YES

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

Proof:
 String Reversal Processor:
  b(a(x1)) -> b(b(b(x1)))
  c(a(x1)) -> b(b(c(x1)))
  a(b(x1)) -> a(a(a(x1)))
  c(b(x1)) -> c(c(c(x1)))
  a(x1) -> x1
  b(x1) -> x1
  c(x1) -> x1
  Bounds Processor:
   bound: 1
   enrichment: match
   automaton:
    final states: {2,11,8,5,1}
    transitions:
     c0(2) -> 6*
     c0(12) -> 11*
     c0(6) -> 12*
     c1(27) -> 28*
     c1(17) -> 18*
     c1(21) -> 22*
     c1(19) -> 20*
     c1(18) -> 19*
     a0(2) -> 9*
     a0(10) -> 8*
     a0(9) -> 10*
     b0(3) -> 4*
     b0(4) -> 1*
     b0(6) -> 7*
     b0(2) -> 3*
     b0(7) -> 5*
     f30() -> 2*
     27 -> 18*
     19 -> 22,12,20
     17 -> 18*
     7 -> 27,5
     12 -> 11*
     28 -> 18*
     4 -> 1*
     11 -> 5*
     2 -> 6,3,9
     22 -> 18*
     6 -> 21,12,7
     21 -> 22*
     8 -> 9,10
     1 -> 7,5,3
     20 -> 28,18,11
     5 -> 6,7,12
     9 -> 10*
     3 -> 17,4
     18 -> 19*
     10 -> 8*
   problem:
    
   Qed