YES

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

Proof:
 String Reversal Processor:
  b(b(a(a(x1)))) -> a(a(c(c(b(b(x1))))))
  c(c(b(b(x1)))) -> b(b(b(b(c(c(x1))))))
  a(a(b(b(x1)))) -> b(b(c(c(a(a(x1))))))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 0]  
    [c](x0) = [0 1 0]x0
              [0 0 0]  ,
    
              [1 0 0]     [0]
    [a](x0) = [0 0 1]x0 + [1]
              [0 1 0]     [0],
    
              [1 0 1]  
    [b](x0) = [1 0 0]x0
              [0 0 1]  
   orientation:
                     [1 0 2]     [2]    [1 0 2]     [0]                       
    b(b(a(a(x1)))) = [1 0 1]x1 + [1] >= [1 0 1]x1 + [1] = a(a(c(c(b(b(x1))))))
                     [0 0 1]     [1]    [0 0 0]     [1]                       
    
                     [1 0 2]      [1 0 0]                         
    c(c(b(b(x1)))) = [1 0 1]x1 >= [1 0 0]x1 = b(b(b(b(c(c(x1))))))
                     [0 0 0]      [0 0 0]                         
    
                     [1 0 2]     [0]    [1 0 0]                         
    a(a(b(b(x1)))) = [1 0 1]x1 + [1] >= [1 0 0]x1 = b(b(c(c(a(a(x1))))))
                     [0 0 1]     [1]    [0 0 0]                         
   problem:
    c(c(b(b(x1)))) -> b(b(b(b(c(c(x1))))))
    a(a(b(b(x1)))) -> b(b(c(c(a(a(x1))))))
   String Reversal Processor:
    b(b(c(c(x1)))) -> c(c(b(b(b(b(x1))))))
    b(b(a(a(x1)))) -> a(a(c(c(b(b(x1))))))
    DP Processor:
     DPs:
      b#(b(c(c(x1)))) -> b#(x1)
      b#(b(c(c(x1)))) -> b#(b(x1))
      b#(b(c(c(x1)))) -> b#(b(b(x1)))
      b#(b(c(c(x1)))) -> b#(b(b(b(x1))))
      b#(b(a(a(x1)))) -> b#(x1)
      b#(b(a(a(x1)))) -> b#(b(x1))
     TRS:
      b(b(c(c(x1)))) -> c(c(b(b(b(b(x1))))))
      b(b(a(a(x1)))) -> a(a(c(c(b(b(x1))))))
     Arctic Interpretation Processor:
      dimension: 1
      interpretation:
       [b#](x0) = x0,
       
       [c](x0) = x0,
       
       [a](x0) = 1x0,
       
       [b](x0) = x0
      orientation:
       b#(b(c(c(x1)))) = x1 >= x1 = b#(x1)
       
       b#(b(c(c(x1)))) = x1 >= x1 = b#(b(x1))
       
       b#(b(c(c(x1)))) = x1 >= x1 = b#(b(b(x1)))
       
       b#(b(c(c(x1)))) = x1 >= x1 = b#(b(b(b(x1))))
       
       b#(b(a(a(x1)))) = 2x1 >= x1 = b#(x1)
       
       b#(b(a(a(x1)))) = 2x1 >= x1 = b#(b(x1))
       
       b(b(c(c(x1)))) = x1 >= x1 = c(c(b(b(b(b(x1))))))
       
       b(b(a(a(x1)))) = 2x1 >= 2x1 = a(a(c(c(b(b(x1))))))
      problem:
       DPs:
        b#(b(c(c(x1)))) -> b#(x1)
        b#(b(c(c(x1)))) -> b#(b(x1))
        b#(b(c(c(x1)))) -> b#(b(b(x1)))
        b#(b(c(c(x1)))) -> b#(b(b(b(x1))))
       TRS:
        b(b(c(c(x1)))) -> c(c(b(b(b(b(x1))))))
        b(b(a(a(x1)))) -> a(a(c(c(b(b(x1))))))
      Arctic Interpretation Processor:
       dimension: 1
       interpretation:
        [b#](x0) = x0,
        
        [c](x0) = 1x0,
        
        [a](x0) = 0,
        
        [b](x0) = x0
       orientation:
        b#(b(c(c(x1)))) = 2x1 >= x1 = b#(x1)
        
        b#(b(c(c(x1)))) = 2x1 >= x1 = b#(b(x1))
        
        b#(b(c(c(x1)))) = 2x1 >= x1 = b#(b(b(x1)))
        
        b#(b(c(c(x1)))) = 2x1 >= x1 = b#(b(b(b(x1))))
        
        b(b(c(c(x1)))) = 2x1 >= 2x1 = c(c(b(b(b(b(x1))))))
        
        b(b(a(a(x1)))) = 0 >= 0 = a(a(c(c(b(b(x1))))))
       problem:
        DPs:
         
        TRS:
         b(b(c(c(x1)))) -> c(c(b(b(b(b(x1))))))
         b(b(a(a(x1)))) -> a(a(c(c(b(b(x1))))))
       Qed