YES

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

Proof:
 String Reversal Processor:
  b(c(a(x1))) -> a(b(a(b(x1))))
  b(x1) -> c(a(c(x1)))
  a(a(x1)) -> a(c(b(a(x1))))
  Bounds Processor:
   bound: 2
   enrichment: match
   automaton:
    final states: {9,6,1}
    transitions:
     b0(4) -> 5*
     b0(2) -> 3*
     b0(10) -> 11*
     c1(13) -> 14*
     c1(31) -> 32*
     c1(39) -> 40*
     c1(27) -> 28*
     c1(15) -> 16*
     c1(29) -> 30*
     c1(25) -> 26*
     c0(2) -> 7*
     c0(8) -> 6*
     c0(11) -> 12*
     a2(48) -> 49*
     a0(3) -> 4*
     a0(12) -> 9*
     a0(2) -> 10*
     a0(7) -> 8*
     a0(5) -> 1*
     f30() -> 2*
     a1(40) -> 41*
     a1(37) -> 38*
     a1(30) -> 31*
     a1(14) -> 15*
     a1(26) -> 27*
     c2(49) -> 50*
     c2(47) -> 48*
     b1(38) -> 39*
     16 -> 3*
     28 -> 11*
     4 -> 29*
     2 -> 13*
     38 -> 47*
     1 -> 3*
     5 -> 37*
     9 -> 10,25
     50 -> 39*
     41 -> 4,29
     10 -> 25*
     32 -> 5*
   problem:
    
   Qed