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:
 String Reversal Processor:
  b(a(x)) -> a(a(b(x)))
  c(b(x)) -> b(b(c(x)))
  a(c(x)) -> c(c(a(x)))
  a(u(x)) -> x
  b(v(x)) -> x
  c(w(x)) -> x
  u(a(x)) -> x
  v(b(x)) -> x
  w(c(x)) -> x
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [v](x0) = x0 + 7,
    
    [a](x0) = x0,
    
    [u](x0) = x0 + 1,
    
    [w](x0) = 8x0 + 6,
    
    [b](x0) = x0,
    
    [c](x0) = x0
   orientation:
    b(a(x)) = x >= x = a(a(b(x)))
    
    c(b(x)) = x >= x = b(b(c(x)))
    
    a(c(x)) = x >= x = c(c(a(x)))
    
    a(u(x)) = x + 1 >= x = x
    
    b(v(x)) = x + 7 >= x = x
    
    c(w(x)) = 8x + 6 >= x = x
    
    u(a(x)) = x + 1 >= x = x
    
    v(b(x)) = x + 7 >= x = x
    
    w(c(x)) = 8x + 6 >= x = x
   problem:
    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(3) -> 4*
      a0(4) -> 1*
      a0(2) -> 9*
      f60() -> 2*
      b0(2) -> 3*
      b0(6) -> 7*
      b0(7) -> 5*
      c0(2) -> 6*
      c0(10) -> 8*
      c0(9) -> 10*
      8 -> 9*
      1 -> 3*
      5 -> 6*
    problem:
     
    Qed