YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [0  0 ]  
   [b](x0) = [-& -&]x0,
   
             [0 0]  
   [a](x0) = [0 2]x0
  orientation:
           [0 0]      [0  0 ]          
   a(x1) = [0 2]x1 >= [-& -&]x1 = b(x1)
   
              [0 2]      [0 2]                
   a(a(x1)) = [2 4]x1 >= [0 2]x1 = a(b(a(x1)))
   
              [0 0]      [0  0 ]                
   a(b(x1)) = [0 0]x1 >= [-& -&]x1 = b(b(b(x1)))
   
                 [2 4]      [2 4]                      
   a(a(a(x1))) = [4 6]x1 >= [4 6]x1 = a(a(b(a(a(x1)))))
   
                 [0 0]      [0 0]                      
   a(a(b(x1))) = [2 2]x1 >= [0 0]x1 = a(b(b(a(b(x1)))))
   
                 [0 2]      [0  2 ]                      
   a(b(a(x1))) = [0 2]x1 >= [-& -&]x1 = b(a(b(b(a(x1)))))
   
                 [0 0]      [0  0 ]                      
   a(b(b(x1))) = [0 0]x1 >= [-& -&]x1 = b(b(b(b(b(x1)))))
   
              [0  2 ]      [0  0 ]                
   b(a(x1)) = [-& -&]x1 >= [-& -&]x1 = b(b(b(x1)))
   
                 [0 2]      [0 0]                      
   a(b(a(x1))) = [0 2]x1 >= [0 0]x1 = a(b(b(a(b(x1)))))
   
                 [2  4 ]      [0  2 ]                      
   b(a(a(x1))) = [-& -&]x1 >= [-& -&]x1 = b(a(b(b(a(x1)))))
   
                 [0  2 ]      [0  0 ]                      
   b(b(a(x1))) = [-& -&]x1 >= [-& -&]x1 = b(b(b(b(b(x1)))))
  problem:
   a(x1) -> b(x1)
   a(a(x1)) -> a(b(a(x1)))
   a(b(x1)) -> b(b(b(x1)))
   a(a(a(x1))) -> a(a(b(a(a(x1)))))
   a(a(b(x1))) -> a(b(b(a(b(x1)))))
   a(b(a(x1))) -> b(a(b(b(a(x1)))))
   a(b(b(x1))) -> b(b(b(b(b(x1)))))
   b(a(x1)) -> b(b(b(x1)))
   a(b(a(x1))) -> a(b(b(a(b(x1)))))
   b(b(a(x1))) -> b(b(b(b(b(x1)))))
  String Reversal Processor:
   a(x1) -> b(x1)
   a(a(x1)) -> a(b(a(x1)))
   b(a(x1)) -> b(b(b(x1)))
   a(a(a(x1))) -> a(a(b(a(a(x1)))))
   b(a(a(x1))) -> b(a(b(b(a(x1)))))
   a(b(a(x1))) -> a(b(b(a(b(x1)))))
   b(b(a(x1))) -> b(b(b(b(b(x1)))))
   a(b(x1)) -> b(b(b(x1)))
   a(b(a(x1))) -> b(a(b(b(a(x1)))))
   a(b(b(x1))) -> b(b(b(b(b(x1)))))
   Matrix Interpretation Processor: dim=2
    
    interpretation:
               [1 0]  
     [b](x0) = [0 0]x0,
     
               [1 2]     [0]
     [a](x0) = [0 1]x0 + [1]
    orientation:
             [1 2]     [0]    [1 0]          
     a(x1) = [0 1]x1 + [1] >= [0 0]x1 = b(x1)
     
                [1 4]     [2]    [1 2]     [0]              
     a(a(x1)) = [0 1]x1 + [2] >= [0 0]x1 + [1] = a(b(a(x1)))
     
                [1 2]      [1 0]                
     b(a(x1)) = [0 0]x1 >= [0 0]x1 = b(b(b(x1)))
     
                   [1 6]     [6]    [1 4]     [4]                    
     a(a(a(x1))) = [0 1]x1 + [3] >= [0 0]x1 + [2] = a(a(b(a(a(x1)))))
     
                   [1 4]     [2]    [1 2]                      
     b(a(a(x1))) = [0 0]x1 + [0] >= [0 0]x1 = b(a(b(b(a(x1)))))
     
                   [1 2]     [0]    [1 0]     [0]                    
     a(b(a(x1))) = [0 0]x1 + [1] >= [0 0]x1 + [1] = a(b(b(a(b(x1)))))
     
                   [1 2]      [1 0]                      
     b(b(a(x1))) = [0 0]x1 >= [0 0]x1 = b(b(b(b(b(x1)))))
     
                [1 0]     [0]    [1 0]                
     a(b(x1)) = [0 0]x1 + [1] >= [0 0]x1 = b(b(b(x1)))
     
                   [1 2]     [0]    [1 2]                      
     a(b(a(x1))) = [0 0]x1 + [1] >= [0 0]x1 = b(a(b(b(a(x1)))))
     
                   [1 0]     [0]    [1 0]                      
     a(b(b(x1))) = [0 0]x1 + [1] >= [0 0]x1 = b(b(b(b(b(x1)))))
    problem:
     a(x1) -> b(x1)
     b(a(x1)) -> b(b(b(x1)))
     a(b(a(x1))) -> a(b(b(a(b(x1)))))
     b(b(a(x1))) -> b(b(b(b(b(x1)))))
     a(b(x1)) -> b(b(b(x1)))
     a(b(a(x1))) -> b(a(b(b(a(x1)))))
     a(b(b(x1))) -> b(b(b(b(b(x1)))))
    Arctic Interpretation Processor:
     dimension: 1
     interpretation:
      [b](x0) = x0,
      
      [a](x0) = 2x0
     orientation:
      a(x1) = 2x1 >= x1 = b(x1)
      
      b(a(x1)) = 2x1 >= x1 = b(b(b(x1)))
      
      a(b(a(x1))) = 4x1 >= 4x1 = a(b(b(a(b(x1)))))
      
      b(b(a(x1))) = 2x1 >= x1 = b(b(b(b(b(x1)))))
      
      a(b(x1)) = 2x1 >= x1 = b(b(b(x1)))
      
      a(b(a(x1))) = 4x1 >= 4x1 = b(a(b(b(a(x1)))))
      
      a(b(b(x1))) = 2x1 >= x1 = b(b(b(b(b(x1)))))
     problem:
      a(b(a(x1))) -> a(b(b(a(b(x1)))))
      a(b(a(x1))) -> b(a(b(b(a(x1)))))
     String Reversal Processor:
      a(b(a(x1))) -> b(a(b(b(a(x1)))))
      a(b(a(x1))) -> a(b(b(a(b(x1)))))
      Bounds Processor:
       bound: 0
       enrichment: match
       automaton:
        final states: {7,1}
        transitions:
         f20() -> 2*
         b0(10) -> 11*
         b0(2) -> 8*
         b0(9) -> 10*
         b0(4) -> 5*
         b0(6) -> 1*
         b0(3) -> 4*
         a0(5) -> 6*
         a0(2) -> 3*
         a0(11) -> 7*
         a0(8) -> 9*
         1 -> 3,9
         7 -> 3,9
       problem:
        
       Qed