YES(?,O(n^2))

Problem:
 f(0()) -> s(0())
 f(s(0())) -> s(0())
 f(s(s(x))) -> f(f(s(x)))

Proof:
 Complexity Transformation Processor:
  strict:
   f(0()) -> s(0())
   f(s(0())) -> s(0())
   f(s(s(x))) -> f(f(s(x)))
  weak:
   
  Matrix Interpretation Processor:
   dimension: 1
   max_matrix:
    1
    interpretation:
     [s](x0) = x0 + 1,
     
     [f](x0) = x0,
     
     [0] = 1
    orientation:
     f(0()) = 1 >= 2 = s(0())
     
     f(s(0())) = 2 >= 2 = s(0())
     
     f(s(s(x))) = x + 2 >= x + 1 = f(f(s(x)))
    problem:
     strict:
      f(0()) -> s(0())
      f(s(0())) -> s(0())
     weak:
      f(s(s(x))) -> f(f(s(x)))
    Matrix Interpretation Processor:
     dimension: 1
     max_matrix:
      1
      interpretation:
       [s](x0) = x0 + 1,
       
       [f](x0) = x0 + 1,
       
       [0] = 1
      orientation:
       f(0()) = 2 >= 2 = s(0())
       
       f(s(0())) = 3 >= 2 = s(0())
       
       f(s(s(x))) = x + 3 >= x + 3 = f(f(s(x)))
      problem:
       strict:
        f(0()) -> s(0())
       weak:
        f(s(0())) -> s(0())
        f(s(s(x))) -> f(f(s(x)))
      Matrix Interpretation Processor:
       dimension: 2
       max_matrix:
        [1 1]
        [0 0]
        interpretation:
                   [1 0]  
         [s](x0) = [0 0]x0,
         
                   [1 1]  
         [f](x0) = [0 0]x0,
         
               [0]
         [0] = [1]
        orientation:
                  [1]    [0]         
         f(0()) = [0] >= [0] = s(0())
         
                     [0]    [0]         
         f(s(0())) = [0] >= [0] = s(0())
         
                      [1 0]     [1 0]              
         f(s(s(x))) = [0 0]x >= [0 0]x = f(f(s(x)))
        problem:
         strict:
          
         weak:
          f(0()) -> s(0())
          f(s(0())) -> s(0())
          f(s(s(x))) -> f(f(s(x)))
        Qed