1 + [0]
                 [0 0]      [1]
         0() = [0]
               [2]
         cons(x1) = [0 0] x1 + [1]
                    [0 0]      [0]
         s(x1) = [1 0] x1 + [0]
                 [0 1]      [2]
         p(x1) = [1 0] x1 + [1]
                 [0 1]      [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs: {f(s(0())) -> f(p(s(0())))}
        Weak Trs:
          {  f(0()) -> cons(0())
           , p(s(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: {f(s(0())) -> f(p(s(0())))}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(f) = {1}, Uargs(cons) = {}, Uargs(s) = {}, Uargs(p) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           f(x1) = [1 0] x1 + [0]
                   [0 0]      [1]
           0() = [0]
                 [0]
           cons(x1) = [0 0] x1 + [0]
                      [0 0]      [1]
           s(x1) = [0 1] x1 + [1]
                   [1 0]      [0]
           p(x1) = [0 1] x1 + [0]
                   [1 0]      [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Weak Trs:
            {  f(s(0())) -> f(p(s(0())))
             , f(0()) -> cons(0())
             , p(s(X)) -> X}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(O(1),O(1))
        
        Proof:
          We consider the following Problem:
          
            Weak Trs:
              {  f(s(0())) -> f(p(s(0())))
               , f(0()) -> cons(0())
               , p(s(X)) -> 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))