We consider the following Problem:

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

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  f(f(a(), a()), x) -> f(a(), f(b(), f(a(), x)))
       , f(x, f(y, z)) -> f(f(x, y), z)}
    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(a(), a()), x) -> f(a(), f(b(), f(a(), x)))}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       f(x1, x2) = [1 2] x1 + [1 0] x2 + [1]
                   [0 0]      [0 0]      [2]
       a() = [1]
             [0]
       b() = [2]
             [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs: {f(x, f(y, z)) -> f(f(x, y), z)}
      Weak Trs: {f(f(a(), a()), x) -> f(a(), f(b(), f(a(), x)))}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      We consider the following Problem:
      
        Strict Trs: {f(x, f(y, z)) -> f(f(x, y), z)}
        Weak Trs: {f(f(a(), a()), x) -> f(a(), f(b(), f(a(), x)))}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The problem is match-bounded by 0.
        The enriched problem is compatible with the following automaton:
        {  f_0(2, 2) -> 1
         , a_0() -> 2
         , b_0() -> 2}

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