YES

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

Proof:
 Matrix Interpretation Processor: dim=2
  
  interpretation:
             [1 0]  
   [b](x0) = [0 0]x0,
   
             [1 1]     [0]
   [a](x0) = [0 0]x0 + [1]
  orientation:
              [1 1]     [1]    [1 0]          
   a(a(x1)) = [0 0]x1 + [1] >= [0 0]x1 = b(x1)
   
                 [1 1]     [2]    [1 1]     [0]              
   a(a(a(x1))) = [0 0]x1 + [1] >= [0 0]x1 + [1] = a(b(a(x1)))
   
                 [1 1]     [0]    [1 0]                
   a(b(a(x1))) = [0 0]x1 + [1] >= [0 0]x1 = b(b(b(x1)))
   
                    [1 1]     [3]    [1 1]     [2]                    
   a(a(a(a(x1)))) = [0 0]x1 + [1] >= [0 0]x1 + [1] = a(a(b(a(a(x1)))))
   
                    [1 1]     [1]    [1 0]     [0]                    
   a(a(b(a(x1)))) = [0 0]x1 + [1] >= [0 0]x1 + [1] = a(b(b(a(b(x1)))))
   
                    [1 1]     [1]    [1 1]                      
   a(b(a(a(x1)))) = [0 0]x1 + [1] >= [0 0]x1 = b(a(b(b(a(x1)))))
   
                    [1 1]     [0]    [1 0]                      
   a(b(b(a(x1)))) = [0 0]x1 + [1] >= [0 0]x1 = b(b(b(b(b(x1)))))
   
                       [1 1]     [4]    [1 1]     [4]                          
   a(a(a(a(a(x1))))) = [0 0]x1 + [1] >= [0 0]x1 + [1] = a(a(a(b(a(a(a(x1)))))))
   
                       [1 1]     [2]    [1 0]     [2]                          
   a(a(a(b(a(x1))))) = [0 0]x1 + [1] >= [0 0]x1 + [1] = a(a(b(b(a(a(b(x1)))))))
   
                       [1 1]     [2]    [1 1]     [0]                          
   a(a(b(a(a(x1))))) = [0 0]x1 + [1] >= [0 0]x1 + [1] = a(b(a(b(a(b(a(x1)))))))
   
                       [1 1]     [1]    [1 0]     [0]                          
   a(a(b(b(a(x1))))) = [0 0]x1 + [1] >= [0 0]x1 + [1] = a(b(b(b(a(b(b(x1)))))))
   
                       [1 1]     [2]    [1 1]     [2]                          
   a(b(a(a(a(x1))))) = [0 0]x1 + [1] >= [0 0]x1 + [0] = b(a(a(b(b(a(a(x1)))))))
   
                       [1 1]     [0]    [1 0]                            
   a(b(a(b(a(x1))))) = [0 0]x1 + [1] >= [0 0]x1 = b(a(b(b(b(a(b(x1)))))))
   
                       [1 1]     [1]    [1 1]                            
   a(b(b(a(a(x1))))) = [0 0]x1 + [1] >= [0 0]x1 = b(b(a(b(b(b(a(x1)))))))
   
                       [1 1]     [0]    [1 0]                            
   a(b(b(b(a(x1))))) = [0 0]x1 + [1] >= [0 0]x1 = b(b(b(b(b(b(b(x1)))))))
  problem:
   a(b(a(x1))) -> b(b(b(x1)))
   a(b(b(a(x1)))) -> b(b(b(b(b(x1)))))
   a(a(a(a(a(x1))))) -> a(a(a(b(a(a(a(x1)))))))
   a(a(a(b(a(x1))))) -> a(a(b(b(a(a(b(x1)))))))
   a(b(a(a(a(x1))))) -> b(a(a(b(b(a(a(x1)))))))
   a(b(a(b(a(x1))))) -> b(a(b(b(b(a(b(x1)))))))
   a(b(b(b(a(x1))))) -> b(b(b(b(b(b(b(x1)))))))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0  -&]  
    [b](x0) = [-& -&]x0,
    
              [0 0]  
    [a](x0) = [0 1]x0
   orientation:
                  [0 0]      [0  -&]                
    a(b(a(x1))) = [0 0]x1 >= [-& -&]x1 = b(b(b(x1)))
    
                     [0 0]      [0  -&]                      
    a(b(b(a(x1)))) = [0 0]x1 >= [-& -&]x1 = b(b(b(b(b(x1)))))
    
                        [3 4]      [2 3]                            
    a(a(a(a(a(x1))))) = [4 5]x1 >= [3 4]x1 = a(a(a(b(a(a(a(x1)))))))
    
                        [1 1]      [0  -&]                            
    a(a(a(b(a(x1))))) = [2 2]x1 >= [1  -&]x1 = a(a(b(b(a(a(b(x1)))))))
    
                        [1 2]      [0  1 ]                            
    a(b(a(a(a(x1))))) = [1 2]x1 >= [-& -&]x1 = b(a(a(b(b(a(a(x1)))))))
    
                        [0 0]      [0  -&]                            
    a(b(a(b(a(x1))))) = [0 0]x1 >= [-& -&]x1 = b(a(b(b(b(a(b(x1)))))))
    
                        [0 0]      [0  -&]                            
    a(b(b(b(a(x1))))) = [0 0]x1 >= [-& -&]x1 = b(b(b(b(b(b(b(x1)))))))
   problem:
    a(b(a(x1))) -> b(b(b(x1)))
    a(b(b(a(x1)))) -> b(b(b(b(b(x1)))))
    a(b(a(b(a(x1))))) -> b(a(b(b(b(a(b(x1)))))))
    a(b(b(b(a(x1))))) -> b(b(b(b(b(b(b(x1)))))))
   Arctic Interpretation Processor:
    dimension: 1
    interpretation:
     [b](x0) = 1x0,
     
     [a](x0) = 3x0
    orientation:
     a(b(a(x1))) = 7x1 >= 3x1 = b(b(b(x1)))
     
     a(b(b(a(x1)))) = 8x1 >= 5x1 = b(b(b(b(b(x1)))))
     
     a(b(a(b(a(x1))))) = 11x1 >= 11x1 = b(a(b(b(b(a(b(x1)))))))
     
     a(b(b(b(a(x1))))) = 9x1 >= 7x1 = b(b(b(b(b(b(b(x1)))))))
    problem:
     a(b(a(b(a(x1))))) -> b(a(b(b(b(a(b(x1)))))))
    Arctic Interpretation Processor:
     dimension: 1
     interpretation:
      [b](x0) = 2x0,
      
      [a](x0) = 9x0
     orientation:
      a(b(a(b(a(x1))))) = 31x1 >= 28x1 = b(a(b(b(b(a(b(x1)))))))
     problem:
      
     Qed