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:
 String Reversal Processor:
  a(a(b(c(x1)))) -> c(b(a(a(x1))))
  a(a(a(b(x1)))) -> b(a(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 1 0]  
    [c](x0) = [0 1 0]x0
              [0 0 1]  ,
    
                   [0]
    [b](x0) = x0 + [0]
                   [1],
    
              [1 0 1]     [0]
    [a](x0) = [0 1 0]x0 + [1]
              [0 0 1]     [0]
   orientation:
                     [1 1 2]     [2]    [1 1 2]     [2]                 
    a(a(b(c(x1)))) = [0 1 0]x1 + [2] >= [0 1 0]x1 + [2] = c(b(a(a(x1))))
                     [0 0 1]     [1]    [0 0 1]     [1]                 
    
                     [1 0 3]     [3]    [1 0 3]     [0]                 
    a(a(a(b(x1)))) = [0 1 0]x1 + [3] >= [0 1 0]x1 + [3] = b(a(a(a(x1))))
                     [0 0 1]     [1]    [0 0 1]     [1]                 
    
                  [1 1 1]     [1]    [1 1 1]     [1]              
    c(b(a(x1))) = [0 1 0]x1 + [1] >= [0 1 0]x1 + [1] = a(b(c(x1)))
                  [0 0 1]     [1]    [0 0 1]     [1]              
    
                     [1 2 0]     [0]    [1 2 0]     [0]                 
    b(b(c(c(x1)))) = [0 1 0]x1 + [0] >= [0 1 0]x1 + [0] = c(c(b(b(x1))))
                     [0 0 1]     [2]    [0 0 1]     [2]                 
   problem:
    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))))
   String Reversal Processor:
    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))))
    Matrix Interpretation Processor: dim=3
     
     interpretation:
                [1 1 0]  
      [c](x0) = [0 0 1]x0
                [0 1 0]  ,
      
                [1 1 0]     [0]
      [b](x0) = [0 0 0]x0 + [0]
                [0 1 1]     [1],
      
                [1 0 0]  
      [a](x0) = [0 1 1]x0
                [0 0 0]  
     orientation:
                       [1 1 1]     [0]    [1 1 1]     [0]                 
      c(b(a(a(x1)))) = [0 1 1]x1 + [1] >= [0 1 1]x1 + [1] = a(a(b(c(x1))))
                       [0 0 0]     [0]    [0 0 0]     [0]                 
      
                    [1 1 1]     [0]    [1 1 1]     [0]              
      a(b(c(x1))) = [0 1 1]x1 + [1] >= [0 1 1]x1 + [1] = c(b(a(x1)))
                    [0 0 0]     [0]    [0 0 0]     [0]              
      
                       [1 2 1]     [2]    [1 2 1]     [0]                 
      c(c(b(b(x1)))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(b(c(c(x1))))
                       [0 1 1]     [2]    [0 1 1]     [2]                 
     problem:
      c(b(a(a(x1)))) -> a(a(b(c(x1))))
      a(b(c(x1))) -> c(b(a(x1)))
     String Reversal Processor:
      a(a(b(c(x1)))) -> c(b(a(a(x1))))
      c(b(a(x1))) -> a(b(c(x1)))
      Matrix Interpretation Processor: dim=2
       
       interpretation:
                  [1 1]  
        [c](x0) = [0 1]x0,
        
                  [1 0]     [1]
        [b](x0) = [0 2]x0 + [2],
        
                  [2 0]     [1]
        [a](x0) = [0 1]x0 + [0]
       orientation:
                         [4 4]     [7]    [4 2]     [6]                 
        a(a(b(c(x1)))) = [0 2]x1 + [2] >= [0 2]x1 + [2] = c(b(a(a(x1))))
        
                      [2 2]     [4]    [2 2]     [3]              
        c(b(a(x1))) = [0 2]x1 + [2] >= [0 2]x1 + [2] = a(b(c(x1)))
       problem:
        
       Qed