YES

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

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