YES

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

Proof:
 String Reversal Processor:
  b(a(a(a(x1)))) -> a(a(a(b(a(b(x1))))))
  b(a(x1)) -> x1
  DP Processor:
   DPs:
    b#(a(a(a(x1)))) -> b#(x1)
    b#(a(a(a(x1)))) -> b#(a(b(x1)))
   TRS:
    b(a(a(a(x1)))) -> a(a(a(b(a(b(x1))))))
    b(a(x1)) -> x1
   Arctic Interpretation Processor:
    dimension: 4
    usable rules:
     b(a(a(a(x1)))) -> a(a(a(b(a(b(x1))))))
     b(a(x1)) -> x1
    interpretation:
     [b#](x0) = [0  0  -& 0 ]x0 + [0],
     
               [-& 0  0  0 ]     [0]
               [-& -& -& 0 ]     [0]
     [a](x0) = [0  1  0  0 ]x0 + [0]
               [-& 0  0  0 ]     [0],
     
               [0  0  0  0 ]     [0 ]
               [-& 0  -& 0 ]     [-&]
     [b](x0) = [-& -& -& 0 ]x0 + [-&]
               [-& 0  -& 0 ]     [-&]
    orientation:
     b#(a(a(a(x1)))) = [0 1 0 1]x1 + [1] >= [0  0  -& 0 ]x1 + [0] = b#(x1)
     
     b#(a(a(a(x1)))) = [0 1 0 1]x1 + [1] >= [-& 0  -& 0 ]x1 + [0] = b#(a(b(x1)))
     
                      [0 1 1 1]     [1]    [0 1 0 1]     [1]                       
                      [0 1 0 1]     [1]    [0 1 0 1]     [1]                       
     b(a(a(a(x1)))) = [0 1 0 1]x1 + [1] >= [0 1 0 1]x1 + [1] = a(a(a(b(a(b(x1))))))
                      [0 1 0 1]     [1]    [0 1 0 1]     [1]                       
     
                [0  1  0  0 ]     [0]           
                [-& 0  0  0 ]     [0]           
     b(a(x1)) = [-& 0  0  0 ]x1 + [0] >= x1 = x1
                [-& 0  0  0 ]     [0]           
    problem:
     DPs:
      b#(a(a(a(x1)))) -> b#(x1)
     TRS:
      b(a(a(a(x1)))) -> a(a(a(b(a(b(x1))))))
      b(a(x1)) -> x1
    Restore Modifier:
     DPs:
      b#(a(a(a(x1)))) -> b#(x1)
     TRS:
      b(a(a(a(x1)))) -> a(a(a(b(a(b(x1))))))
      b(a(x1)) -> x1
     CDG Processor:
      DPs:
       b#(a(a(a(x1)))) -> b#(x1)
      TRS:
       b(a(a(a(x1)))) -> a(a(a(b(a(b(x1))))))
       b(a(x1)) -> x1
      graph:
       
      Qed