YES

Problem:
 a(b(x)) -> b(a(a(x)))
 b(c(x)) -> c(b(b(x)))
 c(a(x)) -> a(c(c(x)))
 u(a(x)) -> x
 v(b(x)) -> x
 w(c(x)) -> x
 a(u(x)) -> x
 b(v(x)) -> x
 c(w(x)) -> x

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [w](x0) = 7x0,
   
   [v](x0) = 10x0,
   
   [u](x0) = 14x0,
   
   [c](x0) = x0,
   
   [a](x0) = x0,
   
   [b](x0) = x0
  orientation:
   a(b(x)) = x >= x = b(a(a(x)))
   
   b(c(x)) = x >= x = c(b(b(x)))
   
   c(a(x)) = x >= x = a(c(c(x)))
   
   u(a(x)) = 14x >= x = x
   
   v(b(x)) = 10x >= x = x
   
   w(c(x)) = 7x >= x = x
   
   a(u(x)) = 14x >= x = x
   
   b(v(x)) = 10x >= x = x
   
   c(w(x)) = 7x >= x = x
  problem:
   a(b(x)) -> b(a(a(x)))
   b(c(x)) -> c(b(b(x)))
   c(a(x)) -> a(c(c(x)))
  String Reversal Processor:
   b(a(x)) -> a(a(b(x)))
   c(b(x)) -> b(b(c(x)))
   a(c(x)) -> c(c(a(x)))
   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 0 1]x0 + [1]
               [0 1 0]     [1],
     
               [1 1 1]  
     [b](x0) = [0 1 1]x0
               [0 1 1]  
    orientation:
               [1 1 1]    [3]    [1 1 1]    [2]             
     b(a(x)) = [0 1 1]x + [2] >= [0 1 1]x + [2] = a(a(b(x)))
               [0 1 1]    [2]    [0 1 1]    [2]             
     
               [1 1 1]     [1 0 0]              
     c(b(x)) = [0 0 0]x >= [0 0 0]x = b(b(c(x)))
               [0 0 0]     [0 0 0]              
     
               [1 0 0]    [1]    [1 0 0]    [1]             
     a(c(x)) = [0 0 0]x + [1] >= [0 0 0]x + [0] = c(c(a(x)))
               [0 0 0]    [1]    [0 0 0]    [0]             
    problem:
     c(b(x)) -> b(b(c(x)))
     a(c(x)) -> c(c(a(x)))
    String Reversal Processor:
     b(c(x)) -> c(b(b(x)))
     c(a(x)) -> a(c(c(x)))
     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(x)) = [0 0 0]x >= [0 0 0]x = c(b(b(x)))
                 [0 1 1]     [0 1 1]              
       
                 [1 2 2]    [2]    [1 2 2]    [1]             
       c(a(x)) = [0 1 1]x + [1] >= [0 1 1]x + [1] = a(c(c(x)))
                 [0 1 1]    [0]    [0 1 1]    [0]             
      problem:
       b(c(x)) -> c(b(b(x)))
      KBO Processor:
       weight function:
        w0 = 1
        w(c) = 1
        w(b) = 0
       precedence:
        b > c
       problem:
        
       Qed