YES

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

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