YES

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

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