YES

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

Proof:
 String Reversal Processor:
  a(c(b(x1))) -> b(a(b(a(x1))))
  b(x1) -> c(c(x1))
  a(a(x1)) -> a(b(c(a(x1))))
  Bounds Processor:
   bound: 3
   enrichment: match
   automaton:
    final states: {4}
    transitions:
     c3(82) -> 83*
     c3(91) -> 92*
     c3(83) -> 84*
     c3(90) -> 91*
     a1(70) -> 71*
     a1(10) -> 11*
     a1(12) -> 13*
     a1(24) -> 25*
     a1(26) -> 27*
     a1(28) -> 29*
     b1(27) -> 28*
     b1(29) -> 30*
     b1(11) -> 12*
     b1(23) -> 24*
     b1(13) -> 14*
     c1(15) -> 16*
     c1(22) -> 23*
     c1(16) -> 17*
     a2(62) -> 63*
     a2(59) -> 60*
     a2(93) -> 94*
     a2(88) -> 89*
     a2(85) -> 86*
     a0(4) -> 4*
     b2(87) -> 88*
     b2(61) -> 62*
     c0(4) -> 4*
     c2(80) -> 81*
     c2(60) -> 61*
     c2(50) -> 51*
     c2(72) -> 73*
     c2(47) -> 48*
     c2(54) -> 55*
     c2(86) -> 87*
     c2(51) -> 52*
     c2(73) -> 74*
     c2(53) -> 54*
     c2(48) -> 49*
     b0(4) -> 4*
     4 -> 15,10
     11 -> 53,22
     12 -> 85*
     13 -> 50,26
     14 -> 11,22,4
     17 -> 4*
     23 -> 47*
     24 -> 59*
     25 -> 11,22,4
     27 -> 72*
     28 -> 93*
     29 -> 80,70
     30 -> 11,4,10,15,22
     49 -> 24*
     52 -> 30,14,4
     55 -> 12*
     61 -> 82*
     63 -> 71,11,22
     71 -> 11*
     74 -> 28*
     81 -> 51*
     84 -> 62*
     87 -> 90*
     89 -> 27*
     92 -> 88*
     94 -> 60*
   problem:
    
   Qed