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:
 String Reversal Processor:
  1(2(1(x1))) -> 2(0(2(x1)))
  1(2(0(x1))) -> 2(0(1(x1)))
  1(2(L(x1))) -> 2(0(1(L(x1))))
  0(2(1(x1))) -> 1(0(2(x1)))
  R(2(1(x1))) -> R(1(0(2(x1))))
  0(2(0(x1))) -> 1(0(1(x1)))
  0(2(L(x1))) -> 1(0(1(L(x1))))
  R(2(0(x1))) -> R(1(0(1(x1))))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 0]  
    [R](x0) = [0 1 0]x0
              [0 0 0]  ,
    
              [1 1 0]     [0]
    [2](x0) = [0 0 0]x0 + [1]
              [0 0 0]     [0],
    
              [1 1 0]     [1]
    [L](x0) = [1 1 1]x0 + [1]
              [0 0 0]     [0],
    
              [1 0 0]  
    [1](x0) = [0 1 0]x0
              [0 1 0]  ,
    
              [1 0 0]  
    [0](x0) = [0 0 0]x0
              [0 0 0]  
   orientation:
                  [1 1 0]     [0]    [1 1 0]     [0]              
    1(2(1(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = 2(0(2(x1)))
                  [0 0 0]     [1]    [0 0 0]     [0]              
    
                  [1 0 0]     [0]    [1 0 0]     [0]              
    1(2(0(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = 2(0(1(x1)))
                  [0 0 0]     [1]    [0 0 0]     [0]              
    
                  [2 2 1]     [2]    [1 1 0]     [1]                 
    1(2(L(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = 2(0(1(L(x1))))
                  [0 0 0]     [1]    [0 0 0]     [0]                 
    
                  [1 1 0]      [1 1 0]                
    0(2(1(x1))) = [0 0 0]x1 >= [0 0 0]x1 = 1(0(2(x1)))
                  [0 0 0]      [0 0 0]                
    
                  [1 1 0]     [0]    [1 1 0]                   
    R(2(1(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 = R(1(0(2(x1))))
                  [0 0 0]     [0]    [0 0 0]                   
    
                  [1 0 0]      [1 0 0]                
    0(2(0(x1))) = [0 0 0]x1 >= [0 0 0]x1 = 1(0(1(x1)))
                  [0 0 0]      [0 0 0]                
    
                  [2 2 1]     [2]    [1 1 0]     [1]                 
    0(2(L(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = 1(0(1(L(x1))))
                  [0 0 0]     [0]    [0 0 0]     [0]                 
    
                  [1 0 0]     [0]    [1 0 0]                   
    R(2(0(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 = R(1(0(1(x1))))
                  [0 0 0]     [0]    [0 0 0]                   
   problem:
    1(2(1(x1))) -> 2(0(2(x1)))
    1(2(0(x1))) -> 2(0(1(x1)))
    0(2(1(x1))) -> 1(0(2(x1)))
    R(2(1(x1))) -> R(1(0(2(x1))))
    0(2(0(x1))) -> 1(0(1(x1)))
    R(2(0(x1))) -> R(1(0(1(x1))))
   String Reversal Processor:
    1(2(1(x1))) -> 2(0(2(x1)))
    0(2(1(x1))) -> 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)))
    0(2(R(x1))) -> 1(0(1(R(x1))))
    WPO Processor:
     algebra: Sum
     weight function:
      w0 = 0
      w(2) = 2
      w(1) = 1
      w(R) = w(0) = 0
     status function:
      st(R) = st(0) = st(2) = st(1) = [0]
     precedence:
      0 > 1 > R ~ 2
     problem:
      
     Qed