We consider the following Problem:

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

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

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

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