YES

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

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