YES

Problem:
 a(b(c(x1))) -> c(b(a(x1)))
 C(B(A(x1))) -> A(B(C(x1)))
 b(a(C(x1))) -> C(a(b(x1)))
 c(A(B(x1))) -> B(A(c(x1)))
 A(c(b(x1))) -> b(c(A(x1)))
 B(C(a(x1))) -> a(C(B(x1)))
 a(A(x1)) -> x1
 A(a(x1)) -> x1
 b(B(x1)) -> x1
 B(b(x1)) -> x1
 c(C(x1)) -> x1
 C(c(x1)) -> x1

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [C](x0) = 8x0,
   
   [B](x0) = x0,
   
   [A](x0) = 4x0,
   
   [a](x0) = 3x0,
   
   [b](x0) = 11x0,
   
   [c](x0) = 5x0
  orientation:
   a(b(c(x1))) = 19x1 >= 19x1 = c(b(a(x1)))
   
   C(B(A(x1))) = 12x1 >= 12x1 = A(B(C(x1)))
   
   b(a(C(x1))) = 22x1 >= 22x1 = C(a(b(x1)))
   
   c(A(B(x1))) = 9x1 >= 9x1 = B(A(c(x1)))
   
   A(c(b(x1))) = 20x1 >= 20x1 = b(c(A(x1)))
   
   B(C(a(x1))) = 11x1 >= 11x1 = a(C(B(x1)))
   
   a(A(x1)) = 7x1 >= x1 = x1
   
   A(a(x1)) = 7x1 >= x1 = x1
   
   b(B(x1)) = 11x1 >= x1 = x1
   
   B(b(x1)) = 11x1 >= x1 = x1
   
   c(C(x1)) = 13x1 >= x1 = x1
   
   C(c(x1)) = 13x1 >= x1 = x1
  problem:
   a(b(c(x1))) -> c(b(a(x1)))
   C(B(A(x1))) -> A(B(C(x1)))
   b(a(C(x1))) -> C(a(b(x1)))
   c(A(B(x1))) -> B(A(c(x1)))
   A(c(b(x1))) -> b(c(A(x1)))
   B(C(a(x1))) -> a(C(B(x1)))
  String Reversal Processor:
   c(b(a(x1))) -> a(b(c(x1)))
   A(B(C(x1))) -> C(B(A(x1)))
   C(a(b(x1))) -> b(a(C(x1)))
   B(A(c(x1))) -> c(A(B(x1)))
   b(c(A(x1))) -> A(c(b(x1)))
   a(C(B(x1))) -> B(C(a(x1)))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [C](x0) = 4x0 + 2,
     
     [B](x0) = x0,
     
     [A](x0) = 2x0,
     
     [a](x0) = x0,
     
     [b](x0) = x0,
     
     [c](x0) = 4x0
    orientation:
     c(b(a(x1))) = 4x1 >= 4x1 = a(b(c(x1)))
     
     A(B(C(x1))) = 8x1 + 4 >= 8x1 + 2 = C(B(A(x1)))
     
     C(a(b(x1))) = 4x1 + 2 >= 4x1 + 2 = b(a(C(x1)))
     
     B(A(c(x1))) = 8x1 >= 8x1 = c(A(B(x1)))
     
     b(c(A(x1))) = 8x1 >= 8x1 = A(c(b(x1)))
     
     a(C(B(x1))) = 4x1 + 2 >= 4x1 + 2 = B(C(a(x1)))
    problem:
     c(b(a(x1))) -> a(b(c(x1)))
     C(a(b(x1))) -> b(a(C(x1)))
     B(A(c(x1))) -> c(A(B(x1)))
     b(c(A(x1))) -> A(c(b(x1)))
     a(C(B(x1))) -> B(C(a(x1)))
    String Reversal Processor:
     a(b(c(x1))) -> c(b(a(x1)))
     b(a(C(x1))) -> C(a(b(x1)))
     c(A(B(x1))) -> B(A(c(x1)))
     A(c(b(x1))) -> b(c(A(x1)))
     B(C(a(x1))) -> a(C(B(x1)))
     Matrix Interpretation Processor: dim=1
      
      interpretation:
       [C](x0) = x0 + 1,
       
       [B](x0) = 3x0,
       
       [A](x0) = 8x0,
       
       [a](x0) = 8x0 + 8,
       
       [b](x0) = x0 + 1,
       
       [c](x0) = x0
      orientation:
       a(b(c(x1))) = 8x1 + 16 >= 8x1 + 9 = c(b(a(x1)))
       
       b(a(C(x1))) = 8x1 + 17 >= 8x1 + 17 = C(a(b(x1)))
       
       c(A(B(x1))) = 24x1 >= 24x1 = B(A(c(x1)))
       
       A(c(b(x1))) = 8x1 + 8 >= 8x1 + 1 = b(c(A(x1)))
       
       B(C(a(x1))) = 24x1 + 27 >= 24x1 + 16 = a(C(B(x1)))
      problem:
       b(a(C(x1))) -> C(a(b(x1)))
       c(A(B(x1))) -> B(A(c(x1)))
      KBO Processor:
       weight function:
        w0 = 1
        w(C) = w(B) = w(A) = w(a) = w(b) = w(c) = 1
       precedence:
        b > c > C ~ B ~ A ~ a
       problem:
        
       Qed