We consider the following Problem:

  Strict Trs:
    {  f(f(X, Y), Z) -> f(X, f(Y, Z))
     , f(X, f(Y, Z)) -> f(Y, Y)}
  StartTerms: basic terms
  Strategy: innermost

Certificate: YES(?,O(n^2))

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  f(f(X, Y), Z) -> f(X, f(Y, Z))
       , f(X, f(Y, Z)) -> f(Y, Y)}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^2))
  
  Proof:
    We consider the following Problem:
    
      Strict Trs:
        {  f(f(X, Y), Z) -> f(X, f(Y, Z))
         , f(X, f(Y, Z)) -> f(Y, Y)}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^2))
    
    Proof:
      The following argument positions are usable:
        Uargs(f) = {2}
      We have the following restricted  polynomial interpretation:
      Interpretation Functions:
       [f](x1, x2) = 1 + 2*x1 + 2*x1*x2 + 2*x1^2 + 2*x2

Hurray, we answered YES(?,O(n^2))