YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [d](x0) = 7x0,
   
   [b](x0) = x0,
   
   [c](x0) = x0,
   
   [a](x0) = x0
  orientation:
   b(c(a(x1))) = x1 >= x1 = a(b(a(b(c(x1)))))
   
   b(x1) = x1 >= x1 = c(c(x1))
   
   c(d(x1)) = 7x1 >= x1 = a(b(c(a(x1))))
   
   a(a(x1)) = x1 >= x1 = a(c(b(a(x1))))
  problem:
   b(c(a(x1))) -> a(b(a(b(c(x1)))))
   b(x1) -> c(c(x1))
   a(a(x1)) -> a(c(b(a(x1))))
  String Reversal Processor:
   a(c(b(x1))) -> c(b(a(b(a(x1)))))
   b(x1) -> c(c(x1))
   a(a(x1)) -> a(b(c(a(x1))))
   Bounds Processor:
    bound: 1
    enrichment: match
    automaton:
     final states: {9,7,1}
     transitions:
      a0(2) -> 3*
      a0(4) -> 5*
      a0(11) -> 9*
      c1(25) -> 26*
      c1(22) -> 23*
      c1(12) -> 13*
      c1(24) -> 25*
      c1(21) -> 22*
      c1(13) -> 14*
      f40() -> 2*
      c0(2) -> 8*
      c0(6) -> 1*
      c0(8) -> 7*
      c0(3) -> 10*
      b0(10) -> 11*
      b0(5) -> 6*
      b0(3) -> 4*
      1 -> 3,21
      3 -> 21*
      5 -> 12*
      9 -> 3,21
      10 -> 24*
      14 -> 6*
      23 -> 4*
      26 -> 11*
    problem:
     
    Qed