YES

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

Proof:
 Matrix Interpretation Processor: dim=2
  
  interpretation:
             [1 0]     [0]
   [c](x0) = [0 2]x0 + [1],
   
             [2 0]  
   [b](x0) = [0 1]x0,
   
             [1 1]     [1]
   [a](x0) = [0 1]x0 + [0]
  orientation:
                    [2 4]     [4]    [2 4]     [4]                 
   c(b(a(a(x1)))) = [0 2]x1 + [1] >= [0 2]x1 + [1] = a(a(b(c(x1))))
   
                    [2 6]     [6]    [2 3]     [3]                 
   b(a(a(a(x1)))) = [0 1]x1 + [0] >= [0 1]x1 + [0] = a(a(a(b(x1))))
   
                 [2 2]     [2]    [2 2]     [2]              
   a(b(c(x1))) = [0 2]x1 + [1] >= [0 2]x1 + [1] = c(b(a(x1)))
   
                    [4 0]     [0]    [4 0]     [0]                 
   c(c(b(b(x1)))) = [0 4]x1 + [3] >= [0 4]x1 + [3] = b(b(c(c(x1))))
  problem:
   c(b(a(a(x1)))) -> a(a(b(c(x1))))
   a(b(c(x1))) -> c(b(a(x1)))
   c(c(b(b(x1)))) -> b(b(c(c(x1))))
  String Reversal Processor:
   a(a(b(c(x1)))) -> c(b(a(a(x1))))
   c(b(a(x1))) -> a(b(c(x1)))
   b(b(c(c(x1)))) -> c(c(b(b(x1))))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
               [1 0 0]  
     [c](x0) = [0 0 1]x0
               [0 1 0]  ,
     
               [1 0 1]     [0]
     [b](x0) = [0 0 0]x0 + [0]
               [0 1 1]     [1],
     
               [1 1 0]  
     [a](x0) = [0 1 1]x0
               [0 0 0]  
    orientation:
                      [1 2 1]     [1]    [1 2 1]     [0]                 
     a(a(b(c(x1)))) = [0 1 1]x1 + [1] >= [0 1 1]x1 + [1] = c(b(a(a(x1))))
                      [0 0 0]     [0]    [0 0 0]     [0]                 
     
                   [1 1 0]     [0]    [1 1 0]     [0]              
     c(b(a(x1))) = [0 1 1]x1 + [1] >= [0 1 1]x1 + [1] = a(b(c(x1)))
                   [0 0 0]     [0]    [0 0 0]     [0]              
     
                      [1 1 2]     [1]    [1 1 2]     [1]                 
     b(b(c(c(x1)))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = c(c(b(b(x1))))
                      [0 1 1]     [2]    [0 1 1]     [2]                 
    problem:
     c(b(a(x1))) -> a(b(c(x1)))
     b(b(c(c(x1)))) -> c(c(b(b(x1))))
    KBO Processor:
     weight function:
      w0 = 1
      w(c) = w(b) = w(a) = 1
     precedence:
      b > c > a
     problem:
      
     Qed