YES

Problem:
 f(0(x1)) -> s(0(x1))
 d(0(x1)) -> 0(x1)
 d(s(x1)) -> s(s(d(x1)))
 f(s(x1)) -> d(f(x1))

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [0  0 ]  
   [d](x0) = [-& 0 ]x0,
   
             [0  0 ]  
   [s](x0) = [-& 0 ]x0,
   
             [2 0]  
   [f](x0) = [0 0]x0,
   
             [0  2 ]  
   [0](x0) = [-& -&]x0
  orientation:
              [2 4]      [0  2 ]             
   f(0(x1)) = [0 2]x1 >= [-& -&]x1 = s(0(x1))
   
              [0  2 ]      [0  2 ]          
   d(0(x1)) = [-& -&]x1 >= [-& -&]x1 = 0(x1)
   
              [0  0 ]      [0  0 ]                
   d(s(x1)) = [-& 0 ]x1 >= [-& 0 ]x1 = s(s(d(x1)))
   
              [2 2]      [2 0]             
   f(s(x1)) = [0 0]x1 >= [0 0]x1 = d(f(x1))
  problem:
   d(0(x1)) -> 0(x1)
   d(s(x1)) -> s(s(d(x1)))
   f(s(x1)) -> d(f(x1))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0  0 ]  
    [d](x0) = [-& 1 ]x0,
    
              [0  -&]  
    [s](x0) = [0  0 ]x0,
    
              [1  0 ]  
    [f](x0) = [-& -&]x0,
    
              [0  -&]  
    [0](x0) = [3  0 ]x0
   orientation:
               [3 0]      [0  -&]          
    d(0(x1)) = [4 1]x1 >= [3  0 ]x1 = 0(x1)
    
               [0 0]      [0 0]                
    d(s(x1)) = [1 1]x1 >= [0 1]x1 = s(s(d(x1)))
    
               [1  0 ]      [1  0 ]             
    f(s(x1)) = [-& -&]x1 >= [-& -&]x1 = d(f(x1))
   problem:
    d(s(x1)) -> s(s(d(x1)))
    f(s(x1)) -> d(f(x1))
   String Reversal Processor:
    s(d(x1)) -> d(s(s(x1)))
    s(f(x1)) -> f(d(x1))
    Matrix Interpretation Processor: dim=3
     
     interpretation:
                [1 0 0]  
      [d](x0) = [0 1 1]x0
                [0 1 1]  ,
      
                [1 1 1]  
      [s](x0) = [0 1 0]x0
                [0 0 1]  ,
      
                [1 0 0]     [0]
      [f](x0) = [0 0 0]x0 + [1]
                [0 0 0]     [0]
     orientation:
                 [1 2 2]      [1 2 2]                
      s(d(x1)) = [0 1 1]x1 >= [0 1 1]x1 = d(s(s(x1)))
                 [0 1 1]      [0 1 1]                
      
                 [1 0 0]     [1]    [1 0 0]     [0]           
      s(f(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = f(d(x1))
                 [0 0 0]     [0]    [0 0 0]     [0]           
     problem:
      s(d(x1)) -> d(s(s(x1)))
     KBO Processor:
      weight function:
       w0 = 1
       w(d) = 1
       w(s) = 0
      precedence:
       s > d
      problem:
       
      Qed