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:
     a1(65) -> 66*
     a1(12) -> 13*
     a1(28) -> 29*
     a1(10) -> 11*
     a1(26) -> 27*
     a1(24) -> 25*
     c1(16) -> 17*
     c1(22) -> 23*
     c1(15) -> 16*
     b1(13) -> 14*
     b1(27) -> 28*
     b1(29) -> 30*
     b1(23) -> 24*
     b1(11) -> 12*
     a0(4) -> 4*
     c3(79) -> 80*
     c3(89) -> 90*
     c3(90) -> 91*
     c3(78) -> 79*
     b0(4) -> 4*
     c0(4) -> 4*
     a2(92) -> 93*
     a2(60) -> 61*
     a2(84) -> 85*
     a2(63) -> 64*
     a2(81) -> 82*
     c2(52) -> 53*
     c2(67) -> 68*
     c2(57) -> 58*
     c2(54) -> 55*
     c2(68) -> 69*
     c2(55) -> 56*
     c2(61) -> 62*
     c2(51) -> 52*
     c2(58) -> 59*
     c2(82) -> 83*
     c2(73) -> 74*
     b2(62) -> 63*
     b2(83) -> 84*
     56 -> 12*
     27 -> 67*
     24 -> 60*
     74 -> 52*
     17 -> 4*
     83 -> 89*
     64 -> 66,11,22
     30 -> 11,4,10,15,22
     69 -> 28*
     85 -> 27*
     12 -> 81*
     93 -> 61*
     14 -> 11,22,4
     66 -> 11*
     28 -> 92*
     4 -> 15,10
     11 -> 54,22
     53 -> 30,14,4
     91 -> 84*
     13 -> 51,26
     62 -> 78*
     59 -> 24*
     80 -> 63*
     29 -> 73,65
     25 -> 11,22,4
     23 -> 57*
   problem:
    
   Qed