YES

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

Proof:
 String Reversal Processor:
  b(a(x1)) -> a(a(b(x1)))
  c(b(x1)) -> b(c(x1))
  a(a(x1)) -> a(c(a(x1)))
  Bounds Processor:
   bound: 2
   enrichment: match
   automaton:
    final states: {7,5,1}
    transitions:
     f30() -> 2*
     a0(2) -> 8*
     a0(9) -> 7*
     a0(4) -> 1*
     a0(3) -> 4*
     b0(2) -> 3*
     b0(6) -> 5*
     c0(2) -> 6*
     c0(8) -> 9*
     a1(14) -> 15*
     a1(26) -> 27*
     a1(16) -> 17*
     c1(15) -> 16*
     a2(22) -> 23*
     a2(24) -> 25*
     a2(31) -> 32*
     c2(30) -> 31*
     c2(23) -> 24*
     1 -> 3,14
     3 -> 14*
     4 -> 26*
     5 -> 6*
     7 -> 8*
     16 -> 22*
     17 -> 27,1
     25 -> 30,4,15
     27 -> 23*
     32 -> 27*
   problem:
    
   Qed