YES

Problem:
 a(l(x1)) -> l(a(x1))
 r(a(a(x1))) -> a(a(r(x1)))
 b(l(x1)) -> b(a(r(x1)))
 r(b(x1)) -> l(b(x1))

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