YES

Problem:
 a(l(x1)) -> l(a(x1))
 r(a(a(x1))) -> a(a(r(x1)))
 b(l(x1)) -> b(a(r(x1)))
 r(b(x1)) -> l(b(x1))

Proof:
 String Reversal Processor:
  l(a(x1)) -> a(l(x1))
  a(a(r(x1))) -> r(a(a(x1)))
  l(b(x1)) -> r(a(b(x1)))
  b(r(x1)) -> b(l(x1))
  Bounds Processor:
   bound: 3
   enrichment: match
   automaton:
    final states: {10,7,4,1}
    transitions:
     b1(45) -> 46*
     b1(20) -> 21*
     b1(95) -> 96*
     l1(37) -> 38*
     l1(19) -> 20*
     l1(28) -> 29*
     r1(47) -> 48*
     r1(13) -> 14*
     a1(12) -> 13*
     a1(29) -> 30*
     a1(46) -> 47*
     a1(31) -> 32*
     a1(11) -> 12*
     a2(40) -> 41*
     a2(97) -> 98*
     a2(67) -> 68*
     a2(57) -> 58*
     a2(88) -> 89*
     a2(68) -> 69*
     l2(59) -> 60*
     l2(39) -> 40*
     l2(56) -> 57*
     f40() -> 2*
     r2(89) -> 90*
     r2(69) -> 70*
     a0(5) -> 6*
     a0(2) -> 5*
     a0(8) -> 9*
     a0(3) -> 1*
     r3(77) -> 78*
     l0(2) -> 3*
     a3(75) -> 76*
     a3(81) -> 82*
     a3(76) -> 77*
     r0(9) -> 7*
     r0(6) -> 4*
     b2(87) -> 88*
     b2(101) -> 102*
     b0(2) -> 8*
     b0(3) -> 10*
     1 -> 3*
     2 -> 45*
     3 -> 95*
     4 -> 5,6
     7 -> 3*
     8 -> 28*
     9 -> 19,11
     10 -> 8*
     11 -> 59*
     12 -> 39*
     13 -> 37,31
     14 -> 1,3
     20 -> 87*
     21 -> 10*
     30 -> 60,57,20
     31 -> 56*
     32 -> 12*
     38 -> 20*
     41 -> 57,38,20
     47 -> 67*
     48 -> 29*
     57 -> 101*
     58 -> 40*
     60 -> 57*
     69 -> 75*
     70 -> 58,40
     77 -> 81*
     78 -> 58,40
     82 -> 76*
     89 -> 97*
     90 -> 29*
     96 -> 46*
     98 -> 68*
     102 -> 96,46
   problem:
    
   Qed