YES

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

Proof:
 String Reversal Processor:
  a(a(b(a(x1)))) -> a(b(a(b(c(x1)))))
  b(c(a(x1))) -> a(b(c(b(a(a(x1))))))
  Bounds Processor:
   bound: 1
   enrichment: match
   automaton:
    final states: {7,1}
    transitions:
     b0(3) -> 4*
     b0(11) -> 12*
     b0(9) -> 10*
     b0(5) -> 6*
     a1(15) -> 16*
     a1(17) -> 18*
     a1(26) -> 27*
     a1(25) -> 26*
     a1(30) -> 31*
     c0(2) -> 3*
     c0(10) -> 11*
     a0(4) -> 5*
     a0(6) -> 1*
     a0(12) -> 7*
     a0(2) -> 8*
     a0(8) -> 9*
     f30() -> 2*
     b1(16) -> 17*
     b1(27) -> 28*
     b1(29) -> 30*
     b1(14) -> 15*
     c1(28) -> 29*
     c1(13) -> 14*
     7 -> 4*
     12 -> 25*
     4 -> 13*
     1 -> 8,9
     31 -> 15*
     18 -> 9*
   problem:
    
   Qed