YES

Problem:
 r1(a(x1)) -> a(a(a(r1(x1))))
 r2(a(x1)) -> a(a(a(r2(x1))))
 a(l1(x1)) -> l1(a(a(a(x1))))
 a(a(l2(x1))) -> l2(a(a(x1)))
 r1(b(x1)) -> l1(b(x1))
 r2(b(x1)) -> l2(a(b(x1)))
 b(l1(x1)) -> b(r2(x1))
 b(l2(x1)) -> b(r1(x1))
 a(a(x1)) -> x1

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
              [1 0 0]     [0]
   [l2](x0) = [0 0 1]x0 + [1]
              [0 0 0]     [0],
   
              [1 0 0]     [1]
   [r1](x0) = [0 0 1]x0 + [0]
              [0 1 0]     [0],
   
              [1 0 0]  
   [l1](x0) = [0 0 1]x0
              [0 1 0]  ,
   
             [1 1 0]     [0]
   [b](x0) = [0 0 0]x0 + [0]
             [0 0 0]     [1],
   
             [1 0 0]  
   [a](x0) = [0 0 1]x0
             [0 1 0]  ,
   
              [1 0 0]  
   [r2](x0) = [0 0 1]x0
              [0 1 0]  
  orientation:
                    [1]         [1]                  
   r1(a(x1)) = x1 + [0] >= x1 + [0] = a(a(a(r1(x1))))
                    [0]         [0]                  
   
                                         
   r2(a(x1)) = x1 >= x1 = a(a(a(r2(x1))))
                                         
   
                                         
   a(l1(x1)) = x1 >= x1 = l1(a(a(a(x1))))
                                         
   
                  [1 0 0]     [0]    [1 0 0]     [0]               
   a(a(l2(x1))) = [0 0 1]x1 + [1] >= [0 0 1]x1 + [1] = l2(a(a(x1)))
                  [0 0 0]     [0]    [0 0 0]     [0]               
   
               [1 1 0]     [1]    [1 1 0]     [0]            
   r1(b(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = l1(b(x1))
               [0 0 0]     [0]    [0 0 0]     [0]            
   
               [1 1 0]     [0]    [1 1 0]     [0]               
   r2(b(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = l2(a(b(x1)))
               [0 0 0]     [0]    [0 0 0]     [0]               
   
               [1 0 1]     [0]    [1 0 1]     [0]            
   b(l1(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(r2(x1))
               [0 0 0]     [1]    [0 0 0]     [1]            
   
               [1 0 1]     [1]    [1 0 1]     [1]            
   b(l2(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(r1(x1))
               [0 0 0]     [1]    [0 0 0]     [1]            
   
                           
   a(a(x1)) = x1 >= x1 = x1
                           
  problem:
   r1(a(x1)) -> a(a(a(r1(x1))))
   r2(a(x1)) -> a(a(a(r2(x1))))
   a(l1(x1)) -> l1(a(a(a(x1))))
   a(a(l2(x1))) -> l2(a(a(x1)))
   r2(b(x1)) -> l2(a(b(x1)))
   b(l1(x1)) -> b(r2(x1))
   b(l2(x1)) -> b(r1(x1))
   a(a(x1)) -> x1
  WPO Processor:
   algebra: Sum
   weight function:
    w0 = 0
    w(b) = 4
    w(l1) = 2
    w(l2) = w(r2) = 1
    w(r1) = w(a) = 0
   status function:
    st(b) = st(l2) = st(l1) = st(r2) = st(r1) = st(a) = [0]
   precedence:
    r1 > r2 > a > b ~ l2 ~ l1
   problem:
    
   Qed