YES

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

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 0 0]  
   [s](x0) = [0 1 0]x0
             [0 0 0]  ,
   
             [1 0 1]  
   [f](x0) = [0 1 0]x0
             [0 0 0]  ,
   
         [0]
   [0] = [1]
         [1]
  orientation:
            [1]    [0]         
   f(0()) = [1] >= [1] = s(0())
            [0]    [0]         
   
               [0]    [0]         
   f(s(0())) = [1] >= [1] = s(0())
               [0]    [0]         
   
                [1 0 0]     [1 0 0]              
   f(s(s(x))) = [0 1 0]x >= [0 1 0]x = f(f(s(x)))
                [0 0 0]     [0 0 0]              
  problem:
   f(s(0())) -> s(0())
   f(s(s(x))) -> f(f(s(x)))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 1]     [0]
    [s](x0) = [0 0 0]x0 + [1]
              [0 1 0]     [0],
    
              [1 0 1]  
    [f](x0) = [0 1 0]x0
              [0 1 0]  ,
    
          [0]
    [0] = [1]
          [0]
   orientation:
                [1]    [0]         
    f(s(0())) = [1] >= [1] = s(0())
                [1]    [1]         
    
                 [1 1 1]    [1]    [1 1 1]    [1]             
    f(s(s(x))) = [0 0 0]x + [1] >= [0 0 0]x + [1] = f(f(s(x)))
                 [0 0 0]    [1]    [0 0 0]    [1]             
   problem:
    f(s(s(x))) -> f(f(s(x)))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
               [1 0 1]     [0]
     [s](x0) = [0 0 0]x0 + [1]
               [0 1 0]     [0],
     
               [1 0 1]  
     [f](x0) = [0 0 0]x0
               [0 0 0]  
    orientation:
                  [1 1 1]    [1]    [1 1 1]              
     f(s(s(x))) = [0 0 0]x + [0] >= [0 0 0]x = f(f(s(x)))
                  [0 0 0]    [0]    [0 0 0]              
    problem:
     
    Qed