YES

Problem:
 1(2(1(x1))) -> 2(0(2(x1)))
 0(2(1(x1))) -> 1(0(2(x1)))
 L(2(1(x1))) -> L(1(0(2(x1))))
 1(2(0(x1))) -> 2(0(1(x1)))
 1(2(R(x1))) -> 2(0(1(R(x1))))
 0(2(0(x1))) -> 1(0(1(x1)))
 L(2(0(x1))) -> L(1(0(1(x1))))
 0(2(R(x1))) -> 1(0(1(R(x1))))

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [0 0]  
   [R](x0) = [1 1]x0,
   
             [1  0 ]  
   [L](x0) = [-& 0 ]x0,
   
             [0  -&]  
   [0](x0) = [-& -&]x0,
   
             [0 0]  
   [2](x0) = [0 0]x0,
   
             [0  -&]  
   [1](x0) = [-& 1 ]x0
  orientation:
                 [0 1]      [0 0]                
   1(2(1(x1))) = [1 2]x1 >= [0 0]x1 = 2(0(2(x1)))
   
                 [0  1 ]      [0  0 ]                
   0(2(1(x1))) = [-& -&]x1 >= [-& -&]x1 = 1(0(2(x1)))
   
                 [1 2]      [1  1 ]                   
   L(2(1(x1))) = [0 1]x1 >= [-& -&]x1 = L(1(0(2(x1))))
   
                 [0  -&]      [0  -&]                
   1(2(0(x1))) = [1  -&]x1 >= [0  -&]x1 = 2(0(1(x1)))
   
                 [1 1]      [0 0]                   
   1(2(R(x1))) = [2 2]x1 >= [0 0]x1 = 2(0(1(R(x1))))
   
                 [0  -&]      [0  -&]                
   0(2(0(x1))) = [-& -&]x1 >= [-& -&]x1 = 1(0(1(x1)))
   
                 [1  -&]      [1  -&]                   
   L(2(0(x1))) = [0  -&]x1 >= [-& -&]x1 = L(1(0(1(x1))))
   
                 [1  1 ]      [0  0 ]                   
   0(2(R(x1))) = [-& -&]x1 >= [-& -&]x1 = 1(0(1(R(x1))))
  problem:
   1(2(1(x1))) -> 2(0(2(x1)))
   0(2(1(x1))) -> 1(0(2(x1)))
   L(2(1(x1))) -> L(1(0(2(x1))))
   1(2(0(x1))) -> 2(0(1(x1)))
   0(2(0(x1))) -> 1(0(1(x1)))
   L(2(0(x1))) -> L(1(0(1(x1))))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0 0]  
    [L](x0) = [1 0]x0,
    
              [0  -&]  
    [0](x0) = [0  -&]x0,
    
              [1 2]  
    [2](x0) = [3 0]x0,
    
              [0  -&]  
    [1](x0) = [1  2 ]x0
   orientation:
                  [3 4]      [3 4]                
    1(2(1(x1))) = [5 5]x1 >= [4 5]x1 = 2(0(2(x1)))
    
                  [3 4]      [1 2]                
    0(2(1(x1))) = [3 4]x1 >= [3 4]x1 = 1(0(2(x1)))
    
                  [3 4]      [3 4]                   
    L(2(1(x1))) = [4 5]x1 >= [3 4]x1 = L(1(0(2(x1))))
    
                  [2  -&]      [2  -&]                
    1(2(0(x1))) = [5  -&]x1 >= [3  -&]x1 = 2(0(1(x1)))
    
                  [2  -&]      [0  -&]                
    0(2(0(x1))) = [2  -&]x1 >= [2  -&]x1 = 1(0(1(x1)))
    
                  [3  -&]      [2  -&]                   
    L(2(0(x1))) = [3  -&]x1 >= [2  -&]x1 = L(1(0(1(x1))))
   problem:
    1(2(1(x1))) -> 2(0(2(x1)))
    0(2(1(x1))) -> 1(0(2(x1)))
    L(2(1(x1))) -> L(1(0(2(x1))))
    1(2(0(x1))) -> 2(0(1(x1)))
    0(2(0(x1))) -> 1(0(1(x1)))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [L](x0) = x0 + 2,
     
     [0](x0) = x0 + 1,
     
     [2](x0) = 4x0 + 1,
     
     [1](x0) = 2x0 + 1
    orientation:
     1(2(1(x1))) = 16x1 + 11 >= 16x1 + 9 = 2(0(2(x1)))
     
     0(2(1(x1))) = 8x1 + 6 >= 8x1 + 5 = 1(0(2(x1)))
     
     L(2(1(x1))) = 8x1 + 7 >= 8x1 + 7 = L(1(0(2(x1))))
     
     1(2(0(x1))) = 8x1 + 11 >= 8x1 + 9 = 2(0(1(x1)))
     
     0(2(0(x1))) = 4x1 + 6 >= 4x1 + 5 = 1(0(1(x1)))
    problem:
     L(2(1(x1))) -> L(1(0(2(x1))))
    Arctic Interpretation Processor:
     dimension: 2
     interpretation:
                [0  1 ]  
      [L](x0) = [-& 0 ]x0,
      
                [0  -&]  
      [0](x0) = [-& 1 ]x0,
      
                [0 1]  
      [2](x0) = [3 0]x0,
      
                [3 2]  
      [1](x0) = [0 0]x0
     orientation:
                    [7 6]      [6 4]                   
      L(2(1(x1))) = [6 5]x1 >= [4 1]x1 = L(1(0(2(x1))))
     problem:
      
     Qed