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]     [0]
    [a](x0) = [0 0 1]x0 + [1]
              [0 1 0]     [0],
    
              [1 0 1]     [0]
    [b](x0) = [1 0 0]x0 + [1]
              [0 0 1]     [0],
    
              [1 0 0]  
    [c](x0) = [0 1 0]x0
              [0 0 0]  
   orientation:
                     [1 0 2]     [2]    [1 0 2]     [0]                       
    b(b(a(a(x1)))) = [1 0 1]x1 + [2] >= [1 0 1]x1 + [2] = a(a(c(c(b(b(x1))))))
                     [0 0 1]     [1]    [0 0 0]     [1]                       
    
                     [1 0 2]     [0]    [1 0 0]     [0]                       
    c(c(b(b(x1)))) = [1 0 1]x1 + [1] >= [1 0 0]x1 + [1] = b(b(b(b(c(c(x1))))))
                     [0 0 0]     [0]    [0 0 0]     [0]                       
    
                     [1 0 2]     [0]    [1 0 0]     [0]                       
    a(a(b(b(x1)))) = [1 0 1]x1 + [2] >= [1 0 0]x1 + [1] = b(b(c(c(a(a(x1))))))
                     [0 0 1]     [1]    [0 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))))))
   WPO Processor:
    algebra: Sum
    weight function:
     w0 = 0
     w(c) = w(a) = w(b) = 0
    status function:
     st(c) = st(a) = st(b) = [0]
    precedence:
     a > c > b
    problem:
     
    Qed