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:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [w](x0) = x0 + 4,
   
   [v](x0) = 12x0,
   
   [u](x0) = 12x0 + 1,
   
   [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)) = 12x + 1 >= x = x
   
   v(b(x)) = 12x >= x = x
   
   w(c(x)) = x + 4 >= x = x
   
   a(u(x)) = 12x + 1 >= x = x
   
   b(v(x)) = 12x >= x = x
   
   c(w(x)) = x + 4 >= x = x
  problem:
   a(b(x)) -> b(a(a(x)))
   b(c(x)) -> c(b(b(x)))
   c(a(x)) -> a(c(c(x)))
   v(b(x)) -> x
   b(v(x)) -> x
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 1 0]     [1]
    [v](x0) = [0 1 1]x0 + [0]
              [0 0 1]     [0],
    
              [1 0 0]     [0]
    [c](x0) = [0 0 0]x0 + [1]
              [0 1 1]     [1],
    
              [1 0 0]  
    [a](x0) = [0 1 0]x0
              [0 1 0]  ,
    
                
    [b](x0) = x0
                
   orientation:
              [1 0 0]     [1 0 0]              
    a(b(x)) = [0 1 0]x >= [0 1 0]x = b(a(a(x)))
              [0 1 0]     [0 1 0]              
    
              [1 0 0]    [0]    [1 0 0]    [0]             
    b(c(x)) = [0 0 0]x + [1] >= [0 0 0]x + [1] = c(b(b(x)))
              [0 1 1]    [1]    [0 1 1]    [1]             
    
              [1 0 0]    [0]    [1 0 0]    [0]             
    c(a(x)) = [0 0 0]x + [1] >= [0 0 0]x + [1] = a(c(c(x)))
              [0 2 0]    [1]    [0 0 0]    [1]             
    
              [1 1 0]    [1]         
    v(b(x)) = [0 1 1]x + [0] >= x = x
              [0 0 1]    [0]         
    
              [1 1 0]    [1]         
    b(v(x)) = [0 1 1]x + [0] >= x = x
              [0 0 1]    [0]         
   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)))
    Bounds Processor:
     bound: 0
     enrichment: match
     automaton:
      final states: {8,5,1}
      transitions:
       a0(2) -> 9*
       a0(4) -> 1*
       a0(3) -> 4*
       b0(7) -> 5*
       b0(2) -> 3*
       b0(6) -> 7*
       c0(10) -> 8*
       c0(2) -> 6*
       c0(9) -> 10*
       f60() -> 2*
       1 -> 3*
       5 -> 6*
       8 -> 9*
     problem:
      
     Qed