YES

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

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