YES

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

Proof:
 String Reversal Processor:
  a(a(x1)) -> b(c(x1))
  b(b(x1)) -> c(a(x1))
  c(c(x1)) -> a(b(x1))
  Bounds Processor:
   bound: 0
   enrichment: match
   automaton:
    final states: {6,4,1}
    transitions:
     c0(2) -> 3*
     c0(5) -> 4*
     a0(2) -> 5*
     a0(7) -> 6*
     b0(3) -> 1*
     b0(2) -> 7*
     f30() -> 2*
     4 -> 7*
     6 -> 3*
     1 -> 5*
   problem:
    
   Qed