We consider the following Problem:

  Strict Trs:
    {  f(0(), y) -> 0()
     , f(s(x), y) -> f(f(x, y), y)}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  f(0(), y) -> 0()
       , f(s(x), y) -> f(f(x, y), y)}
    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(), y) -> 0()}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(f) = {1}, Uargs(s) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       f(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
                   [0 0]      [1 1]      [1]
       0() = [0]
             [0]
       s(x1) = [1 0] x1 + [0]
               [0 0]      [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs: {f(s(x), y) -> f(f(x, y), y)}
      Weak Trs: {f(0(), y) -> 0()}
      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(x), y) -> f(f(x, y), y)}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(f) = {1}, Uargs(s) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         f(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                     [0 0]      [1 1]      [1]
         0() = [0]
               [0]
         s(x1) = [1 0] x1 + [2]
                 [0 0]      [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Weak Trs:
          {  f(s(x), y) -> f(f(x, y), y)
           , f(0(), y) -> 0()}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(O(1),O(1))
      
      Proof:
        We consider the following Problem:
        
          Weak Trs:
            {  f(s(x), y) -> f(f(x, y), y)
             , f(0(), y) -> 0()}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(O(1),O(1))
        
        Proof:
          Empty rules are trivially bounded

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