YES

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

Proof:
 String Reversal Processor:
  a(a(x1)) -> b(b(b(x1)))
  b(b(x1)) -> c(c(c(x1)))
  c(c(c(c(x1)))) -> b(a(x1))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [b](x0) = x0 + 6,
    
    [a](x0) = x0 + 9,
    
    [c](x0) = x0 + 4
   orientation:
    a(a(x1)) = x1 + 18 >= x1 + 18 = b(b(b(x1)))
    
    b(b(x1)) = x1 + 12 >= x1 + 12 = c(c(c(x1)))
    
    c(c(c(c(x1)))) = x1 + 16 >= x1 + 15 = b(a(x1))
   problem:
    a(a(x1)) -> b(b(b(x1)))
    b(b(x1)) -> c(c(c(x1)))
   Bounds Processor:
    bound: 2
    enrichment: match
    automaton:
     final states: {4}
     transitions:
      c1(25) -> 26*
      c1(14) -> 15*
      c1(13) -> 14*
      c1(15) -> 16*
      c2(41) -> 42*
      c2(34) -> 35*
      c2(40) -> 41*
      c2(39) -> 40*
      c2(22) -> 23*
      c2(35) -> 36*
      c2(33) -> 34*
      c2(23) -> 24*
      c2(21) -> 22*
      b1(10) -> 11*
      b1(11) -> 12*
      b1(9) -> 10*
      a0(4) -> 4*
      c0(4) -> 4*
      b0(4) -> 4*
      24 -> 12*
      42 -> 11*
      12 -> 4*
      16 -> 10,4
      4 -> 13,9
      11 -> 33,25
      36 -> 10*
      26 -> 14*
      9 -> 39*
      10 -> 21*
    problem:
     
    Qed